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Title: Curiosa mathematica, part II

Pillow-problems thought out during wakeful hours

Author: Lewis Carroll

Release date: July 12, 2026 [eBook #79080]
Most recently updated: July 28, 2026

Language: English

Original publication: London: Macmillan and Co., 1895

Other information and formats: www.gutenberg.org/ebooks/79080

Credits: Aaron Adrignola, Laura Natal and the Online Distributed Proofreading Team at https://www.pgdp.net (This book was produced from images made available by the HathiTrust Digital Library.)

*** START OF THE PROJECT GUTENBERG EBOOK CURIOSA MATHEMATICA, PART II ***

PILLOW-PROBLEMS

3D coordinate axes X, Y, Z with projected parallelogram and oblique line.

[See p. 100.

Curiosa Mathematica

PART II

PILLOW-PROBLEMS

THOUGHT OUT DURING
WAKEFUL HOURS

BY
CHARLES L. DODGSON, M.A.

Student and late Mathematical Lecturer of Christ Church Oxford

FOURTH EDITION

PRICE TWO SHILLINGS

London
MACMILLAN AND CO.
1895

[All rights reserved]

Oxford
HORACE HART, PRINTER TO THE UNIVERSITY

PREFACE TO FOURTH EDITION.

I TAKE this opportunity of explaining why it is that (as stated in the Note to p. xix) I have used the symbols element of opening downwards and segment to represent the words 'sine' and 'cosine'.

The use of some symbols needs, I suppose, no more justification than the use of + and - to represent 'plus' and 'minus'.

These particular symbols are derived from the old theory of Trigonometry, in which sines, cosines, &c. were actual lines.

A simple geometric diagram of a semicircle with labeled points — P on the arc, O at the center, and N on the diameter — illustrating a classical geometry or trigonometry problem.

In this diagram, upper O upper P being taken as the unit of length, upper P upper N is the sine of the angle upper N upper O upper P, and upper O upper N its cosine.

In each of my two symbols I have retained the semicircle in the symbol sine, I have merely moved upper P upper N to the middle; and, in the symbol cosine, I have lengthened upper O upper N, taking it a little beyond the curve, in order to avoid confusion with the existing symbol for 'semicircle'.

I also take this opportunity of adding a sort of Corollary (lately thought out) to the solution of Problem 59 (see p. 84).

If a, b, c be given lengths, they must, in order that the Tetrahedron may be possible, fulfil certain conditions, as follows:—

(1) they have to form the sides of a Triangle: hence any two of them must be greater than the third;

(2) the three angles of this Triangle have to form a solid angle: hence any two of these angles must be together greater than the third: hence any two of them must be together greater than 90°: hence any one of them must be less than 90°: hence the cosine of any one of them must be greater than 0: i.e. b squared plus c squared minus a squared must be greater than O, &c.: hence a, b, c must be such that the squares of any two of them are together greater than the square of the third.

For example, the lengths 2, 3, 4 would not do as the given lengths, since, although fulfilling the first condition, by having 2 plus 3 greater than 4, they fail to fulfil the second, as 2 squared plus 3 squared is not greater than 4 squared.

C. L. D.

CH. CH., OXFORD.
March, 1895.

PREFACE TO SECOND EDITION.

THE principal changes, made in this Second Edition of "Pillow-Problems", are as follows:—

(1) After the numeral, which precedes each Question, Answer, or Solution, references are given to the pages at which the corresponding matter may be found.

(2) Some of the Solutions have been re-arranged, and duplicate-diagrams have been inserted, in order that every portion of text may have its illustrative diagram visible along with it, and that the reader may thus be saved the trouble, and the strain on his temper, involved in turning a leaf backwards and forwards while referring from the one to the other.

(3) In the title of the book, the words "sleepless nights" have been replaced by "wakeful hours".

This last change has been made in order to allay the anxiety of kind friends, who have written to me to express their sympathy in my broken-down state of health, believing that I am a sufferer from chronic "insomnia", and that it is as a remedy for that exhausting malady that I have recommended mathematical calculation.

The title was not, I fear, wisely chosen; and it certainly was liable to suggest a meaning I did not intend to convey, viz. that my "nights" are very often wholly "sleepless". This is by no means the case: I have never suffered from "insomnia": and the over-wakeful hours, that I have had to spend at night, have often been simply the result of the over-sleepy hours I have spent during the preceding evening! Nor is it as a remedy for wakefulness that I have suggested mathematical calculation; but as a remedy for the harassing thoughts that are apt to invade a wholly-unoccupied mind. I hope the new title will express my meaning more lucidly.

To state the matter logically, the dilemma which my friends suppose me to be in has, for its two horns, the endurance of a sleepless night, and the adoption of some recipe for inducing sleep. Now, so far as my experience goes, no such recipe has any effect, unless when you are sleepy: and mathematical calculation would be more likely to delay, than to hasten, the advent of sleep.

The real dilemma, which I have had to face, is this: given that the brain is in so wakeful a condition that, do what I will, I am certain to remain awake for the next hour or so, I must choose between two courses, viz. either to submit to the fruitless self-torture of going through some worrying topic, over and over again, or else to dictate to myself some topic sufficiently absorbing to keep the worry at bay. A mathematical problem is, for me, such a topic; and is a benefit, even if it lengthens the wakeful period a little. I believe that an hour of calculation is much better for me than half-an-hour of worry.

The reader will, I think, be interested to see a curiously illogical solution which has been proposed, by a correspondent of the Educational Times, for Problem 61, viz. "Prove that, if any 3 Numbers be taken, which cannot be arranged in upper A period upper P period, and whose sum is a multiple of 3, the sum of their squares is also the sum of another set of 3 squares, the 2 sets having no common term."

The proposed solution is as follows:—

"Let 3 m, 21 m, 30 m be the three Numbers; then

3 m plus 21 m plus 30 m equals 3 times 18 m period

Also left parenthesis 3 m right parenthesis squared plus left parenthesis 21 m right parenthesis squared plus left parenthesis 30 m right parenthesis squared equals left parenthesis 6 m right parenthesis squared plus left parenthesis 15 m right parenthesis squared plus left parenthesis 33 m right parenthesis squared

equals left parenthesis 5 m right parenthesis squared plus left parenthesis 13 m right parenthesis squared plus left parenthesis 34 m right parenthesis squared equals left parenthesis 10 m right parenthesis squared plus left parenthesis 17 m right parenthesis squared plus left parenthesis 31 m right parenthesis squared

equals left parenthesis 14 m right parenthesis squared plus left parenthesis 23 m right parenthesis squared plus left parenthesis 25 m right parenthesis squared period"

Now, if we denote, by 'alpha', the property "which cannot be arranged in upper A period upper P period, and whose sum is a multiple of 3," and, by 'beta', the property "the sum of whose squares is also the sum of another set of 3 squares, the 2 sets having no common term," we see that all, that this writer has succeeded in proving, is that certain selected Numbers, which have property 'alpha', have also property 'beta': but this does not prove my Theorem, viz. that any Numbers whatever, which have property 'alpha', have also property 'beta'. If his argument were arranged in a syllogistic form, it would be found to assume a quite untenable Major Premiss, viz. "that, which is true of certain selected Numbers which have property 'alpha', is true of any Numbers whatever which have property 'alpha'."

C. L. D.

CH. CH., OXFORD.
September, 1893.

INTRODUCTION.

NEARLY all of the following seventy-two Problems are veritable "Pillow-Problems", having been solved, in the head, while lying awake at night. (I have put on record the exact dates of some.) No. 37 and one or two others belong to the daylight, having been solved while taking a solitary walk; but every one of them was worked out, to the very end, before drawing any diagram or writing down a single word of the solution. I generally wrote down the answer, first of all: and afterwards the question and its solution. For example, in No. 70, the very first words I wrote down were as follows:—"(1) down back-edge; up again; down again; and so on; (2) about dot 7 of the way down the back-edge; (3) about 18° 18′; (4) about 14°." These answers are not quite correct; but at least they are genuine, as the results of mental work only. "A poor thing, Sir, but mine own!"

My motive, for publishing these Problems, with their mentally-worked solutions, is most certainly not any desire to display powers of mental calculation. Mine, I feel sure, are nothing out-of-the-way; and I have no doubt there are many mathematicians who could produce, mentally, much shorter and better solutions. It is not for such persons that I intend my little book; but rather for the much larger class of ordinary mathematicians, who perhaps have never tried this resource, when mental occupation was needed, and who will, I hope, feel encouraged—by seeing what can be done, after a little practice, by one of average mathematical powers—to try the experiment for themselves, and find in it as much advantage and comfort as I have done.

The word "comfort" may perhaps sound out of place, in connection with so entirely intellectual an occupation; but it will, I think, come home to many who have known what it is to be haunted by some worrying subject of thought, which no effort of will is able to banish. Again and again I have said to myself, on lying down at night, after a day embittered by some vexatious matter, "I will not think of it any more! I have gone through it all, thoroughly. It can do no good whatever to go through it again. I will think of something else!" And in another ten minutes I have found myself, once more, in the very thick of the miserable business, and torturing myself, to no purpose, with all the old troubles.

Now it is not possible—this, I think, all psychologists will admit—by any effort of volition, to carry out the resolution "I will not think of so-and-so." (Witness the common trick, played on a child, of saying "I'll give you a penny, if you'll stand in that corner for five minutes, and not once think of strawberry-jam." No human child ever yet won the tempting wager!) But it is possible—as I am most thankful to know—to carry out the resolution "I will think of so-and-so." Once fasten the attention upon a subject so chosen, and you will find that the worrying subject, which you desire to banish, is practically annulled. It may recur, from time to time—just looking in at the door, so to speak; but it will find itself so coldly received, and will get so little attention paid to it, that it will, after a while, cease to be any worry at all.

Perhaps I may venture, for a moment, to use a more serious tone, and to point out that there are mental troubles, much worse than mere worry, for which an absorbing subject of thought may serve as a remedy. There are sceptical thoughts, which seem for the moment to uproot the firmest faith; there are blasphemous thoughts, which dart unbidden into the most reverent souls; there are unholy thoughts, which torture, with their hateful presence, the fancy that would fain be pure. Against all these some real mental work is a most helpful ally. That "unclean spirit" of the parable, who brought back with him seven others more wicked than himself, only did so because he found the chamber "swept and garnished", and its owner sitting with folded hands: had he found it all alive with the "busy hum" of active work, there would have been scant welcome for him and his seven!

My purpose—of giving this encouragement to others—would not be so well fulfilled had I allowed myself, in writing out my solutions, to improve on the work done in my head. I felt it to be much more important to set down what had actually been done in the head, than to supply shorter or neater solutions, which perhaps would be much harder to do without paper. For example, a Long-Multiplication sum (say the multiplying together of two numbers of 7 digits) is no doubt best done, on paper, by beginning at the unit-end, and writing out 7 rows of figures, and adding up the columns in the usual way. But it would be very difficult indeed—to me quite impossible—to do such a thing in the head. The only chance seems to be to begin with the millions, and get them properly grouped; then the hundred-thousands, adding the results to the previous one; and so on. Very often it seems to happen, that the easiest mental process looks decidedly lengthy and round-about when committed to paper.

When I first tried this plan, easy geometrical problems were all I could manage; and, even in these, I had to pause from time to time, in order to re-draw the diagram, which would persist in getting 'rubbed-out'. Algebraical problems I avoided at first, owing to the provoking fact that, if one single co-efficient escaped the memory, there was no resource but to begin the calculation all over again. But I soon got over both these difficulties, and found myself able to remember fairly large numerical co-efficients, and also to retain, in the mind's eye, fairly complex diagrams, even to the extent of finding my way from one part of the diagram to another. The lettering of the diagrams proved such a troublesome thing to keep in the memory, that I almost gave up using it, and learned to recognise Points by their situation only. In my MS. of No. 53, I find the following memorandum:—

"I had never set myself this Problem before the week ending Ap. 6, 1889. I tried it, two or three nights, lying awake; and finally worked it out on the night of Ap. six sevenths. All the conclusions were worked out mentally before any use was made of pen and paper. While working it, I did not give names to any Points, except upper A, upper B, upper C, and upper P: I merely thought of them by their positions (e. g. 'the foot of the perpendicular from upper P on upper B upper C')."

If any of my readers should feel inclined to reproach me with having worked too uniformly in the region of Common-place, and with never having ventured to wander out of the beaten tracks, I can proudly point to my one Problem in 'Transcendental Probabilities'—a subject in which, I believe, very little has yet been done by even the most enterprising of mathematical explorers. To the casual reader it may seem abnormal, and even paradoxical; but I would have such a reader ask himself, candidly, the question "Is not Life itself a Paradox?"

To give the Reader some idea of the process of construction of these Problems, I will give the biography of No. 63. The history of one is, to a great extent, the history of all.

It was begun during the night of Sept. three fourths, 1890, and completed during the following night. The idea had occurred to me, a short time previously, that something interesting might be found in the subject of what I may call 'partially-regular' Solids. The 'regular' Solids are provokingly few in number; and it would be hopeless to find any question, connected with them, that has not already been exhaustively analysed: some also of the 'partially-regular' Solids (e. g. rhomboidal crystals) have probably been similarly treated; but there seemed to be room for the invention of other such Solids.

Accordingly, I devised a Solid enclosed, above and below, by 2 equal and parallel Squares, having their centres in the same vertical line, and the upper one twisted round so that its sides should be parallel to the diagonals of the lower Square. Then I imagined the upper one raised until its corners formed the vertices of 4 equilateral Triangles, whose bases were the sides of the lower one. The Solid, thus obtained, was evidently enclosed by 2 Squares and 8 equilateral Triangles: and the Problem I set myself was to obtain its Volume.

There was no great difficulty in proving that the distance between the 2 Squares (taking each side as equal to '2') was 2 Superscript three fourths. But, when I looked about for some Trigonometrical method for calculating the Volume, despair soon seized upon me! A calculable Prism could be cut out of the middle of the Solid, I saw: but the outlying projections completely baffled me. After a while, the happy idea occurred to me of trying Algebraical Geometry, and regarding each facet as the base of a Pyramid, having its vertex at the centre of the Solid, which I decided to take as the Origin. I saw at once that I could calculate the co-ordinates of all the vertical Points, thence obtain equations to the Planes containing the facets, and thence calculate their distances from the Origin, which would be the altitudes of the Pyramids. Also it was evident that a sample Pyramid would suffice. I worked out a value for the Volume, that first night; but the thing got into a tangle, and I felt pretty sure I had got it wrong.

The next night I began again, and worked it all through from the beginning. In the morning the answer was clear in my memory, and I wrote it down at once; and did not write out the Problem, and its solution, until later in the day, when I was well pleased to find the written proof confirm the result I had arrived at in the hours of darkness.

It is not, perhaps, much to be wondered at that, when these Problems came to be re-written and arranged for publication, a good many mistakes were discovered. Some were so bad as quite to spoil the solutions in which they occurred: these Problems I have omitted altogether. The others I have corrected, in the solutions as given in Chapter III: but, that I may not be credited with an amount of accuracy, as a computator, which I am well aware I do not possess, I here append a list of them.

In No. 7, in the denominator '2 sine upper A', I forgot the '2'.[1]

In No. 10, I failed to notice that the 3 coins might also be a half-crown and 2 shillings.

In No. 13, in the last line but one, I put '2 b c period c a', instead of '4 b c period c a'.

In No. 32, I brought out the arithmetical value as '358520', instead of '358550'.

In No. 38, I got the decimal wrong, making it dot 476 instead of dot 478, and thus brought out the answer as dot 042 instead of dot 044.

In No. 44, I said that the denominator would be of the form left parenthesis 10 Superscript n Baseline minus 1 right parenthesis period 10 Superscript m. This last factor is superfluous: i. e. m equals 0.

In No. 50, I made a mistake near the end, bringing out StartFraction 41 Over 108 EndFraction, instead of StartFraction 50 Over 108 EndFraction.

In No. 55, I put 'tan' for 'sin'.

In No. 57, in the last paragraph, I replaced the denominator 'a sine upper B sine upper C' by (what I imagined to be its equivalent) '2 m'. Apparently I was under the delusion that 'a sine upper B sine upper C' was the same thing as 'sine upper A period b c'!

In No. 70, section (3), I forgot to add in the dot 45, thus making the answer half a degree wrong. And, in section (4), I forgot to add in the dot 53, thus again making the answer half a degree wrong.

Let me, in conclusion, gratefully acknowledge the valuable assistance I have received from Mr. F. G. Brabant, M.A., of Corpus Christi College, Oxford, who has most patiently and carefully gone through my proofs, first working out each result independently, and has thus detected many mistakes which had escaped my notice. He has also supplied, for No. 59, a much neater answer than mine, viz. StartFraction a b c Over 3 EndFraction dot StartRoot cosine upper A cosine upper B cosine upper C EndRoot.

Other mistakes may perchance, having eluded us both, await the penetrating glance of some critical reader, to whom the joy of discovery, and the intellectual superiority which he will thus discern, in himself, to the author of this little book, will, I hope, repay to some extent the time and trouble its perusal may have cost him!

C. L. D.

CH. CH., OXFORD.
May, 1893.

FOOTNOTES:

[1] In the trigonometrical Problems, I have used the symbols element of opening downwards and segment, to represent the words 'sine' and 'cosine'.

CONTENTS.

| PAGE
CHAPTER I. Questions | 1
CHAPTER II. Answers | 19
CHAPTER III. Solutions | 28

SUBJECTS CLASSIFIED.

Arithmetic. No. 31.

Algebra:—

Equational Problems. Nos. 8, 25, 39, 52, 68.

Series. Nos. 21, 32.

Indeterminate Equations. No. 47.

Properties of Numbers. Nos. 1, 14, 29, 44, 61.

Chances. Nos. 5, 10, 16, 19, 23, 27, 38, 41, 45, 50, 58, 66.

Pure Geometry, Plane. Nos. 2, 3, 9, 15, 17, 18, 20, 24, 26, 30, 34, 35,

36, 40, 46, 51, 57, 62, 64, 71.

Trigonometry:—

Plane. Nos. 4, 6, 7, 11, 12, 13, 18, 22, 28, 37, 42, 43, 48, 54, 55, 56,

57, 60, 65, 69.

Solid. Nos. 49, 59, 63, 70.

Algebraical Geometry:—

Plane. No. 53.

Solid. No. 67.

Differential Calculus:—

Maxima and Minima. No. 33.

Transcendental Probabilities. No. 72.

PILLOW-PROBLEMS.

CHAPTER I.

Questions.

1. (28)[2]

Find a general formula for two squares whose sum = 2.

[24/3/84

2. (29)

In a given Triangle to place a line parallel to the base, such that the portions of sides, intercepted between it and the base, shall be together equal to the base.

3. (30)

If the sides of a Tetragon pass through the vertices of a Parallelogram, and if three of them are bisected at those vertices: prove that the fourth is so also.

4. (30)

In a given acute-angled Triangle inscribe a Triangle, whose sides make, at each of the vertices, equal angles with the sides of the given Triangle.

[19/4/76

5. (19, 31)

A bag contains one counter, known to be either white or black. A white counter is put in, the bag shaken, and a counter drawn out, which proves to be white. What is now the chance of drawing a white counter?

[8/9/87

6. (19, 32)

Given lengths of lines drawn, from vertices of Triangle, to middle points of opposite sides, to find its sides and angles.

7. (19, 33)

Given 2 adjacent sides, and the included angle, of a Tetragon; and that the angles, at the other ends of these 2 sides, are right: find (1) remaining sides, (2) area.

[4 or 5/89

8. (20, 34)

Some men sat in a circle, so that each had 2 neighbours; and each had a certain number of shillings. The first had 1/ more than the second, who had 1/ more than the third, and so on. The first gave 1/ to the second, who gave 2/ to the third, and so on, each giving 1/ more than he received, as long as possible. There were then 2 neighbours, one of whom had 4 times as much as the other. How many men were there? And how much had the poorest man at first?

[3/89

9. (35)

Given two Lines meeting at a Point, and given a Point lying within the angle contained by them: draw, from the given Point, two lines, at right angles to each other, and forming with the given Lines and the line joining their intersection to the given Point, two equal Triangles.

[11/76

10. (20, 36)

A triangular billiard-table has 3 pockets, one in each corner, one of which will hold only one ball, while each of the others will hold two. There are 3 balls on the table, each containing a single coin. The table is tilted up, so that the balls run into one corner, it is not known which. The 'expectation', as to the contents of the pocket, is two sixths. What are the coins?

[8/90

11. (20, 36)

A Triangle upper A upper B upper C has another upper A prime upper B prime upper C prime inscribed in it, so that angle upper B upper A prime upper C prime equals angle upper C upper B prime upper A prime equals angle upper A upper C prime upper B prime equals theta; thus making it similar to the first Triangle. Find ratio between homologous sides. And solve for "theta equals 90 degree".

A geometric diagram of triangle ABC with interior points and a smaller inscribed triangle A'B'C' formed by connecting midpoints or cevians, with labeled intersection point O.

The Triangles can be proved similar thus:—

StartLayout 1st Row angle upper C prime upper A prime upper B prime plus angle upper B prime upper A prime upper C equals supp period of theta comma 2nd Row angle upper B prime upper A prime upper C plus angle upper A prime upper C upper B prime equals supp period of theta semicolon 3rd Row therefore these pairs are equal semicolon therefore angle upper C prime upper A prime upper B prime equals upper C period EndLayout

Hence angle upper A prime upper B prime upper C prime equals upper A comma and angle upper B prime upper C prime upper A prime equals upper B period

Let upper C prime upper A prime equals k a semicolon therefore upper A prime upper B prime equals k b comma and upper B prime upper C prime equals k c. We have to find k.

[31/3/82

12. (20, 37)

Given the semi-perimeter and the area of a Triangle, and also the volume of the cuboid whose edges are equal to the sides of the Triangle: find the sum of the squares of its sides.

[23/1/91

13. (20, 38)

Given the lengths of the radii of two intersecting Circles, and the distance between their centres: find the area of the Tetragon formed by the tangents at the points of intersection.

[3/89

14. (39)

Prove that 3 times the sum of 3 squares is also the sum of 4 squares.

[2/12/81

15. (39)

If a Figure be such that the opposite angles of every inscribed Tetragon are supplementary: the Figure is a Circle.

[3/91

16. (20, 40)

There are two bags, one containing a counter, known to be either white or black; the other containing 1 white and 2 black. A white is put into the first, the bag shaken, and a counter drawn out, which proves to be white. Which course will now give the best chance of drawing a white—to draw from one of the two bags without knowing which it is, or to empty one bag into the other and then draw?

[10/87

17. (40)

In a given Triangle place a line parallel to the base, such that if, from its ends, lines be drawn, parallel to the sides and terminated by the base, they shall be together equal to the first line.

[3/89

18. (21, 41)

Find a Point, in the base of a given Triangle, such that, if from it perpendiculars be dropped upon the sides, the line joining their extremities shall be parallel to the base. (1) Trigonometrically. (2) Geometrically.

[11/89

19. (21, 42)

There are 3 bags; one containing a white counter and a black one, another two white and a black, and the third 3 white and a black. It is not known in what order the bags are placed. A white counter is drawn from one of them, and a black from another. What is the chance of drawing a white counter from the remaining bag?

20. (43)

In the base of a given Triangle find a Point such that if from it two lines be drawn, terminated by the sides, one being perpendicular to the base and one to the left-hand side, they shall be equal.

[5/88

21. (21, 44)

Sum, (1) to n terms, (2) to 100 terms, the series

1 period 3 period 5 plus 2 period 4 period 6 plus ampersand c period

[7/4/89

22. (21, 45)

Given the 3 'altitudes' of a Triangle: find its (1) sides, (2) angles, (3) area.

[4/6/89

23. (21, 46)

A bag contains 2 counters, each of which is known to be black or white. 2 white and a black are put in, and 2 white and a black drawn out. Then a white is put in, and a white drawn out. What is the chance that it now contains 2 white?

[25/9/87

24. (21, 47)

If, from the vertices of a triangle upper A upper B upper C, the lines upper A upper D, upper B upper E, upper C upper F be drawn, intersecting at upper O: find the ratio StartFraction upper D upper O Over upper D upper A EndFraction in terms of the two ratios StartFraction upper E upper O Over upper E upper B EndFraction, StartFraction upper F upper O Over upper F upper C EndFraction.

A geometric diagram of triangle ABC with three cevians meeting at interior point O, creating points D, E, F on the sides.

[5/86

25. (22, 48)

If 'epsilon', 'alpha', 'lamda' represent proper fractions; and if, in a certain hospital, 'epsilon' of the patients have lost an eye, 'alpha' an arm, and 'lamda' a leg: what is the least possible number who have lost all three?

[7/2/76

26. (48)

Within a given Triangle place a similar Triangle, whose area shall have to its area a given ratio less than unity, whose sides shall be parallel to its sides, and whose vertices shall be equidistant from its vertices.

[4/89

27. (22, 50)

There are 3 bags, each containing 6 counters; one contains 5 white and one black; another, 4 white and 2 black; the third, 3 white and 3 black. From two of the bags (it is not known which) 2 counters are drawn, and prove to be black and white. What is the chance of drawing a white counter from the remaining bag?

[4/3/80

28. (22, 50)

If the sides of a given Triangle, taken cyclically, be divided in extreme and mean ratio; and if the Points be joined: find the ratio which the area of the Triangle, so formed, has to the area of the given Triangle.

[12/78

29. (51)

Prove that the sum of 2 different squares, multiplied by the sum of 2 different squares, gives the sum of 2 squares in 2 different ways.

[3/12/81

30. (52)

In a given Triangle, to place a line parallel to the base, such that if from its extremities lines be drawn, to the base, parallel to the sides, they shall be together double of the inscribed Line.

[15/3/89

31. (22, 53)

On July 1, at 8 a.m. by my watch, it was 8h. 4m. by my clock. I took the watch to Greenwich, and, when it said 'noon', the true time was 12h. 5m. That evening, when the watch said '6h.', the clock said '5h. 59m.'

On July 30, at 9 a.m. by my watch, it was 8h. 57m. by my clock. At Greenwich, when the watch said '12h. 10m.', the true time was 12h. 5m. That evening, when the watch said '7h.' the clock said '6h. 58m.'

My watch is only wound up for each journey, and goes uniformly during any one day: the clock is always going, and goes uniformly.

How am I to know when it is true noon on July 31?

[14/3/89

32. (22, 53)

Sum the Series 1 period 5 plus 2 period 6 plus ampersand c period (1) to n terms; (2) to 100 terms.

[7/4/89

33. (54)

Inscribe in a given Circle the maximum Tetragon having 2 parallel sides, one double the other.

34. (55)

From a given Point draw 2 Lines, one to the centre of a given Circle, and the other cutting off from it a Segment containing an angle equal to that between the Lines.

[21/12/74

35. (56)

With a given Triangle, to describe a Circle, cutting each side in two points, such that, if radii be drawn perpendicular to the sides, they are divided by the sides in given ratios.

[11/76

36. (57)

In a given Triangle, to draw a line, from a Point on one side of it, to a Point on the other side, perpendicular to one of these sides, and equal to the sum of the portions, of these sides, intercepted between it and the base.

[3/89

37. (22, 58)

Two given Circles intersect, so that their common chord subtends angles of 30° and 60° at their centres. What fraction of the smaller Circle is within the larger?

[12/91

38. (22, 60)

There are 3 bags, 'upper A', 'upper B', and 'upper C'. 'upper A' contains 3 red counters, 'upper B' 2 red and one white, 'upper C' one red and 2 white. Two bags are taken at random, and a counter drawn from each: both prove to be red. The counters are replaced, and the experiment is repeated with the same two bags: one proves to be red. What is the chance of the other being red?

[3/76

39. (22, 60)

upper A and upper B begin, at 6 a. m. on the same day, to walk along a road in the same direction, upper B having a start of 14 miles, and each walking from 6 a. m. to 6 p. m. daily. upper A walks 10 miles, at a uniform pace, the first day, 9 the second, 8 the third, and so on: upper B walks 2 miles, at a uniform pace, the first day, 4 the second, 6 the third, and so on. When and where are they together?

[16/3/78

40. (61)

In a given Triangle, whose base-angles are acute, draw two lines, at right angles to the base, and together equal to the line drawn, from the vertex, at right angles to the base, and such that

(1) they are equidistant from the line drawn from the vertex;

(2) they are equidistant from the ends of the base.

[5/76

41. (23, 62)

My friend brings me a bag containing four counters, each of which is either black or white. He bids me draw two, both of which prove to be white. He then says "I meant to tell you, before you began, that there was at least one white counter in the bag. However, you know it now, without my telling you. Draw again."

(1) What is now my chance of drawing white?

(2) What would it have been, if he had not spoken?

[9/87

42. (23, 63)

If the angles of a given Triangle be bisected, and if lines be drawn, through its vertices, at right angles to the bisectors, so as to form a fresh Triangle: find the ratio of the area of this Triangle to the area of the given Triangle.

[17/5/78

43. (65)

From the ends of the base of a given Triangle draw two lines, intersecting, terminated by the sides, and forming an isosceles Triangle at the base, and a Tetragon, equal to it, at the vertex.

[2/82

44. (66)

If a, b be two numbers prime to each other, a value may be found for n which will make left parenthesis a Superscript n Baseline minus 1 right parenthesis a multiple of b.

[18/3/81

45. (23, 67)

If an infinite number of rods be broken: find the chance that one at least is broken in the middle.

[5/84

46. (68)

In a given Triangle, whose base is divided at a given Point, inscribe a Triangle, having its angles equal to given angles, and having an assigned vertex at the given Point.

[19/11/87

47. (23, 69)

Solve the 2 Indeterminate Equations

StartLayout 1st Row StartFraction x Over y EndFraction equals x minus z semicolon 2nd Row StartFraction x Over z EndFraction equals x minus y semicolon EndLayout right brace StartLayout 1st Row left parenthesis 1 right parenthesis 2nd Row left parenthesis 2 right parenthesis EndLayout

and find the limits, if any, between which the real values lie.

[12/90

48. (70)

If semicircles be described, externally, on the sides of a given Triangle; and if their common tangents be drawn; and if their lengths be alpha, beta, gamma: prove that

left parenthesis StartFraction beta gamma Over alpha EndFraction plus StartFraction gamma alpha Over beta EndFraction plus StartFraction alpha beta Over gamma EndFraction right parenthesis

is equal to the semiperimeter of the Triangle.

[9/2/81

49. (23, 72)

If four equilateral Triangles be made the sides of a square Pyramid: find the ratio which its volume has to that of a Tetrahedron made of the Triangles.

[16/11/86

50. (23, 72)

There are 2 bags, upper H and upper K, each containing 2 counters: and it is known that each counter is either black or white. A white counter is added to bag upper H, the bag is shaken up, and one counter transferred (without looking at it) to bag upper K, where the process is repeated, a counter being transferred to bag upper H. What is now the chance of drawing a white counter from bag upper H?

51. (74)

From a given Point, in one side of a given Triangle, to draw a line, terminated by the other side, so that, if from its ends lines be drawn at right angles to the base, their sum shall be equal to the first line.

[12/81

52. (23, 75)

Five beggars sat down in a circle, and each piled up, in a heap before him, the pennies he had received that day: and the five heaps were equal.

Then spake the eldest and wisest of them, unfolding, as he spake, an empty sack.

"My friends, let me teach you a pretty little game! First, I name myself 'Number One,' my left-hand neighbour 'Number Two,' and so on to 'Number Five.' I then pour into this sack the whole of my earnings for the day, and hand it on to him who sits next but one on my left, that is, 'Number Three.' His part in the game is to take out of it, and give to his two neighbours, so many pennies as represent their names (that is, he must give four to 'Number Four' and two to 'Number Two'); he must then put into the sack half as much as it contained when he received it; and he must then hand it on just as I did, that is, he must hand it to him who sits next but one on his left—who will of course be 'Number Five.' He must proceed in the same way, and hand it on to 'Number Two,' from whom the sack will find its way to 'Number Four,' and so to me again. If any player cannot furnish, from his own heap, the whole of what he has to put into the sack, he is at liberty to draw upon any of the other heaps, except mine!"

The other beggars entered into the game with much enthusiasm: and in due time the sack returned to 'Number One,' who put into it the two pennies he had received during the game, and carefully tied up the mouth of it with a string. Then, remarking "it is a very pretty little game," he rose to his feet, and hastily quitted the spot. The other four beggars gazed at each other with rueful countenances. Not one of them had a penny left!

How much had each at first?

[16/2/89

53. (24, 76)

In a triangular billiard-table, a Point is given by its trilinear co-ordinates. A ball, starting from the given Point, strikes the three sides, and returns to the starting-point. Find, in terms of the trilinear co-ordinates and of the angles of the Triangle, the Point where the ball strikes the second side.

[6/4/89

54. (24, 78)

Cut off, from a given Triangle, by lines parallel to the sides, 3 Triangles, so that the remaining Hexagon may be equilateral. Also find the lengths of its sides in terms of the sides of the given Triangle: and the ratios in which the sides of the given Triangle are divided.

[18/4/86

55. (79)

Given three cylindrical towers on a Plane: find a Point, on the Plane, from which they shall look the same width.

[20/12/74

56. (24, 80)

Given the 3 altitudes of a Triangle: construct it.

[27/6/84

57. (25, 80)

In a given Triangle describe three Squares, whose bases shall lie along the sides of the Triangle, and whose upper edges shall form a Triangle;

(1) geometrically; (2) trigonometrically.

[27/1/91

58. (25, 83)

Three Points are taken at random on an infinite Plane. Find the chance of their being the vertices of an obtuse-angled Triangle.

[20/1/84

59. (25, 84)

Given a Tetrahedron, having every edge equal to the opposite edge, so that its facets are all (when looked at from the outside) identically equal: find its volume in terms of its edges.

[8/90

60. (25, 87)

Given a Triangle upper A upper B upper C, and that its base upper B upper C is divided at upper D in the ratio m to n: find the angles upper B upper A upper D, upper C upper A upper D.

[21/3/90

61. (89)

Prove that, if any 3 Numbers be taken, which cannot be arranged in A. P., and whose sum is a multiple of 3, the sum of their squares is also the sum of another set of 3 squares, the 2 sets having no common term.

[1/12/81

62. (91)

Given two Lines meeting at a Point, and given a Point lying within the angle contained by them: draw a line, through the given Point, and forming, with the given Lines, the least possible Triangle.

[12/76

63. (26, 92)

Given 2 equal Squares, in different horizontal planes, having their centres in the same vertical line, and so placed that the sides of each are parallel to the diagonals of the other, and at such a distance apart that, by joining neighbouring vertices, 8 equilateral Triangles are formed: find the volume of the solid thus enclosed.

[3, 4/9/90

64. (94)

Given a Triangle, and a Point within it such that its distance from one of the sides is less than its distance from either of the others: describe a Circle, with given Point as centre, such that its intercepts on the sides may be equal to the sides of a right-angled Triangle.

[18/12/74

65. (95)

How many shapes are there for Triangles which have all their angles aliquot parts of 360°?

[5/89

66. (26, 97)

Given that there are 2 counters in a bag, as to which all that was originally known was that each was either white or black. Also given that the experiment has been tried, a certain number of times, of drawing a counter, looking at it, and replacing it; that it has been white every time; and that, as a result, the chance of drawing white, next time, is StartFraction alpha Over alpha plus beta EndFraction. Also given that the same experiment is repeated m times more, and that it still continues to be white every time. What would then be the chance of drawing white?

[9/89

67. (26, 100)

If a regular Tetrahedron be placed, with one vertex downwards, in a socket which exactly fits it, and be turned round its vertical axis, through an angle of 120°, raising it only so much as is necessary, until it again fits the socket: find the Locus of one of the revolving vertices.

[27/1/72

68. (26, 101)

Five friends agreed to form themselves into a Wine-Company (Limited). They contributed equal amounts of wine, which had been bought at the same price. They then elected one of themselves to act as Treasurer; and another of them undertook to act as Salesman, and to sell the wine at 10% over cost-price.

The first day the Salesman drank one bottle, sold some, and handed over the receipts to the Treasurer.

The second day he drank none, but pocketed the profits on one bottle sold, and handed over the rest of the receipts to the Treasurer.

That night the Treasurer visited the Cellars, and counted the remaining wine. "It will fetch just £11," he muttered to himself as he left the Cellars.

The third day the Salesman drank one bottle, pocketed the profits on another, and handed over the rest of the receipts to the Treasurer.

The wine was now all gone: the Company held a Meeting, and found to their chagrin that their profits (i. e. the Treasurer's receipts, less the original value of the wine) only cleared 6d. a bottle on the whole stock. These profits had accrued in 3 equal sums on the 3 days (i. e. the Treasurer's receipts for the day, less the original value of the wine taken out during the day, had come to the same amount every time); but of course only the Salesman knew this.

(1) How much wine had they bought? (2) At what price?

[28/2/89

69. (26, 102)

If, from each of the angles of a given Triangle upper A upper B upper C, taken cyclically, a certain proper fraction of it be cut off, the arithmetical values of the 3 fractions being represented by 'k, l, m'; and if it be given that the Triangle, formed by the lines so drawn, is similar to the given one, the angle, formed by the lines drawn from upper B and upper C, being equal to upper A, and so on: find k, l, m, as similar functions of a single variable. Also find the ratio which each side of the second Triangle bears to the corresponding side of the first.

[8/89

70. (27, 105)

Let an equilateral and equiangular Tetrahedron be placed with one facet in front: and suppose a series of triangles, equal to that facet, constructed in the Plane containing that facet, and having a base common with it; and that they are all wrapped round the Tetrahedron as far as they will go. Find (1) the locus of their vertices; (2) the situation of the vertex of the one whose left-hand base-angle is 15°; (3) the left-hand base-angle of the one which (wrapped round towards the right) covers portions of all four facets of the Tetrahedron, and whose vertex coincides with its vertex; (4) the left-hand base-angle of the one which (similarly treated) occupies all four facets, and then the front and right-hand facet for the second time, and whose vertex coincides with the distal vertex of the base of the Tetrahedron.

71. (108)

In a given Triangle place a Hexagon having its opposite sides equal and parallel, and three of them lying along the sides of the Triangle, and such that its diagonals intersect in a given Point.

[14/12/74

72. (27, 109)

A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colours without taking them out of the bag.

[8/9/87

FOOTNOTES:

[2] The numerals, placed in parentheses, indicate the pages where the corresponding matter may be found.

CHAPTER II.

Answers.

5. (2, 31)

Two-thirds.

6. (2,32)

Calling the sides '2 a', '2 b', '2 c' and the lines 'alpha', 'beta', 'gamma', we have

a squared equals StartFraction minus alpha squared plus 2 beta squared plus 2 gamma squared Over 9 EndFraction comma

cosine upper A equals StartFraction 5 alpha squared minus beta squared minus gamma squared Over 2 StartRoot 2 alpha squared minus beta squared plus 2 gamma squared EndRoot period StartRoot 2 alpha squared plus 2 beta squared minus gamma squared EndRoot EndFraction.

7. (2, 33)

A geometric diagram of triangle ABD with extended points C and E, connected by dashed lines.

Let upper A upper B, upper A upper D be given sides, and upper B, upper D the right angles; and let upper A upper B equals b, upper A upper D equals d.

(1) upper B upper C equals StartFraction d minus b cosine upper A Over sine upper A EndFraction semicolon upper C upper D equals StartFraction b minus d cosine upper A Over sine upper A EndFraction semicolon

(2) area equals StartFraction 2 b d minus left parenthesis b squared plus d squared right parenthesis cosine upper A Over 2 sine upper A EndFraction.

8. (2, 34)

7 men; 2 shillings.

10. (3, 36)

Either 2 florins and a sixpence; or else a half-crown and 2 shillings.

11. (3, 36)

The required ratio is equal to

StartFraction sine upper A sine upper B sine upper C Over sine theta left parenthesis 1 plus cosine upper A cosine upper B cosine upper C right parenthesis plus cosine theta sine upper A sine upper B sine upper C EndFraction period

If theta equals 90 degree, this = StartFraction sine upper A sine upper B sine upper C Over 1 plus cosine upper A cosine upper B cosine upper C EndFraction period

12. (4, 37)

If s = semi-perimeter, m = area, v = volume; then

a squared plus b squared plus c squared equals 2 period left parenthesis s squared minus StartFraction v Over s EndFraction minus StartFraction m squared Over s squared EndFraction right parenthesis period

13. (4, 38)

If '2 upper M' = area of Tetragon whose vertices are the Centres and the Points of intersection; and if its sides be 'a', 'b', and its diagonal, joining the Centres, 'c': required area

equals StartFraction 32 upper M cubed Over left parenthesis b squared plus c squared minus a squared right parenthesis period left parenthesis c squared plus a squared minus b squared right parenthesis EndFraction period

16. (4, 40)

The first course gives chance = one half; the second, five twelfths. Hence the first is best.

18. (5, 41)

(1) Divide base upper B upper C, at upper E, so that StartFraction upper B upper E Over upper E upper C EndFraction equals StartFraction sine 2 upper C Over sine 2 upper B EndFraction.

(2) At upper B, upper C, make right angles upper A upper B upper D, upper A upper C upper D; and join upper A upper D cutting upper B upper C at upper E, which is the required Point.

19. (5, 42)

Eleven-seventeenths.

21. (5, 44)

(1) StartFraction n period ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 4 With quotation dash period ModifyingAbove n plus 5 With quotation dash Over 4 EndFraction; left parenthesis 2 right parenthesis 27 comma 573 comma 000.

22. (5, 45)

Calling the given altitudes 'alpha, beta, gamma'; and the fraction

StartFraction 2 alpha squared beta squared gamma squared period left parenthesis alpha squared plus beta squared plus gamma squared right parenthesis minus left parenthesis beta Superscript 4 Baseline gamma Superscript 4 Baseline plus gamma Superscript 4 Baseline alpha Superscript 4 Baseline plus alpha Superscript 4 Baseline beta Superscript 4 Baseline right parenthesis Over 4 alpha Superscript 4 Baseline beta Superscript 4 Baseline gamma Superscript 4 Baseline EndFraction left single quotation mark k squared right single quotation mark comma

(1) a equals StartFraction 1 Over k alpha EndFraction comma ampersand c period

(2) sine upper A equals k beta gamma comma ampersand c period

(3) area equals StartFraction 1 Over 2 k EndFraction period

23. (5, 46)

Two-fifths.

24. (6, 47)

StartFraction upper D upper O Over upper D upper A EndFraction plus StartFraction upper E upper O Over upper E upper B EndFraction plus StartFraction upper F upper O Over upper F upper C EndFraction equals 1; whence any one can be found in terms of the other two.

25. (6, 48)

epsilon plus alpha plus lamda minus 2 period

27. (6, 50)

Seventeen-twentyfifths.

28. (7, 50)

7 minus 3 StartRoot 5 EndRoot period

31. (7, 53)

When the clock says '12h. 2m. 29 and StartFraction 277 Over 288 EndFraction s e c period'

32. (8, 53)

StartLayout 1st Row 1st Column left parenthesis 1 right parenthesis 2nd Column StartFraction n period left parenthesis n plus 1 right parenthesis period left parenthesis 2 n plus 13 right parenthesis Over 6 EndFraction semicolon 2nd Row 1st Column left parenthesis 2 right parenthesis 2nd Column 358550 period EndLayout

37. (8, 58)

StartFraction 4 plus StartRoot 3 EndRoot Over 12 EndFraction minus StartFraction 1 plus StartRoot 3 EndRoot Over 2 pi EndFraction semicolon i period e period about dot 044

.

38. (9, 60)

Fortynine-seventytwoths.

39. (9, 60)

They meet at end of 2d. 6h., and at end of 4d.: and the distances are 23 miles, and 34 miles.

41. (9, 62)

(1) Seven-twelfths. (2) One-half.

42. (10, 63)

StartFraction a b c Over 2 left parenthesis s minus a right parenthesis period left parenthesis s minus b right parenthesis period left parenthesis s minus c right parenthesis EndFraction period

45. (10, 67)

dot 6321207 ampersand c period

47. (11, 69)

One set of values is 0, 0, 0.

A 2nd set is x equals y equals 0; z has any value.

A 3rd is x equals z equals 0; y has any value.

And the 4th set is x equals StartFraction k squared Over k minus 1 EndFraction, y equals z equals k; where k has any value.

If x has any positive value less than 4, y and z are unreal.

49. (11, 72)

Two.

50. (11, 72)

Seventeen-twentysevenths.

52. (12, 75)

2l. 18s. 0d.

53. (13, 76)

The portion, cut off from the second side, is equal to

StartFraction left parenthesis alpha sine upper C plus gamma sine upper A right parenthesis left parenthesis 2 gamma cosine upper A plus beta right parenthesis Over alpha cosine upper C plus gamma cosine upper A plus beta EndFraction plus StartFraction beta cosine upper A plus gamma cosine 2 upper A Over sine upper A EndFraction period

54. (13,78)

Side upper A upper B must be divided at upper D, upper G, so that

upper A upper D colon upper D upper G colon upper G upper B colon colon StartFraction 1 Over a EndFraction colon StartFraction 1 Over c EndFraction colon StartFraction 1 Over b EndFraction semicolon

and similarly for the other sides. Also each side of the Hexagon = StartStartFraction 1 OverOver StartFraction 1 Over a EndFraction plus StartFraction 1 Over b EndFraction plus StartFraction 1 Over c EndFraction EndEndFraction.

56. (13, 80)

A geometric diagram of two vertical parallel lines with points D/F and E/G, connected by multiple crossing lines through interior points A, B, C.

Draw upper B upper C, upper C upper E, upper B upper D equal to the given altitudes, so as to form right angles at upper B and upper C: and produce upper D upper B, upper E upper C. Join upper D upper C, and draw upper C upper F perpendicular to it. Join upper E upper B, and draw upper B upper G perpendicular to it. With centre upper B, and distance upper B upper F, describe a circle: with centre upper C, and distance upper C upper G describe another: let them meet at upper A: and join upper A upper B, upper A upper C. Triangle upper A upper B upper C may be proved to be similar to required Triangle. The rest of the construction is obvious.

57. (14, 80)

(1) Geometrically.

If Squares be described externally on the sides of the given Triangle; and if their outer edges be produced to form a new Triangle; and if the sides of the given Triangle be divided similarly to those of the new Triangle: their central portions will be the bases of the required Squares.

(2) Trigonometrically.

If a, b, c be the sides of the given Triangle, and m its area; and if x, y, z be the sides of the required Squares: then

StartFraction a Over x EndFraction equals StartFraction b Over y EndFraction equals StartFraction c Over z EndFraction equals StartFraction a squared plus b squared plus c squared Over 2 m EndFraction plus 1 period

58. (14, 83)

StartStartFraction 3 OverOver 8 minus StartFraction 6 StartRoot 3 EndRoot Over pi EndFraction EndEndFraction period

59. (14, 84)

Calling lengths of the 3 pairs of edges 'a, b, c', and the corresponding angles, in each facet, 'upper A, upper B, upper C'; volume =

StartFraction a b c Over 6 EndFraction period StartRoot 1 minus left parenthesis cosine squared upper A plus cosine squared upper B plus cosine squared upper C right parenthesis plus 2 cosine upper A cosine upper B cosine upper C period EndRoot

60. (14, 87)

cotangent upper B upper A upper D equals StartFraction left parenthesis m plus n right parenthesis cotangent upper A plus n cotangent upper B Over m EndFraction;

similarly, cotangent upper C upper A upper D equals StartFraction left parenthesis m plus n right parenthesis cotangent upper A plus m cotangent upper C Over n EndFraction.

63. (15, 92)

If each side of each Square = 2, the volume =

StartFraction 8.2 Superscript one fourth Baseline period left parenthesis StartRoot 2 EndRoot plus 1 right parenthesis Over 3 EndFraction period

66. (15, 97)

StartFraction 2 Superscript m Baseline period left parenthesis alpha minus beta right parenthesis plus beta Over 2 Superscript m Baseline period left parenthesis alpha minus beta right parenthesis plus 2 beta EndFraction period

67. (16, 100)

If the centre of the horizontal facet be taken as the Origin, and if the upper X-axis pass through one of the vertices of that facet, and the upper Y-axis be parallel to the opposite edge of that facet, and the upper Z-axis be perpendicular to that facet: and if the altitude (measured downwards) of the Tetrahedron be called 'h', and the intercept on the upper X-axis be called 'a': the Equations to the Locus are

StartLayout 1st Row 1st Column left parenthesis x plus StartRoot 3 EndRoot period y right parenthesis period left parenthesis h minus z right parenthesis 2nd Column equals a h semicolon 2nd Row 1st Column x squared plus y squared 2nd Column equals a squared period EndLayout

68. (16, 101)

(1) 5 dozen; (2) 8/4 a bottle.

69. (17, 102)

(1) k equals StartFraction theta minus upper B Over upper A EndFraction; l equals StartFraction theta minus upper C Over upper B EndFraction; m equals StartFraction theta minus upper A Over upper C EndFraction.

(2) Calling new Triangle 'upper A prime upper B prime upper C prime',

StartFraction a prime Over a EndFraction equals StartFraction b prime Over b EndFraction equals StartFraction c prime Over c EndFraction equals 2 cosine theta period

70. (17, 105)

(1) Down the back-edge; up again; and so on. (2) about dot 7 of the way down the back-edge. (3) About 18 dot 65 degree. (4) About 14 dot 53 degree.

72. (18, 109)

One is black, and the other white.

CHAPTER III.

Solutions.

1. (1)

Let u, v be the Nos.

Then u squared plus v squared equals 2.

Evidently 'left parenthesis 1 plus k right parenthesis comma left parenthesis 1 minus k right parenthesis' is a form for the squares.

Also, if we write '2 m squared' for '2' (which will not interfere with the problem, as we can divide by m squared, and get StartFraction u squared Over m squared EndFraction plus StartFraction v squared Over m squared EndFraction equals 2), the above form becomes 'left parenthesis m squared plus k right parenthesis comma left parenthesis m squared minus k right parenthesis'.

Now, as these are squares, their resemblance to

left single quotation mark left parenthesis a squared plus b squared plus 2 a b right parenthesis comma left parenthesis a squared plus b squared minus 2 a b right parenthesis right single quotation mark

at once suggests itself; so that the problem depends on the known one of finding a, b, such that left parenthesis a squared plus b squared right parenthesis is a square; and we can then take 2 a b as k.

A general form for this is

StartLayout 1st Row StartLayout 1st Row 1st Column a 2nd Column equals x squared minus y squared comma 2nd Row 1st Column b 2nd Column equals 2 x y semicolon 3rd Row 1st Column therefore a squared plus b squared 2nd Column equals left parenthesis x squared plus y squared right parenthesis squared semicolon EndLayout 2nd Row therefore the formula left single quotation mark u squared plus v squared equals 2 m squared right single quotation mark becomes 3rd Row left parenthesis x squared minus y squared plus 2 x y right parenthesis squared plus left parenthesis x squared minus y squared minus 2 x y right parenthesis squared equals 2 left parenthesis x squared plus y squared right parenthesis squared semicolon 4th Row i period e period left parenthesis StartFraction x squared minus y squared plus 2 x y Over x squared plus y squared EndFraction right parenthesis squared plus left parenthesis StartFraction x squared minus y squared minus 2 x y Over x squared plus y squared EndFraction right parenthesis squared equals 2 period EndLayout

Q. E. F.

2. (1)

A geometric diagram of triangle ABC with multiple lines radiating from base point F to vertices and interior points A', B', C', D, E.

(Analysis.)

Let upper A upper B upper C be the Triangle, and upper D upper E the required line, so that upper B upper D plus upper C upper E equals upper B upper C.

From upper B upper C cut off upper B upper F equal to upper B upper D; then upper C upper F equals upper C upper E.

Join upper D upper F, upper E upper F.

Now angle upper B upper D upper F equals angle upper B upper F upper D equals left bracket by upper I period 29 right bracket angle upper F upper D upper E;

Similarly angle upper C upper E upper F equals angle upper F upper E upper D;

therefore angle s upper B upper D upper E comma upper C upper E upper D, are bisected by upper D upper F, upper E upper F, and upper F is centre of circled dot escribed to white up pointing triangle upper A upper D upper E.

Drop, from upper F, perpendiculars on upper B upper D, upper D upper E, upper E upper C; then these perpendiculars are equal.

Hence, if upper A upper F be joined, it bisects angle upper A.

Hence construction.

(Synthesis.)

Bisect angle upper A by upper A upper F: from upper F draw upper F upper B prime, upper F upper C prime, perpendicular upper A upper C, upper A upper B: also draw upper F upper A prime perpendicular upper B upper C and equal to upper F upper B prime: and through upper A prime draw upper D upper E perpendicular upper F upper A prime, i. e. parallel to upper B upper C. Then upper D upper E shall be line required.

because angles at upper A prime, upper B prime, upper C prime, are right, and upper F upper A Superscript prime Baseline equals upper F upper B Superscript prime Baseline equals upper F upper C prime,

therefore angles upper B upper D upper E, upper C upper E upper D, are bisected by upper D upper F, upper E upper F.

Now angle upper B upper F upper D equals angle upper F upper D upper A prime; therefore it equals angle upper B upper D upper F; therefore upper B upper F equals upper B upper D;

Similarly upper C upper F equals upper C upper E; therefore upper B upper C equals upper B upper D plus upper C upper E.

Q. E. F.

3. (1)

A geometric diagram of quadrilateral ABCD with interior points E, F, G, H forming an inscribed quadrilateral, connected by crossing diagonals.

Let upper A upper B upper C upper D be the Tetragon; and let the 3 sides, upper A upper B, upper B upper C, upper C upper D, be bisected by vertices of the Parallelogram upper E upper F upper G upper H.

Join upper B upper D.

because, in Triangle upper B upper C upper D, sides upper B upper C, upper C upper D are bisected at upper F and upper G,

therefore upper F upper G is parallel to upper B upper D;

but upper E upper H is parallel to upper F upper G;

therefore upper E upper H is parallel to upper B upper D;

therefore Triangles upper A upper E upper H, upper A upper B upper D are similar;

now upper A upper E is half of upper A upper B;

therefore upper A upper H is half of upper A upper D.

Q. E. D.

4. (1)

A geometric diagram of triangle ABC with an inscribed triangle A'B'C' and central point O, connected by multiple cevians.

Let upper A upper B upper C be the given Triangle, and upper A prime upper B prime upper C prime the required Triangle, so that angle upper B upper A prime upper C Superscript prime Baseline equals angle upper C upper A prime upper B prime, &c.

Evidently upper A prime upper C prime, upper A prime upper B prime are equally inclined to a line drawn, from upper A prime, perpendicular upper B upper C; and so of the others: i. e. these perpendiculars bisect the angles at upper A prime, upper B prime, upper C prime;

therefore they meet in the same Point. Draw them; let them meet at upper O; and call the angle upper C prime upper A prime upper B prime '2 alpha', and so on.

Now left parenthesis beta plus gamma right parenthesis equals pi minus angle upper B prime upper O upper C Superscript prime Baseline equals upper A;

therefore 2 upper A equals 2 left parenthesis beta plus gamma right parenthesis equals pi minus 2 alpha;

therefore alpha equals 90 degree negative upper A;

therefore angle upper B upper A prime upper C Superscript prime Baseline equals upper A.

Similarly, angle upper B upper C prime upper A Superscript prime Baseline equals upper C.

therefore Triangle upper B upper C prime upper A prime is similar to Triangle upper B upper C upper A; and so of the others;

StartLayout 1st Row 1st Column therefore upper B upper A prime 2nd Column equals StartFraction c Over a EndFraction dot upper B upper C Superscript prime Baseline equals StartFraction c Over a EndFraction dot left parenthesis c minus upper A upper C Superscript prime Baseline right parenthesis comma 2nd Row 1st Column Blank 2nd Column equals StartFraction c Over a EndFraction dot left parenthesis c minus StartFraction b Over c EndFraction dot upper A upper B Superscript prime Baseline right parenthesis comma 3rd Row 1st Column Blank 2nd Column equals StartFraction c squared Over a EndFraction minus StartFraction b Over a EndFraction dot left parenthesis b minus upper C upper B Superscript prime Baseline right parenthesis comma 4th Row 1st Column Blank 2nd Column equals StartFraction c squared Over a EndFraction minus StartFraction b squared Over a EndFraction plus StartFraction b Over a EndFraction dot StartFraction a Over b EndFraction dot upper C upper A Superscript prime Baseline comma 5th Row 1st Column Blank 2nd Column equals StartFraction c squared Over a EndFraction minus StartFraction b squared Over a EndFraction plus a minus upper B upper A Superscript prime Baseline semicolon 6th Row 1st Column therefore 2 upper B upper A prime 2nd Column equals StartFraction c squared plus a squared minus b squared Over a EndFraction equals StartFraction 2 c a cosine upper B Over a EndFraction semicolon 7th Row 1st Column therefore upper B upper A prime 2nd Column equals c cosine upper B semicolon EndLayout

therefore upper A prime is foot of perpendicular drawn, from upper A, to upper B upper C. Hence the construction is obvious.

Q. E. F.

5. (2, 19)

At first sight, it would appear that, as the state of the bag, after the operation, is necessarily identical with its state before it, the chance is just what it then was, viz. one half. This, however, is an error.

The chances, before the addition, that the bag contains (a) 1 white (b) 1 black, are (a) one half (b) one half. Hence the chances, after the addition, that it contains (a) 2 white (b) 1 white, 1 black, are the same, viz. (a) one half (b) one half. Now the probabilities, which these 2 states give to the observed event, of drawing a white counter, are (a) certainty (b) one half. Hence the chances, after drawing the white counter, that the bag, before drawing, contained (a) 2 white, (b) 1 white, 1 black, are proportional to (a) one half dot 1 (b) one half dot one half; i. e. (a) one half (b) one fourth; i. e. (a) 2 (b) 1. Hence the chances are (a) two thirds (b) one third. Hence, after the removal of a white counter, the chances, that the bag now contains (a) 1 white (b) 1 black, are for (a) two thirds and for (b) one third.

Thus the chance, of now drawing a white counter, is two thirds.

Q. E. F.

6. (2, 19)

Call sides '2 a, 2 b, 2 c', and lines in question 'alpha, beta, gamma'.

A triangle with vertices A (apex, top right), B (bottom left), and C (bottom right). Point D lies on base BC between B and C, with line segment AD drawn from apex A down to D.

Now cosine upper A upper D upper B plus cosine upper A upper D upper C equals 0;

StartLayout 1st Row 1st Column therefore 2nd Column StartFraction alpha squared plus a squared minus 4 c squared Over 2 alpha a EndFraction plus StartFraction alpha squared plus a squared minus 4 b squared Over 2 alpha a EndFraction equals 0 semicolon 2nd Row 1st Column therefore 2 alpha squared 2nd Column plus 2 a squared minus 4 b squared minus 4 c squared equals 0 semicolon 3rd Row 1st Column therefore alpha squared 2nd Column equals minus a squared plus 2 b squared plus 2 c squared period 4th Row 1st Column Similarly comma beta squared 2nd Column equals 2 a squared minus b squared plus 2 c squared semicolon 5th Row 1st Column gamma squared 2nd Column equals 2 a squared plus 2 b squared minus c squared period EndLayout

To eliminate b, c, let us multiply by k, l, m, so taken that

StartLayout 1st Row 1st Column Blank 2nd Column 2 k minus l plus 2 m equals 0 comma 2nd Row 1st Column and 2nd Column 2 k plus 2 l minus m equals 0 semicolon 3rd Row 1st Column therefore 2nd Column 3 left parenthesis l minus m right parenthesis equals 0 semicolon i period e period l equals m semicolon 4th Row 1st Column therefore 2nd Column 2 k equals negative l equals negative m semicolon EndLayout

hence we may make k equals negative 1 comma l equals 2 comma m equals 2;

StartLayout 1st Row 1st Column Blank 2nd Column therefore minus alpha squared plus 2 beta squared plus 2 gamma squared equals 9 a squared semicolon 2nd Row 1st Column Blank 2nd Column i period e period a squared equals StartFraction minus alpha squared plus 2 beta squared plus 2 gamma squared Over 9 EndFraction semicolon 3rd Row 1st Column Blank 2nd Column therefore upper B upper C left parenthesis which equals 2 a right parenthesis equals two thirds StartRoot minus alpha squared plus 2 beta squared plus 2 gamma squared EndRoot comma ampersand c period comma EndLayout

which gives lengths of sides.

StartLayout 1st Row 1st Column Also cosine upper A 2nd Column equals StartFraction b squared plus c squared minus a squared Over 2 b c EndFraction 2nd Row 1st Column Blank 2nd Column equals StartFraction 2 alpha squared minus beta squared plus 2 gamma squared plus 2 alpha squared plus 2 beta squared minus gamma squared plus alpha squared minus 2 beta squared minus 2 gamma squared Over 2 period StartRoot 2 alpha squared minus beta squared plus 2 gamma squared EndRoot period StartRoot 2 alpha squared plus 2 beta squared minus gamma squared EndRoot EndFraction 3rd Row 1st Column Blank 2nd Column equals StartFraction 5 alpha squared minus beta squared minus gamma squared Over den period EndFraction semicolon and so for other angles period EndLayout

Q. E. F.

7. (2, 19)

Let upper A upper B, upper A upper D be given sides, and upper B, upper D the right angles; and let upper A upper B equals b, upper A upper D equals d.

A triangle with vertices A (bottom left), B (bottom right), and D (top right). Point C lies on side BD near D, and point E lies on the base outside the triangle to the right of B, with dashed lines from C and D down to E.

Produce upper D upper C to meet upper A upper B-produced at upper E.

StartLayout 1st Row 1st Column Now upper A upper E 2nd Column equals upper A upper D period secant upper A equals d secant upper A semicolon 2nd Row 1st Column therefore upper B upper E 2nd Column equals d secant upper A minus b period 3rd Row 1st Column Also upper B upper C 2nd Column equals upper B upper E period tangent upper E equals left parenthesis d secant upper A minus b right parenthesis period cotangent upper A comma 4th Row 1st Column Blank 2nd Column equals StartFraction d minus b cosine upper A Over sine upper A EndFraction semicolon EndLayout

similarly, upper C upper D equals StartFraction b minus d cosine upper A Over sine upper A EndFraction; which answers (1).

Triangle ABD with point C on side DB and external point E with dashed lines.

StartLayout 1st Row 1st Column Also area 2nd Column equals one half dot left parenthesis upper A upper B period upper B upper C plus upper A upper D period upper D upper C right parenthesis comma 2nd Row 1st Column Blank 2nd Column equals one half dot StartFraction b period left parenthesis d minus b cosine upper A right parenthesis plus d period left parenthesis b minus d cosine upper A right parenthesis Over sine upper A EndFraction comma 3rd Row 1st Column Blank 2nd Column equals StartFraction 2 b d minus left parenthesis b squared plus d squared right parenthesis cosine upper A Over 2 sine upper A EndFraction semicolon which answers left parenthesis 2 right parenthesis period EndLayout

Q. E. F.

8. (2, 20)

Let m = No. of men, k = No. of shillings possessed by the last (i. e. the poorest) man. After one circuit, each is a shilling poorer, and the moving heap contains m shillings. Hence, after k circuits, each is k shillings poorer, the last man now having nothing, and the moving heap contains m k shillings. Hence the thing ends when the last man is again called on to hand on the heap, which then contains left parenthesis m k plus m minus 1 right parenthesis shillings, the penultimate man now having nothing, and the first man having left parenthesis m minus 2 right parenthesis shillings.

It is evident that the first and last man are the only 2 neighbours whose possessions can be in the ratio '4 to 1'. Hence either

StartLayout 1st Row 1st Column m k plus m minus 1 2nd Column equals 4 left parenthesis m minus 2 right parenthesis comma 2nd Row 1st Column or else 4 left parenthesis m k plus m minus 1 right parenthesis 2nd Column equals m minus 2 period EndLayout

The first equation gives m k equals 3 m minus 7, i. e. k equals 3 minus StartFraction 7 Over m EndFraction, which evidently gives no integral values other than m equals 7, k equals 2.

The second gives 4 m k equals 2 minus 3 m, which evidently gives no positive integral values.

Hence the answer is '7 men; 2 shillings'.

9. (2)

Let upper A upper B, upper A upper C, be the given Lines, and upper P the given Point; and join upper A upper P.

Parallelogram AEGF (left) and parallelogram AKPH (right) overlap near center points G, K, H, M. Line AC runs horizontally through base; line AB crosses diagonally upper right through point L and P, with interior intersection points labeled K, H, M.

Through upper A draw upper E upper A upper F, perpendicular upper A upper P, and bisected at upper A; from upper E, upper F, draw upper E upper G, upper F upper H, parallel to upper A upper P, and meeting upper A upper B, upper A upper C, at upper G, upper H; join upper G upper H, and on it describe a semicircle cutting upper A upper P at upper K; and join upper K upper G, upper K upper H. Then angle upper G upper K upper H is a right angle. From upper P draw upper P upper L, upper P upper M, parallel to upper K upper G, upper K upper H.

Now Triangle upper A upper P upper L has, to Triangle upper A upper K upper G, the duplicate ratio of upper A upper P to upper A upper K;

but so also has triangle upper A upper P upper M to Triangle upper A upper K upper H;

also Triangles upper A upper K upper G, upper A upper K upper H, are equal, being on the same base upper A upper K, and having equal altitudes upper A upper E, upper A upper F;

therefore Triangles upper A upper P upper L, upper A upper P upper M are equal: and angle upper L upper P upper M is evidently equal to angle upper G upper K upper H; therefore it is a right angle.

Q. E. F.

10. (3, 20)

Call them x, y, z; and let x plus y plus z equals s.

The chance, that the pocket contains 2 balls, is two thirds; and, if it does, the 'expectation' is the average value of

left parenthesis y plus z right parenthesis comma left parenthesis z plus x right parenthesis comma left parenthesis x plus y right parenthesis semicolon i period e period it is StartFraction 2 s Over 3 EndFraction period

Also the chance, that it contains only one, is one third; and, if it does, the 'expectation' is StartFraction s Over 3 EndFraction.

Hence total 'expectation' equals StartFraction 4 s Over 9 EndFraction plus StartFraction s Over 9 EndFraction equals StartFraction 5 s Over 9 EndFraction.

therefore StartFraction 5 s Over 9 EndFraction equals 30 d period semicolon therefore s equals 54 d period equals 4 divided by 6 period

Hence the coins must be 2 florins and a sixpence; or else a half-crown and 2 shillings.

Q. E. F.

11. (3, 20)

Large triangle with vertices A (top), B (bottom left), and C (bottom right). Midpoints A' (on BC), B' (on AC), and C' (on AB) form an inscribed medial triangle A'B'C', with all six connecting segments drawn, creating four smaller congruent triangles inside.

Now StartFraction upper B upper A Superscript prime Baseline Over upper A prime upper C Superscript prime Baseline EndFraction equals StartFraction sine left parenthesis upper B plus theta right parenthesis Over sine upper B EndFraction semicolon and StartFraction upper A prime upper C Over upper A prime upper B Superscript prime Baseline EndFraction equals StartFraction sine theta Over sine upper C EndFraction period

therefore upper B upper A equals StartFraction sine left parenthesis upper B plus theta right parenthesis Over sine upper B EndFraction period k a semicolon and upper A prime upper C equals StartFraction sine theta Over sine upper C EndFraction period k b

but upper B upper A Superscript prime Baseline plus upper A prime upper C equals a semicolon therefore k

equals StartStartFraction a OverOver StartFraction a period sine left parenthesis upper B plus theta right parenthesis Over sine upper B EndFraction plus StartFraction b sine theta Over sine upper C EndFraction EndEndFraction equals StartStartFraction sine upper A OverOver StartFraction sine upper A sine left parenthesis upper B plus theta right parenthesis Over sine upper B EndFraction plus StartFraction sine upper B sine theta Over sine upper C EndFraction EndEndFraction

StartLayout 1st Row 1st Column equals 2nd Column StartFraction sine upper A sine upper B sine upper C Over sine upper A sine left parenthesis upper B plus theta right parenthesis sine upper C plus sine squared upper B sine theta EndFraction 2nd Row 1st Column equals 2nd Column StartFraction sine upper A sine upper B sine upper C Over sine upper A sine upper C left parenthesis sine upper B cosine theta plus cosine upper B sine theta right parenthesis plus left parenthesis 1 minus cosine squared upper B right parenthesis sine theta EndFraction 3rd Row 1st Column equals 2nd Column StartFraction sine upper A sine upper B sine upper C Over StartLayout 1st Row StartLayout 1st Row sine theta plus sine theta left parenthesis sine upper A sine upper C cosine upper B minus cosine squared upper B right parenthesis 2nd Row plus cosine theta sine upper A sine upper B sine upper C EndLayout EndLayout right brace EndFraction 4th Row 1st Column equals 2nd Column StartFraction sine upper A sine upper B sine upper C Over StartLayout 1st Row StartLayout 1st Row sine theta plus sine theta cosine upper B left parenthesis sine upper A sine upper C plus cosine left parenthesis upper A plus upper C right parenthesis right parenthesis 2nd Row plus cosine theta sine upper A sine upper B sine upper C EndLayout EndLayout right brace EndFraction 5th Row 1st Column equals 2nd Column StartFraction sine upper A sine upper B sine upper C Over sine theta left parenthesis 1 plus cosine upper A cosine upper B cosine upper C right parenthesis plus cosine theta sine upper A sine upper B sine upper C EndFraction period EndLayout

Q. E. F.

Cor. Let theta equals 90 degree; then k equals StartFraction sine upper A sine upper B sine upper C Over 1 plus cosine upper A cosine upper B cosine upper C EndFraction.

12. (4, 20)

Let s equals semi-perimeter, m equals area, v equals volume.

We know that m equals StartRoot s period left parenthesis s minus a right parenthesis period left parenthesis s minus b right parenthesis period left parenthesis s minus c right parenthesis EndRoot;

StartLayout 1st Row 1st Column therefore m squared 2nd Column equals s period left parenthesis s minus a right parenthesis period left parenthesis s minus b right parenthesis period left parenthesis s minus c right parenthesis semicolon 2nd Row 1st Column therefore StartFraction m squared Over s EndFraction 2nd Column equals s cubed minus s squared period left parenthesis a plus b plus c right parenthesis plus s period left parenthesis b c plus c a plus a b right parenthesis minus a b c semicolon 3rd Row 1st Column Blank 2nd Column equals s cubed minus 2 s cubed plus s period left parenthesis b c plus c a plus a b right parenthesis minus v semicolon 4th Row 1st Column therefore 2nd Column StartFraction m squared Over s squared EndFraction plus StartFraction v Over s EndFraction plus s squared equals b c plus c a plus a b semicolon 5th Row 1st Column therefore 2nd Column 2 period left parenthesis StartFraction m squared Over s squared EndFraction plus StartFraction v Over s EndFraction plus s squared right parenthesis equals left parenthesis a plus b plus c right parenthesis squared minus left parenthesis a squared plus b squared plus c squared right parenthesis semicolon 6th Row 1st Column Blank 2nd Column equals 4 s squared minus left parenthesis a squared plus b squared plus c squared right parenthesis semicolon 7th Row 1st Column therefore 2nd Column a squared plus b squared plus c squared equals 2 period left parenthesis s squared minus StartFraction v Over s EndFraction minus StartFraction m squared Over s squared EndFraction right parenthesis period EndLayout

Q. E. F.

13. (4, 20)

A rhombus with vertices A (left), B (right), C (top), and D (bottom), with diagonals AB and CD intersecting at center. Points F and E lie on diagonal AB. An ellipse is inscribed touching all four sides, with tick marks at C and D indicating equal segments.

Let upper A, upper B, be the centres of the Circles; upper C, upper D, their points of intersection; and upper C upper F upper D upper E the Tetragon whose area is required.

Let the sides of the Triangle upper A upper B upper C be a, b, c; and its angles alpha, beta, gamma.

Then upper C upper E equals b period tangent alpha, and upper C upper F equals a period tangent beta.

Also angle upper F upper C upper E equals angle upper A upper C upper E plus angle upper F upper C upper B minus gamma equals pi minus gamma;

therefore sine upper F upper C upper E equals sine gamma.

Hence area of Triangle upper F upper C upper E equals one half period a b period tangent alpha period tangent beta period sine gamma;

therefore area of Tetragon = StartFraction a b sine alpha sine beta sine gamma Over cosine alpha cosine beta EndFraction.

Now, writing 'upper M' for area of Triangle upper A upper B upper C, we have

sine alpha equals StartFraction 2 upper M Over b c EndFraction comma sine beta equals StartFraction 2 upper M Over c a EndFraction comma sine gamma equals StartFraction 2 upper M Over a b EndFraction semicolon

StartLayout 1st Row 1st Column therefore area of Tetragon 2nd Column equals a b dot StartFraction 8 upper M cubed Over a squared b squared c squared EndFraction dot StartFraction 4 b c period c a Over left parenthesis b squared plus c squared minus a squared right parenthesis left parenthesis c squared plus a squared minus b squared right parenthesis EndFraction semicolon 2nd Row 1st Column Blank 2nd Column equals StartFraction 32 upper M cubed Over left parenthesis b squared plus c squared minus a squared right parenthesis period left parenthesis c squared plus a squared minus b squared right parenthesis EndFraction period EndLayout

Q. E. F.

14. (4)

This simply expresses the identity

StartLayout 1st Row 1st Column Blank 2nd Column 3 left parenthesis a squared plus b squared plus c squared right parenthesis 2nd Row 1st Column Blank 2nd Column equals left parenthesis a plus b plus c right parenthesis squared plus left parenthesis b squared minus 2 b c plus c squared right parenthesis plus left parenthesis c squared minus 2 c a plus a squared right parenthesis plus left parenthesis a squared minus 2 a b plus b squared right parenthesis semicolon 3rd Row 1st Column Blank 2nd Column equals left parenthesis a plus b plus c right parenthesis squared plus left parenthesis b minus c right parenthesis squared plus left parenthesis c minus a right parenthesis squared plus left parenthesis a minus b right parenthesis squared period EndLayout

Q. E. D.

Numerical Examples (not thought out).

StartLayout 1st Row 1st Column 3 left parenthesis 1 squared plus 2 squared plus 3 squared right parenthesis 2nd Column equals 6 squared plus 1 squared plus 2 squared plus 1 squared period 2nd Row 1st Column 3 left parenthesis 1 squared plus 3 squared plus 7 squared right parenthesis 2nd Column equals 11 squared plus 4 squared plus 6 squared plus 2 squared period EndLayout

15. (4)

Rectangle ADCD with vertices A (top left), D (top right), C (bottom right), and B' (bottom, left of C). Point B lies on the base between B″ and B', with dashed lines from A to B″ (external left point) and solid cevians from A to B and B', forming overlapping triangles on a shared base.

Let upper A upper B upper C upper D be an inscribed Tetragon. Join upper A upper C: and about Triangle upper A upper C upper D describe a Circle.

Now, if this Circle does not pass through upper B, let it cut upper C upper B, or upper C upper B produced, in upper B prime or upper B double prime. Join upper A upper B prime, upper A upper B double prime.

Then angle upper A upper B prime upper C, or angle upper A upper B double prime upper C, is supplementary to angle upper A upper D upper C;

therefore it equals angle upper A upper B upper C; which is absurd;

therefore this Circle does pass through upper B.

The same thing may be proved for any other Point on that portion, of the perimeter of the given Figure, which lies on the same side of upper A upper C as the Point upper D.

Similarly for the other portion.

Hence the Figure is a Circle.

Q. E. D.

16. (4, 20)

The 'a priori' chances of possible states of first bag are 'upper W, one half; upper B, one half'. Hence chances, after putting upper W in, are 'upper W upper W, one half; upper W upper B, one half'. The chances, which these give to the 'observed event', are 1, one half. Hence chances of possible states 'upper W, upper B', after the event, are proportional to 1, one half; i. e. to 2, 1; i. e. their actual values are two thirds, one third.

Now, in first course, chance of drawing upper W is one half dot two thirds plus one half dot one third; i. e. one half.

And, in second course, chances of possible states 'upper W upper W upper B upper B, upper W upper B upper B upper B' are two thirds, one third: hence chance of drawing upper W is two thirds dot one half plus one third dot one fourth; i. e. five twelfths.

Hence first course gives best chance.

Q. E. F.

17. (4)

(Analysis.)

Let upper A upper B upper C be the given Triangle, and upper D upper E the line required.

Large triangle with vertices A (top), B (bottom left), C (bottom right). Segment DE is parallel to BC with D on AB and E on AC. Points F and G lie on BC, H on DE. Solid and dashed cevians cross between D, E, F, G, H, forming an inner quadrilateral with intersecting diagonals.

From upper D, upper E comma draw upper D upper F, upper E upper G, parallel to the sides. Then upper D upper F plus upper E upper G equals upper D upper E.

Because upper B upper E is a Parallelogram, therefore upper D upper B equals upper E upper G;

similarly upper E upper C equals upper D upper F;

therefore upper D upper B plus upper E upper C equals upper D upper E.

Hence construction.

(Synthesis.)

Bisect angles upper B, upper C, by upper B upper H, upper C upper H, meeting at upper H; through upper H draw upper D upper E parallel to upper B upper C; and from upper D, upper E, draw upper D upper F, upper E upper G, parallel to upper A upper C, upper A upper B.

Because upper D upper E is parallel to upper B upper C,

therefore angle upper D upper H upper B equals alternate angle upper H upper B upper F equals angle upper D upper B upper H semicolon

therefore upper D upper B equals upper D upper H period

Similarly upper E upper C equals upper E upper H.

therefore upper D upper B plus upper E upper C equals upper D upper E.

Because upper B upper E, upper D upper C are Parallelograms,

therefore upper E upper G equals upper D upper B comma and upper D upper F equals upper E upper C semicolon

therefore upper D upper F plus upper E upper G equals upper D upper E period

Q. E. F.

18. (5, 21)

Large triangle with vertices A (top), B (bottom left), C (bottom right). Segment FG is parallel to BC with F on AB and G on AC. Points H, E, K lie on BC between B and C.

(1) Call required Point upper E. From upper E draw upper E upper F, upper F upper G perpendicular sides. Join upper F upper G. From upper F, upper G, draw upper F upper H, upper G upper K perpendicular upper B upper C. Call upper B upper E 'x', and upper E upper C 'y'.

StartLayout 1st Row 1st Column Now upper F upper H must equals upper G upper K period 2nd Row 1st Column Also upper E upper F equals x sine upper B semicolon and upper F upper H 2nd Column equals upper E upper F sine upper F upper E upper H comma 3rd Row 1st Column Blank 2nd Column equals upper E upper F cosine upper B comma 4th Row 1st Column Blank 2nd Column equals x sine upper B cosine upper B semicolon 5th Row 1st Column similarly comma upper G upper K 2nd Column equals y sine upper C cosine upper C period EndLayout

StartLayout 1st Row 1st Column But upper F upper H equals upper G upper K semicolon 2nd Column therefore x sine upper B cosine upper B equals y sine upper C cosine upper C semicolon 2nd Row 1st Column Blank 2nd Column therefore StartFraction x Over y EndFraction equals StartFraction sine 2 upper C Over sine 2 upper B EndFraction period EndLayout

Q. E. F.

A geometric figure with apex A (top), base vertices B (left), C (right), D (bottom), and midpoints F (upper left), G (upper right), E (center). Multiple triangulating segments connect all vertices, forming a faceted diamond-like shape with a small triangle marked at apex A.

(2) At upper B, upper C, make right angles upper A upper B upper D, upper A upper C upper D; and join upper A upper D, cutting upper B upper C at upper E. From upper E draw upper E upper F, upper E upper G perpendicular sides; and join upper F upper G.

because upper B upper D, upper F upper E are perpendicular upper A upper B comma therefore they are parallel to; therefore upper A upper F colon upper F upper B colon colon upper A upper E colon upper E upper D;

because upper C upper D, upper G upper E are perpendicular upper A upper C comma therefore they are parallel to; therefore upper A upper G colon upper G upper C colon colon upper A upper E colon upper E upper D;

therefore upper A upper F colon upper F upper B colon colon upper A upper G colon upper G upper C;

therefore upper F upper G is parallel to upper B upper C.

Q. E. F.

19. (5, 21)

Call the bags upper A, upper B, upper C; so that upper A contains a white counter and a black one; &c.

The chances of the orders upper A upper B upper C, upper A upper C upper B, upper B upper A upper C, upper B upper C upper A, upper C upper A upper B, upper C upper B upper A, are, a priori, one sixth each. Since they are equal, we may, instead of multiplying each by the probability it gives to the observed event, simply assume those probabilities as being proportional to the chances after the observed event.

These probabilities are:—

StartLayout 1st Row 1st Column for 2nd Column upper A upper B upper C comma one half times one third semicolon i period e period one sixth period 2nd Row 1st Column Blank 2nd Column upper A upper C upper B comma one half times one fourth semicolon i period e period one eighth period 3rd Row 1st Column Blank 2nd Column upper B upper A upper C comma two thirds times one half semicolon i period e period one third period 4th Row 1st Column Blank 2nd Column upper B upper C upper A comma two thirds times one fourth semicolon i period e period one sixth period 5th Row 1st Column Blank 2nd Column upper C upper A upper B comma three fourths times one half semicolon i period e period three eighths period 6th Row 1st Column Blank 2nd Column upper C upper B upper A comma three fourths times one third semicolon i period e period one fourth period EndLayout

Hence the chances are proportional to 4, 3, 8, 4, 9, 6; i. e. they are these Nos. divided by 34.

Hence the chance, of drawing a white counter from the remaining bag; is

StartLayout 1st Row 1st Column Blank 2nd Column one thirty fourth dot left brace 4 times three fourths plus 3 times two thirds plus 8 times three fourths plus 4 times one half plus 9 times two thirds plus 6 times one half right brace semicolon 2nd Row 1st Column i period e period 2nd Column one thirty fourth times left brace 3 plus 2 plus 6 plus 2 plus 6 plus 3 right brace semicolon i period e period StartFraction 22 Over 34 EndFraction semicolon i period e period StartFraction 11 Over 17 EndFraction period EndLayout

20. (5)

(Analysis.)

Let upper A upper B upper C be the given Triangle, and upper P the required Point. Draw upper P upper Q perpendicular upper B upper C, and upper P upper R perpendicular upper A upper B. Then upper P upper Q equals upper P upper R.

Triangle ABC (lower, vertices B bottom-left, C bottom-right, A upper-middle) overlaps with triangle BEC (larger, apex E upper-right). Interior points R (on AB), D, P (on BC), and Q (on AC/AE) mark cevian intersections, with multiple segments drawn through A creating a dense central crossing.

Hence upper P upper C tangent upper C equals upper P upper B sine upper B;

therefore upper P upper C colon upper P upper B colon colon sine upper B colon tangent upper C, (draw upper A upper D perpendicular upper B upper C,)

StartLayout 1st Row colon colon StartFraction upper A upper D Over upper A upper B EndFraction colon StartFraction upper A upper D Over upper D upper C EndFraction comma 2nd Row colon colon upper D upper C colon upper A upper B period EndLayout

Hence construction.

(Synthesis.)

From upper A draw upper A upper D perpendicular upper B upper C. Produce upper B upper A to upper E, making upper A upper E equal to upper D upper C. Join upper E upper C. From upper A draw upper A upper P parallel to upper E upper C; and from upper P draw upper P upper Q perpendicular upper B upper C, and upper P upper R perpendicular upper A upper B.

StartLayout 1st Row 1st Column Then StartFraction upper P upper Q Over upper P upper C EndFraction equals StartFraction upper A upper D Over upper D upper C EndFraction 2nd Column equals StartFraction upper A upper D Over upper A upper B EndFraction dot StartFraction upper A upper B Over upper D upper C EndFraction comma 2nd Row 1st Column Blank 2nd Column equals StartFraction upper P upper R Over upper P upper B EndFraction dot StartFraction upper A upper B Over upper A upper E EndFraction comma 3rd Row 1st Column Blank 2nd Column equals StartFraction upper P upper R Over upper P upper B EndFraction dot StartFraction upper P upper B Over upper P upper C EndFraction equals StartFraction upper P upper R Over upper P upper C EndFraction semicolon 4th Row 1st Column therefore upper P upper Q 2nd Column equals upper P upper R period EndLayout

Q. E. F.

21. (5, 21)

(1) The nth term is n period ModifyingAbove n plus 2 With quotation dash period ModifyingAbove n plus 4 With quotation dash;

therefore the left parenthesis n plus 1 right parenthesisth term is ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 3 With quotation dash period ModifyingAbove n plus 5 With quotation dash;

equals left parenthesis n plus 1 right parenthesis period left parenthesis ModifyingAbove n plus 2 With quotation dash plus 1 right parenthesis period left parenthesis n plus 5 right parenthesis semicolon

equals ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash period ModifyingAbove n plus 5 With quotation dash plus ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 5 With quotation dash

equals ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash period left parenthesis ModifyingAbove n plus 3 With quotation dash plus 2 right parenthesis plus ModifyingAbove n plus 1 With quotation dash period left parenthesis ModifyingAbove n plus 2 With quotation dash plus 3 right parenthesis semicolon

equals ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash period ModifyingAbove n plus 3 With quotation dash plus 2 period ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash plus ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash plus 3 period ModifyingAbove n plus 1 With quotation dash semicolon

equals ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash period ModifyingAbove n plus 3 With quotation dash plus 3 period ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash plus 3 period ModifyingAbove n plus 1 With quotation dash period

therefore upper S equals StartFraction n period ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash period ModifyingAbove n plus 3 With quotation dash Over 4 EndFraction plus n period ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 2 With quotation dash plus three halves period n period ModifyingAbove n plus 1 With quotation dash plus upper C semicolon

and upper C equals 0 period

therefore upper S equals n period ModifyingAbove n plus 1 With quotation dash period left parenthesis StartFraction n squared plus 5 n plus 6 Over 4 EndFraction plus n plus 2 plus three halves right parenthesis semicolon

equals n period ModifyingAbove n plus 1 With quotation dash period StartFraction n squared plus 9 n plus 20 Over 4 EndFraction equals StartFraction n period ModifyingAbove n plus 1 With quotation dash period ModifyingAbove n plus 4 With quotation dash period ModifyingAbove n plus 5 With quotation dash Over 4 EndFraction period

Q. E. F.

(2) upper S, to 100 terms,

equals StartFraction 100.101 .104 .105 Over 4 EndFraction equals 100.101 .26 .105 semicolon

now 101.105 equals 10 comma 605;

therefore 101.105 .13 equals 130 comma 000 plus 7800 plus 65 equals 137 comma 865;

and twice this equals 274 comma 000 plus 1730 equals 275 comma 730;

therefore upper S equals 27 comma 573 comma 000.

Q. E. F.

22. (5, 21)

A triangle with vertices A (top), B (bottom left), and C (bottom right). A vertical altitude is drawn from apex A perpendicular to base BC, with the altitude labeled α, dividing the base into two segments.

Call given altitudes 'alpha, beta, gamma'.

Now a alpha equals b beta equals c gamma;

therefore alpha sine upper A equals beta sine upper B equals gamma sine upper C;

therefore StartFraction sine upper A Over beta gamma EndFraction equals StartFraction sine upper B Over gamma alpha EndFraction equals StartFraction sine upper C Over alpha beta EndFraction equals k (say);

therefore sine upper A equals k beta gamma comma sine upper B equals k gamma alpha comma sine upper C equals k alpha beta.

Now sine left parenthesis upper A plus upper B right parenthesis equals sine upper C;

therefore sine upper A cosine upper B plus cosine upper A sine upper B equals sine upper C;

therefore sine upper A cosine upper B equals sine upper C minus cosine upper A sine upper B;

therefore sine squared upper A left parenthesis 1 minus sine squared upper B right parenthesis equals sine squared upper C plus sine squared upper B left parenthesis 1 minus sine squared upper A right parenthesis minus 2 sine upper C cosine upper A sine upper B;

therefore sine squared upper A minus sine squared upper A sine squared upper B equals sine squared upper C plus sine squared upper B minus sine squared upper A sine squared upper B minus 2 sine upper B sine upper C cosine upper A;

therefore sine squared upper A minus sine squared upper B minus sine squared upper C equals minus 2 sine upper B sine upper C cosine upper A;

therefore comma squaring comma left parenthesis sine Superscript 4 Baseline upper A plus ampersand c period right parenthesis minus 2 sine squared upper A sine squared upper B minus 2 sine squared upper A sine squared upper C plus 2 sine squared upper B sine squared upper C equals 4 sine squared upper B sine squared upper C left parenthesis 1 minus sine squared upper A right parenthesis;

therefore left parenthesis sine Superscript 4 Baseline upper A plus ampersand c period right parenthesis minus 2 left parenthesis sine squared upper B sine squared upper C plus ampersand c period right parenthesis plus 4 sine squared upper A sine squared upper B sine squared upper C equals 0;

therefore, substituting for sine upper A, &c., and dividing by k Superscript 4, left parenthesis beta Superscript 4 Baseline gamma Superscript 4 Baseline plus ampersand c period right parenthesis minus 2 alpha squared beta squared gamma squared period left parenthesis alpha squared plus ampersand c period right parenthesis plus 4 k squared alpha Superscript 4 Baseline beta Superscript 4 Baseline gamma Superscript 4 Baseline equals 0;

Triangle ABC with altitude from A to base BC labeled α.

therefore k squared equals StartFraction 2 alpha squared beta squared gamma squared left parenthesis alpha squared plus ampersand c period right parenthesis minus left parenthesis beta Superscript 4 Baseline gamma Superscript 4 Baseline plus ampersand c period right parenthesis Over 4 alpha Superscript 4 Baseline beta Superscript 4 Baseline gamma Superscript 4 Baseline EndFraction.

Now sine upper A equals k beta gamma, &c.; which answers (2).

Also alpha equals b sine upper C; and similarly gamma equals a sine upper B;

therefore a equals StartFraction gamma Over sine upper B EndFraction equals StartFraction gamma Over k gamma alpha EndFraction equals StartFraction 1 Over k alpha EndFraction, \text{&c.}; which answers (1).

Also area equals StartFraction b c sine upper A Over 2 EndFraction equals one half dot StartFraction 1 Over k beta EndFraction dot StartFraction 1 Over k gamma EndFraction dot k beta gamma equals StartFraction 1 Over 2 k EndFraction;

which answers (3).

Q.E.F.

23. (5, 21)

The original chances, as to states of bag, are

StartLayout 1st Row 1st Column for 2nd Column 2 upper W 3rd Column one fourth semicolon 2nd Row 1st Column Blank 2nd Column 1 upper W comma 1 upper B 3rd Column one half semicolon 3rd Row 1st Column Blank 2nd Column 2 upper B 3rd Column one fourth period EndLayout

therefore the chances, after adding 2 W and 1 B, are

StartLayout 1st Row 1st Column for 2nd Column 4 upper W comma 1 upper B 3rd Column one fourth semicolon 2nd Row 1st Column Blank 2nd Column 3 upper W comma 2 upper B 3rd Column one half semicolon 3rd Row 1st Column Blank 2nd Column 2 upper W comma 3 upper B 3rd Column one fourth period EndLayout

Now the chances, which these give to the observed event, drawing 2 W and 1 B, are three fifths, three fifths, three tenths.

therefore the chances, after this event, are proportional to three twentieths, three tenths, three fortieths; i. e. to 2, 4, 1. Hence they are two sevenths, four sevenths, one seventh.

Hence the chances, as to states, now are

StartLayout 1st Row 1st Column for 2nd Column 2 upper W 3rd Column two sevenths semicolon 2nd Row 1st Column Blank 2nd Column 1 upper W comma 1 upper B 3rd Column four sevenths semicolon 3rd Row 1st Column Blank 2nd Column 2 upper B 3rd Column one seventh period EndLayout

therefore the chances, after adding 1 upper W, are

StartLayout 1st Row 1st Column for 2nd Column 3 upper W 3rd Column two sevenths semicolon 2nd Row 1st Column Blank 2nd Column 2 upper W comma 1 upper B 3rd Column four sevenths semicolon 3rd Row 1st Column Blank 2nd Column 1 upper W comma 2 upper B 3rd Column one seventh period EndLayout

Now the chances, which these give to the observed event, of drawing 1 upper W, are 1, two thirds, one third.

therefore the chances, after this event, are proportional to two sevenths, eight twenty firsts, one twenty first;

i. e. to 6, 8, 1. Hence they are six fifteenths, eight fifteenths, one fifteenth.

Hence the chance, that the bag now contains 2 white, is six fifteenths; i. e. two fifths.

Q. E. F.

24. (6, 21)

A triangle with vertices A (top), B (bottom left), C (bottom right). Three cevians AD, BE, and CF intersect at interior point O (centroid), where D lies on BC, E on AC, and F on AB, dividing the triangle into six smaller triangles meeting at O.

StartLayout 1st Row 1st Column Because 2nd Column StartFraction upper D upper O Over upper O upper A EndFraction equals StartFraction white up pointing triangle upper D upper O upper C Over white up pointing triangle upper O upper A upper C EndFraction equals StartFraction white up pointing triangle upper D upper O upper B Over white up pointing triangle upper O upper A upper B EndFraction equals StartFraction white up pointing triangle upper O upper B upper C Over white up pointing triangle upper O upper C upper A plus white up pointing triangle upper O upper A upper B EndFraction semicolon 2nd Row 1st Column therefore 2nd Column StartFraction upper D upper O Over upper D upper A EndFraction equals StartFraction white up pointing triangle upper O upper B upper C Over white up pointing triangle upper A upper B upper C EndFraction period 3rd Row 1st Column Similarly comma 2nd Column StartFraction upper E upper O Over upper E upper B EndFraction equals StartFraction white up pointing triangle upper O upper C upper A Over white up pointing triangle upper A upper B upper C EndFraction comma and StartFraction upper F upper O Over upper F upper C EndFraction equals StartFraction white up pointing triangle upper O upper A upper B Over white up pointing triangle upper A upper B upper C EndFraction period 4th Row 1st Column Hence 2nd Column StartFraction upper D upper O Over upper D upper A EndFraction plus StartFraction upper E upper O Over upper E upper B EndFraction plus StartFraction upper F upper O Over upper F upper C EndFraction equals 1 period EndLayout

Q. E. F.

25. (6, 22)

Let 'upper E' mean 'having lost an eye', 'upper A' 'having lost an arm', and 'upper L' 'having lost a leg'.

A horizontal line segment labeled ε (total length) above, subdivided into two portions: left part labeled (1−a) and right part labeled a, each marked with a brace beneath, showing a partition of the whole segment.

Then the state of things which gives the least possible number of those who, being upper E and upper A, are also upper L, may evidently be found by arranging the patients in a row, so that the upper E upper A-class may begin from one end of the row, and the upper L-class from the other end, and counting the portion where they overlap; and, the smaller the upper E upper A-class, the smaller will be this common portion: hence we must make the upper E upper A-class a minimum.

A horizontal line segment with total length labeled (ε+a−1) above, subdivided into two portions: left part labeled (1−λ) and right part labeled λ, each marked with a brace beneath, showing a complementary partition of the whole segment.

This may be done by re-arranging the patients, so that the upper E-class may begin from one end of the row, and the upper A-class from the other: and the least possible number for the upper E upper A-class is the common portion, i. e. left parenthesis epsilon minus ModifyingAbove 1 minus alpha With quotation dash right parenthesis, i. e. left parenthesis epsilon plus alpha minus 1 right parenthesis.

Then, as already shown, the least possible number for the upper E upper A upper L-class is the common portion, i. e. left parenthesis epsilon plus alpha minus 1 minus ModifyingAbove 1 minus lamda With quotation dash right parenthesis, i. e. left parenthesis epsilon plus alpha plus lamda minus 2 right parenthesis.

Q. E. F.

26. (6)

(Analysis.)

Let upper A upper B upper C be the given Triangle, and upper A prime upper B prime upper C prime the required one; and let the ratio, which upper B prime upper C prime has to upper B upper C, be 'k'; so that k is less than 1.

Since upper B upper B Superscript prime Baseline equals upper C upper C prime, and that upper B upper C, upper B prime upper C prime, are parallel, it may easily be proved, by dropping perpendiculars from upper B prime, upper C prime, upon upper B upper C, which must necessarily be equal, that angle s upper B prime upper B upper C, upper C prime upper C upper B, are equal.

Similarly, angle s upper A prime upper A upper C, upper C prime upper C upper A, are equal; and so are angle s upper A prime upper A upper B, upper B prime upper B upper A.

A triangle with vertices A (top), B (bottom left), C (bottom right). Interior triangle A'B'C' is inscribed with A' (upper middle), B' (lower left), C' (lower right). Point D lies at the center of A'B'C', with cevians from each outer vertex through the inner triangle converging near D.

Call angle upper B prime upper B upper C 'theta'; then angle upper C prime upper C upper B equals theta;

therefore angle upper C prime upper C upper A equals upper C minus theta equals angle upper A prime upper A upper C;

therefore angle upper A prime upper A upper B equals upper A minus left parenthesis upper C minus theta right parenthesis equals angle upper B prime upper B upper A.

Now angle s upper B prime upper B upper C comma upper B prime upper B upper A, together equals upper B;

therefore theta plus upper A minus left parenthesis upper C minus theta right parenthesis equals upper B;

therefore 2 theta equals upper B plus upper C minus upper A equals 180 degree minus 2 upper A;

therefore theta equals 90 degree negative upper A.

Hence, if upper B upper B prime, upper C upper C prime, be produced to meet at upper D, Triangle upper D upper B upper C will be isosceles, with a vertical angle equal to 2 upper A.

Now, if a Circle be drawn about upper A upper B upper C, and its centre joined to upper B and upper C, the Triangle, so formed, will fulfil the same conditions;

hence the centre of this Circle will be upper D;

hence the construction.

(Synthesis.)

Bisect the sides, and draw perpendiculars, meeting at upper D. Join upper D to the vertices upper B, upper C. From upper D upper B cut off upper D upper B Superscript prime Baseline equals k period upper D upper B. From upper B prime draw upper B prime upper C prime parallel to upper B upper C.

Then upper B prime upper C prime is easily proved equal to k period upper B upper C.

And if, from upper B prime, upper C prime, parallels to upper A upper B, upper A upper C, be drawn, it may easily be proved that they meet on upper D upper A, and that they are respectively equal to k period upper A upper B, k period upper A upper C.

Q. E. F.

27. (6, 22)

Call the bags upper A, upper B, upper C.

If remaining bag be upper A, chance of observed event = one half chance of drawing white from upper B and black from upper C + one half chance of drawing black from upper B and white from upper C:

i. e. it equals one half period left brace two thirds times one half plus one third times one half right brace equals one fourth.

Similarly, if remaining bag be upper B, it is one half period left brace five sixths period one half plus one sixth period one half right brace equals one fourth; and, if it be upper C, it is one half period left brace five sixths period one third plus one sixth period two thirds right brace equals seven thirty sixths.

therefore chances of remaining bag being upper A, upper B, or upper C, are as one fourth to one fourth to seven thirty sixths; i. e. as 9 to 9 to 7. therefore they are, in value, StartFraction 9 comma 9 comma 7 Over 25 EndFraction.

Now, if remaining bag be upper A, chance of drawing white from it is five sixths; therefore chance, on this issue, is five sixths period nine twenty fifths equals three tenths; similarly, for upper B, it is two thirds period nine twenty fifths equals six twenty fifths; and, for upper C, one half period seven twenty fifths equals seven fiftieths. And entire chance of drawing white from the remaining bag is the sum of these; i. e. StartFraction 15 plus 12 plus 7 Over 50 EndFraction equals StartFraction 34 Over 50 EndFraction equals StartFraction 17 Over 25 EndFraction.

28. (7, 22)

Let upper A upper B upper C be the given Triangle; and let its sides be divided internally at upper A prime, upper B prime, upper C prime, in extreme and mean ratio.

A triangle with vertices A (top), B (bottom left), C (bottom right). Midpoints A' (on BC), B' (on AC), and C' (on AB) form inscribed medial triangle A'B'C'.

And let upper M be the area of upper A upper B upper C.

Let upper B upper A Superscript prime Baseline equals x; then x squared equals a period left parenthesis a minus x right parenthesis;

i. e. x squared plus a x minus a squared equals 0;

therefore x equals StartFraction negative a plus or minus a StartRoot 5 EndRoot Over 2 EndFraction equals StartFraction a Over 2 EndFraction period left parenthesis StartRoot 5 EndRoot minus 1 right parenthesis, the other sign being excluded by the terms of the question.

Then area of Triangle upper A upper B prime upper C prime

StartLayout 1st Row 1st Column Blank 2nd Column equals one half period StartFraction c Over 2 EndFraction period left parenthesis StartRoot 5 EndRoot minus 1 right parenthesis period left brace b minus StartFraction b Over 2 EndFraction period left parenthesis StartRoot 5 EndRoot minus 1 right parenthesis right brace period sine upper A comma 2nd Row 1st Column Blank 2nd Column equals one eighth period left parenthesis StartRoot 5 EndRoot minus 1 right parenthesis left parenthesis 3 minus StartRoot 5 EndRoot right parenthesis b c period sine upper A comma 3rd Row 1st Column Blank 2nd Column equals one fourth period left parenthesis 4 StartRoot 5 EndRoot minus 8 right parenthesis period upper M equals left parenthesis StartRoot 5 EndRoot minus 2 right parenthesis period upper M period EndLayout

Similarly for upper B upper C prime upper A prime and upper C upper A prime upper B prime.

Hence the sum of these 3 Triangles equals 3 period left parenthesis StartRoot 5 EndRoot minus 2 right parenthesis period upper M, and area of Triangle upper A prime upper B prime upper C Superscript prime Baseline equals left parenthesis 7 minus 3 StartRoot 5 EndRoot right parenthesis period upper M.

Q. E. F.

29. (7)

This may be deduced from the identity

left parenthesis a squared plus b squared right parenthesis period left parenthesis c squared plus d squared right parenthesis equals a squared c squared plus b squared d squared plus a squared d squared plus b squared c squared period

StartLayout 1st Row 1st Column left parenthesis a squared plus b squared right parenthesis period left parenthesis c squared plus d squared right parenthesis 2nd Column equals a squared c squared plus b squared d squared plus a squared d squared plus b squared c squared semicolon 2nd Row 1st Column StartLayout 1st Row or else EndLayout 2nd Column StartLayout 1st Row equals a squared c squared plus b squared d squared plus 2 a c b d plus a squared d squared plus b squared c squared minus 2 a d b c comma 2nd Row equals a squared c squared plus b squared d squared minus 2 a c b d plus a squared d squared plus b squared c squared plus 2 a d b c semicolon EndLayout right brace 3rd Row 1st Column StartLayout 1st Row i period e period 2nd Row or else EndLayout 2nd Column StartLayout 1st Row equals left parenthesis a c plus b d right parenthesis squared plus left parenthesis a d minus b c right parenthesis squared comma 2nd Row equals left parenthesis a c minus b d right parenthesis squared plus left parenthesis a d plus b c right parenthesis squared period EndLayout right brace EndLayout

Now, if these last 2 sets are identical, left parenthesis a c plus b d right parenthesis must = left parenthesis a d plus b c right parenthesis; for it cannot = left parenthesis a c minus b d right parenthesis;

i. e., a left parenthesis c minus d right parenthesis minus b left parenthesis c minus d right parenthesis must = 0;

i. e., left parenthesis a minus b right parenthesis period left parenthesis c minus d right parenthesis must = 0;

i. e., one or other of the first 2 sets is the sum of 2 identical squares.

Hence, contranominally, if each of the original sets consists of 2 different squares, their product gives the sum of 2 squares in 2 different ways.

Q. E. D.

30. (7)

Large triangle with vertices A (top), B (bottom left), C (bottom right). Midpoints B' (upper right) and C' (upper left) with points D, E on BC. Interior points G, K, L, H, F, R mark cevian intersections, with solid and dashed segments forming a dense network of crossing lines and a central quadrilateral region.

(Analysis.)

Let upper A upper B upper C be the Triangle: and suppose upper B prime upper C prime so placed that upper B prime upper D, upper C prime upper E, drawn parallel to the sides, shall together = 2 upper B prime upper C prime.

By Euc. I. 34, upper B prime upper D equals upper C prime upper B, and upper C prime upper E equals upper B prime upper C:

therefore upper B prime upper C plus upper C prime upper B equals 2 upper B prime upper C Superscript prime Baseline period

Hence, if upper B prime upper L be cut off equal to half upper B prime upper C, upper C prime upper L equals half upper C prime upper B period

Hence construction.

(Synthesis.)

In upper B upper C prime take any point upper F: draw upper F upper G, parallel to upper B upper C, and equal to half upper B upper F: and join upper B upper G.

Similarly, in upper C upper B prime take any point upper H: draw upper H upper K, parallel to upper B upper C, and equal to half upper H upper C: and join upper C upper K.

Produce upper B upper G, upper C upper K, to meet at upper L: and through upper L draw upper C prime upper B Superscript prime Baseline parallel to upper B upper C: and from upper B prime, upper C prime, draw upper B prime upper D, upper C prime upper E, parallel to the sides.

because upper F upper G equals half upper F upper B semicolon therefore, by similar Triangles, upper C prime upper L equals half upper C prime upper B;

Similarly upper B prime upper L equals half upper B prime upper C;

therefore upper C prime upper B Superscript prime Baseline equals half sum of upper C prime upper B comma upper B prime upper C; i. e. upper C prime upper B plus upper B prime upper C equals 2 upper B prime upper C prime.

But, by Euc. I. 34, upper C prime upper B equals upper B prime upper D comma and upper B prime upper C equals upper C prime upper E;

therefore upper B prime upper D plus upper C prime upper E equals 2 upper B prime upper C prime.

Q. E. F.

31. (7, 22)

On July 1, watch gained on clock 5m. in 10h.; i. e. one halfm. per hour; i. e. 2m. in 4h. Hence, when watch said 'noon', clock said '12h. 2m.'; i. e. clock was 3m. slow of true time, when true time was 12h. 5m.

On July 30, watch lost on clock 1m. in 10h.; i. e. 6sec. per hour; i. e. 19sec. in 3h. 10m. Hence, when watch said '12h. 10m.', clock said '12h. 7m. 19sec.'; i. e. clock was 2m. 19sec. fast of true time, when true time was 12h. 5m.

Hence clock gains, on true time, 5m. 19sec. in 29 days; i. e. 319sec. in 29 days; i. e. 11sec. per day; i. e. StartFraction 11 Over 24 times 12 EndFractionsec. in 5m.

Hence, while true time goes 5m., watch goes 5m. StartFraction 11 Over 288 EndFractionsec.

Now, when true time is 12h. 5m. on July 31, clock is (2m. 19sec. + 11sec.) fast of it; i. e. says '12h. 7 and one halfm.' Hence, if true time be put 5m. back, clock must be put 5m. StartFraction 11 Over 288 EndFractionsec. back; i. e. must be put back to 12h. 2m. 29 and StartFraction 277 Over 288 EndFractionsec.

Hence, on July 31, when clock indicates this time, it is true noon.

Q. E. F.

32. (8, 22)

The nth term is n period left parenthesis n plus 4 right parenthesis;

StartLayout 1st Row 1st Column Blank 2nd Column therefore the left parenthesis n plus 1 right parenthesis th term is left parenthesis n plus 1 right parenthesis period left parenthesis n plus 5 right parenthesis equals left parenthesis n plus 1 right parenthesis period left brace left parenthesis n plus 2 right parenthesis plus 3 right brace comma 2nd Row 1st Column Blank 2nd Column equals left parenthesis n plus 1 right parenthesis period left parenthesis n plus 2 right parenthesis plus 3 left parenthesis n plus 1 right parenthesis semicolon 3rd Row 1st Column Blank 2nd Column therefore upper S Subscript n Baseline equals StartFraction n period left parenthesis n plus 1 right parenthesis period left parenthesis n plus 2 right parenthesis Over 3 EndFraction plus 3 dot StartFraction n period left parenthesis n plus 1 right parenthesis Over 2 EndFraction plus upper C semicolon and upper C equals 0 semicolon 4th Row 1st Column Blank 2nd Column therefore upper S Subscript n Baseline equals n period left parenthesis n plus 1 right parenthesis period left parenthesis StartFraction n plus 2 Over 3 EndFraction plus three halves right parenthesis equals StartFraction n period left parenthesis n plus 1 right parenthesis period left parenthesis 2 n plus 13 right parenthesis Over 6 EndFraction period EndLayout

Q. E. F.

Also upper S 100 equals StartFraction 100.101 .213 Over 6 EndFraction equals StartFraction 100.101 .71 Over 2 EndFraction equals StartFraction 100.7171 Over 2 EndFraction equals StartFraction 717100 Over 2 EndFraction equals 358550.

Q. E. F.

33. (8)

A circle centered at origin O with horizontal X-axis and vertical Y-axis. Points B (upper left), C (upper right), D (lower left), and E (lower right) lie on the circle, forming an inscribed trapezoid with BC parallel to DE, symmetrically placed about the Y-axis.

Let upper D upper E equals x; therefore upper B upper C equals 2 x.

Area equals 3 x period left parenthesis StartRoot r squared minus x squared EndRoot plus StartRoot r squared minus 4 x squared EndRoot right parenthesis equals max.

let v equals x period left parenthesis StartRoot r squared minus x squared EndRoot plus StartRoot r squared minus 4 x squared EndRoot right parenthesis equals max.

StartLayout 1st Row therefore StartFraction d v Over d x EndFraction equals StartRoot r squared minus x squared EndRoot plus StartRoot r squared minus 4 x squared EndRoot minus x squared period left parenthesis StartFraction 1 Over StartRoot r squared minus x squared EndRoot EndFraction plus StartFraction 4 Over StartRoot r squared minus 4 x squared EndRoot EndFraction right parenthesis equals 0 semicolon 2nd Row therefore left parenthesis r squared minus x squared right parenthesis period StartRoot r squared minus 4 x squared EndRoot plus left parenthesis r squared minus 4 x squared right parenthesis period StartRoot r squared minus x squared EndRoot equals x squared period left parenthesis 4 StartRoot r squared minus x squared EndRoot plus StartRoot r squared minus 4 x squared EndRoot right parenthesis semicolon 3rd Row therefore left parenthesis r squared minus 2 x squared right parenthesis period StartRoot r squared minus 4 x squared EndRoot equals minus left parenthesis r squared minus 8 x squared right parenthesis period StartRoot r squared minus x squared EndRoot semicolon 4th Row therefore r Superscript 4 Baseline minus 4 left parenthesis r squared x squared plus 4 x Superscript 4 Baseline right parenthesis period left parenthesis r squared minus 4 x squared right parenthesis equals left parenthesis r Superscript 4 Baseline minus 16 r squared x squared plus 64 x Superscript 4 Baseline right parenthesis period left parenthesis r squared minus x squared right parenthesis semicolon 5th Row therefore r Superscript 6 Baseline minus 8 r Superscript 4 Baseline x squared plus 20 r squared x Superscript 4 Baseline minus 16 x Superscript 6 Baseline equals r Superscript 6 Baseline minus 17 r Superscript 4 Baseline x squared plus 80 r squared x Superscript 4 Baseline minus 64 x Superscript 6 Baseline semicolon EndLayout

therefore, omitting r Superscript 6, and dividing by x squared,

StartLayout 1st Row 1st Column 48 x Superscript 4 minus 60 r squared x squared plus 9 r Superscript 4 2nd Column equals 0 semicolon 2nd Row 1st Column i period e period 16 x Superscript 4 Baseline minus 20 r squared x squared plus 3 r Superscript 4 Baseline 2nd Column equals 0 semicolon EndLayout

therefore StartFraction x squared Over r squared EndFraction equals StartFraction 20 plus or minus StartRoot 208 EndRoot Over 32 EndFraction equals StartFraction 5 minus StartRoot 13 EndRoot Over 8 EndFraction (upper sign being inadmissible, though this was not thought out.)

Q. E. F.

34. (8)

A circle with external point A to the left. Two secants are drawn from A: one passing through chord points B and E (lower), and another through B and D (upper), with C marked on the lower secant inside the circle.

(Analysis.)

Let upper A be the given Point, and upper C the centre of the given Circle. Join upper A upper C, and let upper A upper B upper D be the required Line. From upper B draw the Chord upper B upper E parallel to upper A upper C. Then angle upper D upper B upper E equals angle upper A. Hence Arc upper D upper E = Arc upper B upper D; i. e. Arc upper B upper E is bisected by upper D; i. e. upper D is on perpendicular from upper C.

A circle with external point A to the left. A secant from A passes through chord points B and E (horizontal diameter line), with C marked at center. A second line from A meets the circle at D (top).

(Synthesis.)

Join upper A upper C. From upper C draw upper C upper D perpendicular to upper A upper C. Join upper A upper D cutting Circle at upper B. From upper B draw upper B upper E parallel to upper A upper C.

It is easily proved that Arc upper B upper D = Arc upper D upper E. Hence Arc upper B upper D subtends, in the Circle, an angle = angle upper D upper B upper E = angle upper A.

Q. E. F.

35. (8)

A regular pentagon with vertices A (top), B (left), C (right), E (upper left), L (upper right) inscribed in a circle. All diagonals drawn, intersecting at center O and interior points F, G, H, A', B', C', M, D, K, forming a pentagram with a central inner pentagon and dense crossing segments.

Let upper A upper B upper C be the given Triangle; and let the portions of the radii, outside the Triangle, have to the radius the given ratios k colon 1, l colon 1, m colon 1. (N.B. k, l, m, are supposed to be proper fractions.)

From upper B draw upper B upper D perpendicular upper B upper A, and upper B upper E perpendicular upper B upper C; and make upper B upper D have, to upper B upper E, the ratio 1 minus m colon 1 minus k. Through upper D draw upper D upper F parallel to upper B upper A, and upper E upper F parallel to upper B upper C; and join upper B upper F. From upper F draw upper F upper G perpendicular upper B upper A, and upper F upper H perpendicular upper B upper C.

Then upper F upper G colon upper F upper H colon colon 1 minus m colon 1 minus k.

Similarly, draw upper C upper O so that the perpendiculars, drawn from any Point of it to upper C upper A and upper C upper B, are in the ratio 1 minus l colon 1 minus k; and produce upper B upper F to meet it at upper O.

From upper O draw upper O upper A acute, upper O upper B acute, upper O upper C acute, perpendicular the sides.

Then upper O upper A acute colon upper O upper B acute colon upper O upper C acute colon colon 1 minus k colon 1 minus l colon 1 minus m.

Produce upper O upper A acute to upper K, so that upper O upper K colon upper O upper A acute colon colon 1 colon 1 minus k.

With centre upper O, and distance upper O upper K, describe a Circle; and produce upper O upper B prime, upper O upper C prime, to meet it at upper L, upper M.

StartLayout 1st Row 1st Column Now 2nd Column upper O upper K colon upper O upper A Superscript prime Baseline colon colon 1 colon 1 minus k semicolon 2nd Row 1st Column Blank 2nd Column upper O upper A Superscript prime Baseline colon upper O upper B Superscript prime Baseline colon colon 1 minus k colon 1 minus l semicolon 3rd Row 1st Column therefore 2nd Column upper O upper K colon upper O upper B Superscript prime Baseline colon colon 1 colon 1 minus l semicolon EndLayout

Similarly, upper O upper K colon upper O upper C Superscript prime Baseline colon colon 1 colon 1 minus m.

But upper A prime upper K colon upper O upper K colon colon upper O upper K minus upper O upper A Superscript prime Baseline colon upper O upper K colon colon k colon 1.

Similarly upper B prime upper L colon radius colon colon l colon 1, and upper C prime upper M colon radius colon colon m colon 1.

Q. E. F.

36. (8)

A triangle with vertices A (top left), B (bottom left), C (bottom right). Interior point D lies centrally with cevians from each vertex. Points C' (on AB, upper left), B' (on AC, middle right), and E (on AB below C') are marked, with dashed lines from D and B extending toward C.

(Analysis.)

Let upper B prime upper C prime be required Line: and let angle at upper C prime be right.

Cut off upper C prime upper D equal to upper C prime upper B: then upper D upper B Superscript prime Baseline equals upper B prime upper C.

Join upper D upper B, upper D upper C: then angle upper D upper B upper C Superscript prime Baseline equals 45 degree, and angle upper B prime upper D upper C equals angle upper B prime upper C upper D period

From upper C draw upper C upper E perpendicular upper A upper B.

Then angle upper B prime upper D upper C equals angle upper D upper C upper E; therefore angle upper B prime upper C upper D equals angle upper D upper C upper E.

(Synthesis.)

Hence construction. Draw upper C upper E perpendicular upper A upper B: bisect angle upper A upper C upper E: at upper B make angle upper A upper B upper D equals 45 degree. Let these lines meet at upper D. Through upper D draw upper B prime upper D upper C prime perpendicular upper A upper B.

Then angle upper C prime upper D upper B equals pi minus left parenthesis angle upper D upper C prime upper B plus angle upper C prime upper B upper D right parenthesis equals 45 degree equals angle upper C prime upper B upper D;

therefore upper C prime upper D equals upper C prime upper B.

Triangle ABC with cevians concurrent at D and harmonic points B', C', E.

StartLayout 1st Row 1st Column Blank 2nd Column Also angle upper B prime upper D upper C equals angle upper D upper C upper E equals angle upper D upper C upper B Superscript prime Baseline semicolon 2nd Row 1st Column Blank 2nd Column therefore upper D upper B Superscript prime Baseline equals upper B prime upper C semicolon 3rd Row 1st Column Blank 2nd Column therefore upper C prime upper B Superscript prime Baseline equals sum of upper B upper C Superscript prime Baseline comma upper C upper B Superscript prime Baseline period EndLayout

Q.E.F.

Limits of possibility:—

angle upper A must not be greater than 90 degree;

angle upper B must not be less than 45 degree;

angle upper C must not be less than half complement of upper A,

i. e. not less than left parenthesis 45 degree minus StartFraction upper A Over 2 EndFraction right parenthesis.

37. (8, 22)

Let upper B upper C be the common chord, and upper A, upper D, the centres.

Let angle upper A equals 30°, and angle upper D equals 60°.

And let upper B upper C (which = upper D upper B equals upper D upper C right parenthesis equals 1.

And let upper A upper B equals x.

Now cosine upper A equals StartFraction StartRoot 3 EndRoot Over 2 EndFraction equals StartFraction 2 x squared minus 1 Over 2 x squared EndFraction;

therefore StartFraction StartRoot 3 EndRoot Over 2 EndFraction equals 1 minus StartFraction 1 Over 2 x squared EndFraction; therefore StartFraction 1 Over 2 x squared EndFraction equals StartFraction 2 minus StartRoot 3 EndRoot Over 2 EndFraction;

therefore x squared equals StartFraction 1 Over 2 minus StartRoot 3 EndRoot EndFraction equals 2 plus StartRoot 3 EndRoot;

therefore areas of Circles are pi period left parenthesis 2 plus StartRoot 3 EndRoot right parenthesis and pi;

therefore areas of Sectors are pi period StartFraction 2 plus StartRoot 3 EndRoot Over 12 EndFraction and StartFraction pi Over 6 EndFraction;

therefore their sum equals pi period StartFraction 4 plus StartRoot 3 EndRoot Over 12 EndFraction.

Again, area of Triangle upper A upper B upper C equals one half period left parenthesis 2 plus StartRoot 3 EndRoot right parenthesis period one half,

equals StartFraction 2 plus StartRoot 3 EndRoot Over 4 EndFraction;

also area of Triangle upper D upper B upper C equals StartFraction StartRoot 3 EndRoot Over 4 EndFraction;

therefore their sum equals StartFraction 2 plus 2 StartRoot 3 EndRoot Over 4 EndFraction equals StartFraction 1 plus StartRoot 3 EndRoot Over 2 EndFraction.

Now the portion, of the smaller Circle, that is within the larger one, is the difference between these two sums;

therefore it equals pi period StartFraction 4 plus StartRoot 3 EndRoot Over 12 EndFraction minus StartFraction 1 plus StartRoot 3 EndRoot Over 2 EndFraction.

Hence its ratio, to the area of the smaller Circle, is this sum divided by pi;

therefore it equals StartFraction 4 plus StartRoot 3 EndRoot Over 12 EndFraction minus StartFraction 1 plus StartRoot 3 EndRoot Over 2 pi EndFraction,

equals StartFraction 5 dot 732 Over 12 EndFraction minus StartStartFraction 2 dot 732 OverOver left parenthesis StartFraction 44 Over 7 EndFraction right parenthesis EndEndFraction equals dot 478 minus StartFraction dot 248 Over left parenthesis four sevenths right parenthesis EndFraction,

equals dot 478 minus StartFraction 1 dot 736 Over 4 EndFraction equals dot 478 minus dot 434 equals dot 044.

Q. E. F.

38. (9, 22)

Taking, in order, the bag from which this unknown counter is drawn, the bag from which a red one was twice drawn, and the remaining bag, we see that there are six possible arrangements of 'upper A', 'upper B', and 'upper C': viz.—

(1) upper A upper B upper C,      | (4) upper B upper C upper A,
(2) upper A upper C upper B, | (5) upper C upper A upper B,
(3) upper B upper A upper C, | (6) upper C upper B upper A.

Now the chance of the observed event is, in case (1), 1 times four ninths equals four ninths; in case (2), 1 times one ninth equals one ninth; in case (3), two thirds times 1 equals two thirds; in case (4), two thirds times one ninth equals two twenty sevenths; in case (5), one third times 1 equals one third; and in case (6), one third times four ninths equals four twenty sevenths.

Hence the chances of existence, for these 6 states, are proportional to '12, 3, 18, 2, 9, 4'. Hence their actual values are 'one fourth comma one sixteenth comma three eighths comma one twenty fourth comma three sixteenths comma one twelfth'.

Hence the chance of the unknown counter being red is the sum of one fourth times 1 comma one sixteenth times 1 comma three eighths times two thirds comma one twenty fourth times two thirds comma three sixteenths times one third comma one twelfth times one third;

i. e. it is StartFraction 36 plus 9 plus 36 plus 4 plus 9 plus 4 Over 9 times 16 EndFraction; which equals StartFraction 98 Over 9 times 16 EndFraction equals StartFraction 49 Over 72 EndFraction.

Q. E. F.

39. (9, 22)

Let x = no. of days.

Then left parenthesis 2 times 10 minus ModifyingAbove x minus 1 With quotation dash right parenthesis period StartFraction x Over 2 EndFraction equals 14 plus left brace 2 times 2 plus ModifyingAbove x minus 1 With quotation dash period 2 right brace dot StartFraction x Over 2 EndFraction;

i. e. StartFraction 21 x Over 2 EndFraction minus StartFraction x squared Over 2 EndFraction equals 14 plus x plus x squared;

therefore 3 x squared minus 19 x plus 28 equals 0; therefore x equals StartFraction 19 plus or minus 5 Over 6 EndFraction equals 4 or seven thirds.

Now the above solution has taken no account of the discontinuity of increase, or decrease of pace, and is the true solution only on the supposition that the increase or decrease is continuous, and such as to coincide with the above data at the end of each day. Hence '4' is a correct answer; but 'seven thirds' only indicates that a meeting occurs during the third day. To find the hour of this, let y equals no. of hours.

Now in 2 days upper A has got to the end of 19 miles, upper B to the end of left parenthesis 14 plus 6 right parenthesis, i. e. 20.

therefore 19 plus y period eight twelfths equals 20 plus y period six twelfths.

i. e. y period two thirds equals 1 plus y period one half semicolon therefore y equals 6.

Hence they meet at end of 2d. 6h., and at end of 4d.: and the distances are 23 miles, and 34 miles.

Q. E. F.

40. (9)

(1) Let upper A upper B upper C be the given Triangle, and upper A upper D the line from the vertex.

A triangle with vertices A (top), B (bottom left), C (bottom right). Vertical altitude AD drops from A to base point D. Points F (on AB) and E (on AC) connect down to H and G on BC respectively, forming two smaller symmetric triangles BFH and EGC flanking the central altitude, with H and G marked on the base.

From upper D draw upper D upper E, upper D upper F, parallel to the sides; and from upper E and upper F draw upper E upper G, upper F upper H, perpendicular upper B upper C.

Then Triangles upper F upper B upper D, upper E upper D upper C, are similar to upper A upper B upper C;

therefore upper F upper H colon upper A upper D colon colon upper B upper D colon upper B upper C comma

and upper E upper G colon upper A upper D colon colon upper D upper C colon upper B upper C semicolon

therefore left parenthesis upper F upper H plus upper E upper G right parenthesis colon upper A upper D colon colon upper B upper C colon upper B upper C semicolon

therefore upper F upper H plus upper E upper G equals upper A upper D.

Also, because Triangles upper A upper E upper D, upper A upper F upper D, are equal and on the same base upper A upper D,

therefore their altitudes are equal; i. e. upper D upper H equals upper D upper G.

Q. E. F.

(2) Let upper A upper B upper C be the given Triangle, and upper A upper D the line from the vertex.

A triangle with vertices A (top), B (bottom left), C (bottom right). Vertical altitude AD drops to base with E just right of D. Points G (on AB) and F (on AC) drop vertical altitudes to K and H on BC respectively, forming two smaller triangles BGK and FHC symmetrically flanking the central altitude, with K, D, E, H marked along the base.

Make upper C upper E equals upper B upper D; from upper E draw upper E upper F, upper E upper G, parallel to the sides; and from upper F, upper G, draw upper F upper H, upper G upper K, perpendicular upper B upper C.

Then Triangles upper G upper B upper E, upper F upper E upper C, are similar to upper A upper B upper C;

therefore upper G upper K colon upper A upper D colon colon upper B upper E colon upper B upper C comma

and upper F upper H colon upper A upper D colon colon upper E upper C colon upper B upper C semicolon

therefore left parenthesis upper G upper K plus upper F upper H right parenthesis colon upper A upper D colon colon upper B upper C colon upper B upper C semicolon

therefore upper G upper K plus upper F upper H equals upper A upper D.

Also upper B upper K colon upper B upper E colon colon upper B upper D colon upper B upper C semicolon

therefore upper B upper K colon upper D upper C colon colon upper E upper C colon upper B upper C semicolon

colon colon upper H upper C colon upper D upper C semicolon

therefore upper B upper K equals upper H upper C.

Q. E. F.

41. (9, 23)

(1) As there was certainly at least one upper W in the bag at first, the 'a priori' chances for the various states of the bag, 'upper W upper W upper W upper W, upper W upper W upper W upper B, upper W upper W upper B upper B, upper W upper B upper B upper B comma' were 'one eighth, three eighths, three eighths, one eighth'.

These would have given, to the observed event, the chances '1, one half, one sixth, 0'.

Hence the chances, after the event, for the various states, are proportional to 'one eighth dot 1, three eighths dot one half, three eighths dot one sixth'; i. e. to 'one eighth, three sixteenths, one sixteenth'; i.e. to '2, 3, 1'. Hence their actual values are 'one third, one half, one sixth'.

Hence the chance, of now drawing upper W, is 'one third dot 1 plus one half dot one half'; i. e. it is seven twelfths.

Q. E. F.

(2) If he had not spoken, the 'a priori' chances for the states 'upper W upper W upper W upper W, upper W upper W upper W upper B, upper W upper W upper B upper B, upper W upper B upper B upper B, upper B upper B upper B upper B', would have been 'StartFraction 1 comma 4 comma 6 comma 4 comma 1 Over 16 EndFraction'.

These would have given, to the observed event, the chances '1, one half, one sixth, 0, 0'.

Hence the chances, after the event, for the various states, are proportional to 'one sixteenth dot 1, one fourth dot one half, one sixth dot three eighths'; i. e. to '1, 2, 1'. Hence their actual values are 'one fourth, one half, one fourth'.

Hence the chance, of now drawing upper W, is 'one fourth dot 1 plus one half dot one half'; i. e. it is one half.

Q. E. F.

42. (10, 23)

Let upper A upper B upper C be the given Triangle. Bisect its angles, and draw perpendiculars to them, forming the Triangle upper A prime upper B prime upper C prime.

A large inverted triangle with vertices C' (top left), B' (top right), and A' (bottom). Inner triangle ABC is inscribed with A (top center), B (middle left), and C (middle right).

Now angle upper C upper B upper A prime equals 90 degree minus StartFraction upper B Over 2 EndFraction; and so of the others.

therefore upper A prime equals 180 degree minus left parenthesis upper C upper B upper A prime plus upper B upper C upper A prime right parenthesis equals StartFraction upper B plus upper C Over 2 EndFraction equals 90 degree minus StartFraction upper A Over 2 EndFraction semicolon therefore upper B upper A prime equals a period StartStartFraction cosine StartFraction upper C Over 2 EndFraction OverOver cosine StartFraction upper A Over 2 EndFraction EndEndFraction period

Similarly, upper B upper C Superscript prime Baseline equals c period StartStartFraction cosine StartFraction upper A Over 2 EndFraction OverOver cosine StartFraction upper C Over 2 EndFraction EndEndFraction;

StartLayout 1st Row 1st Column therefore upper A prime upper C prime 2nd Column equals StartStartFraction a cosine squared StartFraction upper C Over 2 EndFraction plus c cosine squared StartFraction upper A Over 2 EndFraction OverOver cosine StartFraction upper A Over 2 EndFraction cosine StartFraction upper C Over 2 EndFraction EndEndFraction equals StartStartFraction a period StartFraction s period left parenthesis s minus c right parenthesis Over a b EndFraction plus c period StartFraction s period left parenthesis s minus a right parenthesis Over b c EndFraction OverOver StartFraction s Over b EndFraction dot StartRoot StartFraction left parenthesis s minus a right parenthesis period left parenthesis s minus c right parenthesis Over a c EndFraction EndRoot EndEndFraction comma 2nd Row 1st Column Blank 2nd Column equals StartStartFraction s minus c plus s minus a OverOver sine StartFraction upper B Over 2 EndFraction EndEndFraction equals StartStartFraction b OverOver sine StartFraction upper B Over 2 EndFraction EndEndFraction period EndLayout

Inverted triangle A'B'C' with inscribed inner triangle ABC and cevians from A.

Similarly, upper A prime upper B Superscript prime Baseline equals StartStartFraction c OverOver sine StartFraction upper C Over 2 EndFraction EndEndFraction;

therefore area of upper A prime upper B prime upper C Superscript prime Baseline equals StartStartFraction b c cosine StartFraction upper A Over 2 EndFraction OverOver 2 sine StartFraction upper B Over 2 EndFraction sine StartFraction upper C Over 2 EndFraction EndEndFraction;

StartLayout 1st Row 1st Column therefore StartFraction area of upper A prime upper B prime upper C Superscript prime Baseline Over area of upper A upper B upper C EndFraction 2nd Column equals StartStartFraction b c cosine StartFraction upper A Over 2 EndFraction OverOver 2 sine StartFraction upper B Over 2 EndFraction sine StartFraction upper C Over 2 EndFraction EndEndFraction dot StartFraction 2 Over b c sine upper A EndFraction comma 2nd Row 1st Column Blank 2nd Column equals StartStartFraction cosine StartFraction upper A Over 2 EndFraction OverOver sine StartFraction upper B Over 2 EndFraction sine StartFraction upper C Over 2 EndFraction .2 sine StartFraction upper A Over 2 EndFraction cosine StartFraction upper A Over 2 EndFraction EndEndFraction comma 3rd Row 1st Column Blank 2nd Column equals StartStartFraction 1 OverOver 2 sine StartFraction upper A Over 2 EndFraction sine StartFraction upper B Over 2 EndFraction sine StartFraction upper C Over 2 EndFraction EndEndFraction comma 4th Row 1st Column Blank 2nd Column equals StartFraction a b c Over 2 left parenthesis s minus a right parenthesis period left parenthesis s minus b right parenthesis period left parenthesis s minus c right parenthesis EndFraction period EndLayout

Q. E. F.

43. (10)

A triangle with vertices A (top), B (bottom left), C (bottom right). Cevian BD runs from B up to point D on AC, and cevian CE runs from C up to point E on AB.

Let upper A upper B upper C be the given Triangle; and let upper B upper F upper D, upper C upper F upper E, be the required lines, so that upper F upper B equals upper F upper C, and Tetragon upper A upper E upper F upper D = Triangle upper F upper B upper C. And call the angle upper F upper B upper C 'theta'. It will suffice to calculate this angle.

Because Triangle upper F upper B upper C equals Tetragon upper A upper E upper F upper D,

StartLayout 1st Row 1st Column therefore Triangle upper D upper B upper C 2nd Column equals Triangle upper A upper E upper C comma 2nd Row 1st Column Blank 2nd Column equals Triangle upper A upper B upper C minus Triangle upper E upper B upper C semicolon EndLayout

therefore Triangles upper D upper B upper C comma upper E upper B upper C comma together equals Triangle upper A upper B upper C semicolon

therefore one half dot StartFraction a squared Over cotangent theta plus cotangent upper C EndFraction plus one half dot StartFraction a squared Over cotangent theta plus cotangent upper B EndFraction equals one half dot StartFraction a squared Over cotangent upper B plus cotangent upper C EndFraction semicolon

therefore StartFraction 1 Over cotangent theta plus cotangent upper C EndFraction plus StartFraction 1 Over cotangent theta plus cotangent upper B EndFraction equals StartFraction 1 Over cotangent upper B plus cotangent upper C EndFraction semicolon

therefore StartFraction 2 cotangent theta plus left parenthesis cotangent upper B plus cotangent upper C right parenthesis Over cotangent squared theta plus cotangent theta dot left parenthesis cotangent upper B plus cotangent upper C right parenthesis plus cotangent upper B cotangent upper C EndFraction equals do period semicolon

therefore cotangent squared theta plus cotangent theta dot left parenthesis cotangent upper B plus cotangent upper C right parenthesis plus cotangent upper B cotangent upper C equals 2 cotangent theta dot left parenthesis cotangent upper B plus cotangent upper C right parenthesis plus left parenthesis cotangent upper B plus cotangent upper C right parenthesis squared semicolon

therefore cotangent squared theta minus cotangent theta period left parenthesis cotangent upper B plus cotangent upper C right parenthesis minus left parenthesis cotangent squared upper B plus cotangent upper B cotangent upper C plus cotangent squared upper C right parenthesis equals 0 semicolon

therefore cotangent theta equals one half period left brace cotangent upper B plus cotangent upper C plus or minus square root left parenthesis 5 cotangent squared upper B plus 6 cotangent upper B cotangent upper C plus 5 cotangent squared upper C right parenthesis right brace.

Q. E. F.

44. (10)

Let k be a No. not containing 2 or 5 as a factor, i. e. let it be prime to 10. Then, if StartFraction 1 Over k EndFraction be reduced to a circulating decimal, and that to a vulgar fraction, the digits of the denominator will be a certain number of 9's; i. e. it will be of the form left parenthesis 10 Superscript n Baseline minus 1 right parenthesis.

And since this fraction equals StartFraction 1 Over k EndFraction, and that k is prime to 10, and so prime to 10 Superscript m, the factor left parenthesis 10 Superscript n Baseline minus 1 right parenthesis must be a multiple of k.

This evidently holds good in any other scale of notation. Hence, if a be the radix of the scale of notation, and b a No. prime to a, a value may be found for n, which will make left parenthesis a Superscript n Baseline minus 1 right parenthesis a multiple of b.

Q. E. D.

EXAMPLES (not thought out).

(1) With radix 10, find a value, for n, which will make left parenthesis 10 Superscript n Baseline minus 1 right parenthesis a multiple of 7.

one seventh equals dot ModifyingAbove 1 With dot 4285 ModifyingAbove 7 With dot equals StartFraction 142857 Over 10 Superscript 6 Baseline minus 1 EndFraction period

Ans. n equals 6.

(2) Let the two given Nos. be 8, 9.

Taking 8 as radix, we get one ninth equals dot ModifyingAbove 0 With dot ModifyingAbove 7 With dot equals StartFraction 7 Over 8 squared minus 1 EndFraction.

Ans. n equals 2.

(3) Let the two given Nos. be 7, 13.

Taking 7 as radix, we get

one thirteenth equals dot ModifyingAbove 0 With dot 3524563142 ModifyingAbove 1 With dot equals StartFraction 35245631421 Over 7 Superscript 12 Baseline minus 1 EndFraction period

Ans. n equals 12.

45. (10, 23)

Divide each rod into left parenthesis n plus 1 right parenthesis parts, where n is assumed to be odd, and the n points of division are assumed to be the only points where the rod will break, and to be equally frangible.

StartLayout 1st Row 1st Column The chance of one failure 2nd Column is StartFraction n minus 1 Over n EndFraction semicolon 2nd Row 1st Column therefore right double quotation mark right double quotation mark n failures 2nd Column is left parenthesis StartFraction n minus 1 Over n EndFraction right parenthesis Superscript n Baseline 3rd Row 1st Column Blank 2nd Column equals left parenthesis 1 minus StartFraction 1 Over n EndFraction right parenthesis Superscript n Baseline period EndLayout

Now, if m equals StartFraction 1 Over n EndFraction; then, when n equals StartFraction 1 Over 0 EndFraction, m equals 0;

therefore the chance that no rod is broke in the middle equals left parenthesis 1 minus m right parenthesis Superscript StartFraction 1 Over m EndFraction, when m equals 0;

i. e. it approaches the limit left parenthesis 1 minus 0 right parenthesis Superscript StartFraction 1 Over 0 EndFraction.

And Ans. equals 1 minus left parenthesis 1 minus 0 right parenthesis Superscript StartFraction 1 Over 0 EndFraction.

Now left parenthesis 1 plus 0 right parenthesis Superscript StartFraction 1 Over 0 EndFraction Baseline equals e. Hence if, in the series for e, we call the sum of the odd terms 'a', and of the even terms 'b'; then e equals a plus b; and left parenthesis 1 minus 0 right parenthesis Superscript StartFraction 1 Over 0 EndFraction Baseline equals a minus b equals 2 a minus e.

Q. E. F.

[N. B. What follows here was not thought out.]

StartLayout 1st Row 1st Column Now a 2nd Column equals 1 plus StartFraction 1 Over vertical bar ModifyingBelow 2 With quotation dash EndFraction plus StartFraction 1 Over vertical bar ModifyingBelow 4 With quotation dash EndFraction plus ampersand c period 2nd Row 1st Column 1 2nd Column equals 1 3rd Row 1st Column StartFraction 1 Over vertical bar ModifyingBelow 2 With quotation dash EndFraction 2nd Column equals dot 5 4th Row 1st Column StartFraction 1 Over vertical bar ModifyingBelow 4 With quotation dash EndFraction 2nd Column equals dot 04166666 ampersand c period 5th Row 1st Column StartFraction 1 Over vertical bar ModifyingBelow 6 With quotation dash EndFraction 2nd Column equals dot 00138888 ampersand c period 6th Row 1st Column StartFraction 1 Over vertical bar ModifyingBelow 8 With quotation dash EndFraction 2nd Column equals dot 00002480 ampersand c period 7th Row 1st Column StartFraction 1 Over vertical bar ModifyingBelow 10 With quotation dash EndFraction 2nd Column equals dot 00000027 8th Row 1st Column Blank 2nd Column c period 9th Row 1st Column Blank 2nd Column em dash em dash em dash em dash em dash em dash em dash em dash em dash em dash 10th Row 1st Column therefore a 2nd Column equals 1 dot 5430806 ampersand c period 11th Row 1st Column therefore 2 a 2nd Column equals 3 dot 0861612 ampersand c period 12th Row 1st Column e 2nd Column equals 2 dot 7182818 ampersand c period 13th Row 1st Column Blank 2nd Column em dash em dash em dash em dash em dash em dash em dash em dash em dash em dash 14th Row 1st Column therefore left parenthesis 1 minus 0 right parenthesis Superscript StartFraction 1 Over 0 EndFraction 2nd Column equals dot 3678793 ampersand c period 15th Row 1st Column therefore Ans period 2nd Column equals 1 minus left parenthesis 1 minus 0 right parenthesis Superscript StartFraction 1 Over 0 EndFraction Baseline equals dot 6321207 ampersand c period EndLayout

46. (10)

Let upper A upper B upper C be the given Triangle, and upper D the given Point.

Left: triangle ABC with interior points F, E, D marking cevian intersections. Right: larger triangle A'B'C' with three circles centered at F', E', D', each passing through two vertices, overlapping at interior point and illustrating the corresponding circle construction for the cevian configuration.

If we make a Triangle upper D prime upper E prime upper F prime, having its angles equal to the given angles, and having upper D prime as its assigned vertex, the Problem may be solved, if we can circumscribe, about the Triangle upper D prime upper E prime upper F prime, a Triangle similar to upper A upper B upper C.

Now we can construct, on upper E prime upper F prime, upper F prime upper D prime, upper D prime upper E prime, segments of Circles containing angles equal to upper A, upper B, upper C. Hence the Problem may be solved, if we can place, in these Circles, a line upper B prime upper D prime upper C prime, divided in the same proportion as upper B upper D upper C.

This Lemma may be solved as follows. Let upper G, upper H, be the centres of the Circles. Join upper G upper H, and divide it, at upper K, proportionally to upper B upper D upper C.

Two circles of equal size intersecting at two points, with the intersection region overlapping centrally. A horizontal baseline runs through B' (far left) to C' (far right). Points L and M lie on the baseline, with vertical segments LG and MH and horizontal connector GKH above, all meeting near central intersection point D'.

Join upper K upper D prime; through upper D prime draw upper B prime upper D prime upper C prime perpendicular upper K upper D prime; and from upper G, upper H, draw upper G upper L, upper H upper M, perpendicular upper B prime upper C prime.

Now it may be easily proved that

upper L upper D Superscript prime Baseline colon upper D prime upper M colon colon upper G upper K colon upper K upper H colon colon upper B upper D colon upper D upper C period

But upper B prime upper D prime, upper D prime upper C prime, are doubles of upper L upper D prime, upper D prime upper M;

therefore upper B prime upper D prime colon upper D prime upper C prime colon colon upper B upper D colon upper D upper C period

Q. E. F.

[The construction is now obvious, viz. to join upper B prime upper F prime, upper C prime upper E prime, and produce them to meet, on the third Circle (as they may be easily proved to do), at upper A prime; then to divide upper A upper B, upper A upper C, at upper F and upper E, proportionally to upper A prime upper F prime upper B prime, upper A prime upper E prime upper C prime; and then to join upper D upper E, upper D upper F.]

47. (11, 23)

By inspection, '0 comma 0 comma 0' are one set of values.

Subtracting, we get x period left parenthesis StartFraction 1 Over y EndFraction minus StartFraction 1 Over z EndFraction right parenthesis equals y minus z;

therefore x equals y z period StartFraction y minus z Over z minus y EndFraction equals minus y z, unless y equals z, in which case x equals StartFraction 0 Over 0 EndFraction.

Now, by (1), x equals x y minus y z;

therefore, when y not equals z, x equals x y plus x;

therefore x y equals 0, unless x be infinite.

Similarly, by (2), x z equals 0, unless x be infinite.

Hence, if x be finite, and if y not equals z, either x or y equals 0, and also either x or z equals 0; i. e. either x equals 0, or else y equals z equals 0. But the latter is excluded by our hypothesis. Hence x equals 0. Hence y z equals 0; i. e. either y or z equals 0, and the other may take any value.

This gives us 2 more sets of values, viz.

StartLayout 1st Row 1st Column x 2nd Column equals y equals 0 semicolon z has any value semicolon 2nd Row 1st Column x 2nd Column equals z equals 0 semicolon y has any value period EndLayout

We have now to ascertain what happens when y equals z.

By (1), StartFraction x Over y EndFraction equals x minus y;

therefore y squared equals x period left parenthesis y minus 1 right parenthesis semicolon i period e period x equals StartFraction y squared Over y minus 1 EndFraction.

Similarly, by (2), x equals StartFraction z squared Over z minus 1 EndFraction.

This gives us a 4th set of values, viz. x equals StartFraction k squared Over k minus 1 EndFraction comma y equals z equals k; where k has any value.

Now y and z may evidently have any real values, but x is restricted by the equation

y squared minus x y plus x equals 0 comma

in which y cannot be real, unless left parenthesis x squared minus 4 x right parenthesis greater than 0. Hence x may have any negative value, and any positive value that is not less than 4; but it cannot have any positive value, less than 4, without making y unreal.

Q. E. F.

48. (11)

Let upper A upper B upper C be the given Triangle, upper A prime, upper B prime, upper C prime, the centres of the semicircles, and upper D upper E, upper F upper G, upper H upper J, the common tangents; so that upper D upper E equals alpha, upper F upper G equals beta, and upper H upper J equals gamma.

Three large circles mutually intersecting, forming a curved triangular outer boundary with points F, G (left), H, J (right), and D, E (top). Inner triangle B'C'A' is inscribed with numerous cevians and chords connecting A, B, C, D, E, K, creating a dense triangulated network at the center.

Join upper B prime upper D, upper C prime upper E; and from upper C prime draw upper C prime upper K perpendicular upper B prime upper D. Hence upper C upper K equals alpha.

Call sides of given Triangle '2 a, 2 b, 2 c'.

Then upper B prime upper C prime equals a, and upper B prime upper K equals b minus c;

therefore upper C prime upper K equals square root left brace a squared minus left parenthesis b minus c right parenthesis squared right brace semicolon

i period e period alpha equals square root left brace left parenthesis a minus b plus c right parenthesis period left parenthesis a plus b minus c right parenthesis right brace semicolon

similarly comma beta equals square root left brace left parenthesis a plus b minus c right parenthesis period left parenthesis negative a plus b plus c right parenthesis right brace comma

and gamma equals square root left brace left parenthesis negative a plus b plus c right parenthesis period left parenthesis a minus b plus c right parenthesis right brace semicolon

therefore StartFraction beta gamma Over alpha EndFraction equals negative a plus b plus c semicolon

similarly comma StartFraction gamma alpha Over beta EndFraction equals a minus b plus c comma

and StartFraction alpha beta Over gamma EndFraction equals a plus b minus c semicolon

therefore their sum equals a plus b plus c comma

equals semi hyphen perimeter of upper A upper B upper C period

Q.E.D.

49. (11, 23)

Take, as unit, a side of one of the Triangles.

If the Tetrahedron be cut by a vertical Plane containing one of the slant edges, the section is a Triangle whose base is StartFraction StartRoot 3 EndRoot Over 2 EndFraction, and whose sides are StartFraction StartRoot 3 EndRoot Over 2 EndFraction comma 1;

hence cosine of smaller base-angle

equals left parenthesis three fourths plus 1 minus three fourths right parenthesis dot StartFraction 1 Over StartRoot 3 EndRoot EndFraction equals StartFraction 1 Over StartRoot 3 EndRoot EndFraction semicolon

therefore its sine equals StartFraction StartRoot 2 EndRoot Over StartRoot 3 EndRoot EndFraction equals its altitude;

and this is the altitude of the Tetrahedron;

therefore volume of Tetrahedron equals one third dot StartFraction StartRoot 2 EndRoot Over StartRoot 3 EndRoot EndFraction dot StartFraction StartRoot 3 EndRoot Over 4 EndFraction equals StartFraction StartRoot 2 EndRoot Over 12 EndFraction.

Also altitude of Pyramid = altitude of Triangle whose base is StartRoot 2 EndRoot, and whose sides are 1, 1;

i.e. it = StartFraction StartRoot 2 EndRoot Over 2 EndFraction;

therefore volume of Pyramid = one third dot StartFraction StartRoot 2 EndRoot Over 2 EndFraction equals StartFraction StartRoot 2 EndRoot Over 6 EndFraction.

Hence required ratio = StartFraction StartRoot 2 EndRoot Over 6 EndFraction dot StartFraction 12 Over StartRoot 2 EndRoot EndFraction equals 2.

Q. E. F.

50. (11,23)

At first, the chance that bag upper H shall contain

StartLayout 1st Row 1st Column 2 upper W counters comma 2nd Column is one fourth period 2nd Row 1st Column 1 upper W and 1 upper B comma 2nd Column is one half period 3rd Row 1st Column 2 upper B comma 2nd Column is one fourth period EndLayout

therefore, after adding a upper W, the chance that it shall contain

StartLayout 1st Row 1st Column Blank 2nd Column 3 upper W comma 3rd Column is one fourth period 2nd Row 1st Column Blank 2nd Column 2 upper W comma 1 upper B comma 3rd Column is one half period 3rd Row 1st Column Blank 2nd Column 1 upper W comma 2 upper B comma 3rd Column is one fourth period EndLayout

hence the chance of drawing a W from it is

one fourth times 1 plus one half times two thirds plus one fourth times one third colon i period e period two thirds period

therefore the chance of drawing a upper B is one third.

After transferring this (unseen) counter to bag upper K, the chance that it shall contain

StartLayout 1st Row 1st Column 3 upper W comma 2nd Column Blank 3rd Column is two thirds times one fourth semicolon 4th Column i period e period one sixth period 2nd Row 1st Column 2 upper W comma and 2nd Column 1 upper B comma 3rd Column is two thirds times one half plus one third times one fourth semicolon 4th Column i period e period five twelfths period 3rd Row 1st Column 1 upper W comma 2nd Column 2 upper B comma 3rd Column is two thirds times one fourth plus one third times one half semicolon 4th Column i period e period one third period 4th Row 1st Column 3 upper B comma 2nd Column Blank 3rd Column is one third times one fourth semicolon 4th Column i period e period one twelfth semicolon EndLayout

therefore the chance of drawing a upper W from it is

one sixth times 1 plus five twelfths times two thirds plus one third times one third semicolon i period e period five ninths period

therefore the chance of drawing a upper B is four ninths.

Before transferring this to bag upper H, the chance that bag upper H shall contain

StartLayout 1st Row 1st Column 2 upper W comma 2nd Column is 3rd Column one fourth times 1 plus one half times one third semicolon 4th Column i period e period five twelfths period 2nd Row 1st Column 1 upper W comma 1 upper B comma 2nd Column Blank 3rd Column one half times two thirds plus one fourth times two thirds semicolon 4th Column i period e period one half period 3rd Row 1st Column 2 upper B comma 2nd Column Blank 3rd Column one fourth times one third semicolon 4th Column i period e period one twelfth period EndLayout

therefore, after transferring it, the chance that bag upper H shall contain

StartLayout 1st Row 1st Column 3 upper W comma 2nd Column is 3rd Column five twelfths times five ninths semicolon 4th Column i period e period StartFraction 25 Over 108 EndFraction period 2nd Row 1st Column 2 upper W comma 1 upper B comma 2nd Column Blank 3rd Column five twelfths times four ninths plus one half times five ninths semicolon 4th Column i period e period StartFraction 50 Over 108 EndFraction period 3rd Row 1st Column 1 upper W comma 2 upper B comma 2nd Column Blank 3rd Column one half times four ninths plus one twelfth times five ninths semicolon 4th Column i period e period StartFraction 29 Over 108 EndFraction period 4th Row 1st Column 3 upper B comma 2nd Column Blank 3rd Column one twelfth times four ninths semicolon 4th Column i period e period StartFraction 4 Over 108 EndFraction period EndLayout

Hence the chance of drawing a upper W is

StartFraction 1 Over 108 EndFraction times left brace 25 times 1 plus 50 times two thirds plus 29 times one third right brace semicolon i period e period StartFraction 17 Over 27 EndFraction period

i. e. the odds are 17 to 10 on its happening.

Q. E. F.

51. (12)

Let upper A upper B upper C be the given Triangle, and upper D the given Point.

Large triangle with vertices A (top), B (lower left), C (lower right). Multiple cevians and horizontal transversals create interior intersection points D, L, N, R, H, E. Base points S, B, F, K, M, G lie along BC, with vertical segments from D and E to the base, forming a dense grid of crossing lines and quadrilateral regions inside.

(Analysis.)

Let upper D upper E be the line required. Draw upper D upper F, upper E upper G, perpendicular the base. Then their sum is equal to upper D upper E.

Bisect upper D upper E at upper H, and draw upper H upper K perpendicular the base: then it is evident that upper H upper K is the upper A period upper M period of upper D upper F, upper E upper G, and is equal to half their sum; i. e. it is equal to half of upper D upper E. Hence a Circle, drawn with centre upper H and at distance upper H upper D, will pass through upper E and upper K, and will touch the base at upper K.

Through upper H draw upper L upper H upper M parallel to upper A upper C. Then upper D upper A is evidently bisected at upper L. Also upper L upper M passes through the centre of the Circle. Hence, if upper D upper N be drawn perpendicular upper L upper M (or upper C upper A), it is a chord of the Circle, and is bisected at upper R. Produce upper N upper D to meet the base produced at upper S. Hence upper S upper D upper N cuts the Circle, and upper S upper K touches it at upper K. But upper S can be found, and upper S upper K can then be taken, so that sq. of upper S upper K may be equal to rect. of upper S upper D, upper S upper N.

(Synthesis.)

From upper D draw upper D upper N perpendicular upper A upper C, and produce it to meet the base produced at upper S. Take upper S upper K, so that its square may be equal to rect. of upper S upper D, upper S upper N.

Bisect upper D upper A at upper L, and from upper L draw upper L upper M parallel to upper A upper C; and from upper K draw upper K upper H perpendicular the base, to meet upper L upper M at upper H. Join upper D upper H, and produce it to meet upper A upper C at upper E, and draw upper D upper F, upper E upper G, perpendicular the base.

Because upper D upper L equals upper L upper A, and that upper L upper M is parallel to upper A upper C,

therefore upper D upper H equals upper H upper E equals upper H upper K semicolon therefore upper D upper E equals 2 upper H upper K.

But upper D upper F plus upper E upper G equals 2 upper H upper K semicolon therefore upper D upper F plus upper E upper G equals upper D upper E.

Q. E. F.

[N. B. This proof is incomplete. I have assumed, without proving it, that upper D upper H equals upper H upper K. It may be proved thus. Because sq. of upper S upper K = rect. of upper S upper D, upper S upper N, therefore upper D upper N is a chord of a Circle which touches the base at upper K; therefore upper L upper M, which bisects it at right angles, passes through the centre. But upper K upper H also passes through the centre; therefore upper H is the centre; therefore upper H upper D equals upper H upper K.]

52. (12, 23)

Let x be the number of pennies each had at first.

No. (3) received x, took out (2 + 4), and put in StartFraction x Over 2 EndFraction; so that the sack then contained left parenthesis x period three halves minus 6 right parenthesis. Let us write 'a' for 'three halves.'

No. (5) received left parenthesis x a minus 6 right parenthesis, took out (4 + 1), and put in enough to multiply, by a, its contents when he received it. The sack now contained left parenthesis x a squared minus 6 a minus 5 right parenthesis.

No. (2) took out (1 + 3), and handed on left parenthesis x a cubed minus 6 a squared minus 5 a minus 4 right parenthesis.

No. (4) took out (3 + 5), and handed on

left parenthesis x a Superscript 4 Baseline minus 6 a cubed minus 5 a squared minus 4 a minus 8 right parenthesis period

No. (1) put in 2. The sack now contained 5 x.

Hence x a Superscript 4 Baseline minus 6 a cubed minus 5 a squared minus 4 a minus 6 equals 5 x;

therefore x equals StartFraction 6 a cubed plus 5 a squared plus 4 a plus 6 Over a Superscript 4 Baseline minus 5 EndFraction semicolon

equals StartFraction left parenthesis 6.3 cubed plus 5.3 squared .2 plus 4.3 .2 squared plus 6.2 cubed right parenthesis .2 Over 3 Superscript 4 Baseline minus 5.2 Superscript 4 Baseline EndFraction semicolon

equals StartFraction left parenthesis 162 plus 90 plus 48 plus 48 right parenthesis .2 Over 81 minus 80 EndFraction equals 696 equals 2 l period 18 s period 0 d period

Q. E. F.

53. (13,24)

Let upper A upper B upper C be the given Triangle, and upper P the given Point; and call its trilinear co-ordinates 'alpha, beta, gamma'.

Three overlapping triangles sharing a central region, with outer vertices A (top), B (bottom left), C (bottom right), E (far upper right), C'' (far left), A'' (bottom center). Interior points D, U, V, Q, W, R, P, S, F, C', B', A' mark cevian and transversal intersections, forming a dense star-like projective configuration.

From upper P draw upper P upper A prime, upper P upper B prime, upper P upper C prime, perpendicular the sides, and therefore equal to alpha, beta, gamma. Produce upper P upper A prime and upper P upper C prime to upper A double prime and upper C double prime, making upper A prime upper A double prime equals upper P upper A prime, and upper C prime upper C double prime equals upper P upper C prime. From upper C double prime draw upper C double prime upper D perpendicular upper A upper C, and produce it to upper E, making upper D upper E equals upper C double prime upper D. Join upper E upper A double prime, cutting upper A upper C in upper R, and upper B upper C in upper S. Join upper C double prime upper R, cutting upper A upper B in upper Q. Join upper P upper Q, upper P upper S.

The path of the ball is evidently upper P upper Q upper R upper S upper P; and we have to calculate the length of upper A upper R.

Now upper A upper R equals upper D upper R plus upper A upper D equals upper D upper R plus upper A upper B prime minus upper D upper B prime.

First, to calculate upper D upper R.

From upper P draw upper P upper U, upper P upper V, parallel to upper A upper B, upper A upper C; from upper C prime draw upper C prime upper W perpendicular upper P upper V; and from upper A double prime draw upper A double prime upper F perpendicular upper A upper C.

By similar Triangles, upper D upper R colon upper R upper F colon colon upper D upper E colon upper A double prime upper F colon colon upper C double prime upper D colon upper A double prime upper F;

therefore upper D upper R colon upper D upper F colon colon upper C double prime upper D colon left parenthesis upper C double prime upper D plus upper A double prime upper F right parenthesis;

therefore upper D upper R equals StartFraction upper D upper F dot upper C double prime upper D Over upper C double prime upper D plus upper A upper F EndFraction.

Now angle upper C prime upper V upper P equals upper A semicolon therefore angle upper C prime upper P upper V equals 90 degree negative upper A;

therefore upper W upper P equals gamma sine upper A;

therefore upper D upper B Superscript prime Baseline comma which equals 2 upper W upper P comma equals 2 gamma sine upper A.

Similarly, upper B prime upper F equals 2 alpha sine upper C;

therefore upper D upper F equals 2 left parenthesis alpha sine upper C plus gamma sine upper A right parenthesis.

Again, upper C prime upper W equals gamma cosine upper A;

therefore upper C double prime upper D comma which equals 2 upper C prime upper W plus upper P upper B Superscript prime Baseline comma equals 2 gamma cosine upper A plus beta.

Similarly, upper A double prime upper F equals 2 alpha cosine upper C plus beta;

therefore upper C double prime upper D plus upper A double prime upper F equals 2 left parenthesis alpha cosine upper C plus gamma cosine upper A plus beta right parenthesis;

therefore upper D upper R equals StartFraction left parenthesis alpha sine upper C plus gamma sine upper A right parenthesis dot left parenthesis 2 gamma cosine upper A plus beta right parenthesis Over alpha cosine upper C plus gamma cosine upper A plus beta EndFraction.

Now upper A upper B Superscript prime Baseline equals upper B prime upper U plus upper U upper A equals upper B prime upper U plus upper P upper V,

equals beta cotangent upper A plus gamma c o s e c upper A equals StartFraction beta cosine upper A plus gamma Over sine upper A EndFraction;

therefore upper A upper B Superscript prime Baseline minus upper D upper B Superscript prime Baseline equals StartFraction beta cosine upper A plus gamma Over sine upper A EndFraction minus 2 gamma sine upper A,

equals StartFraction beta cosine upper A plus gamma left parenthesis 1 minus 2 sine squared upper A right parenthesis Over sine upper A EndFraction

equals StartFraction beta cosine upper A plus gamma cosine 2 upper A Over sine upper A EndFraction.

Now upper A upper R equals upper D upper R plus upper A upper B Superscript prime Baseline minus upper D upper B prime;

therefore upper A upper R equals StartFraction left parenthesis alpha sine upper C plus gamma sine upper A right parenthesis dot left parenthesis 2 gamma cosine upper A plus beta right parenthesis Over alpha cosine upper C plus gamma cosine upper A plus beta EndFraction plus StartFraction beta cosine upper A plus gamma cosine 2 upper A Over sine upper A EndFraction.

Q. E. F.

54. (13, 24)

It is evident that Triangle upper A upper D upper E is similar to upper A upper B upper C.

A triangle with vertices A (top right), B (bottom left), C (bottom right). A horizontal transversal cuts across the interior with points D (on AB) and E (on AC), and intermediate points G and H. Points F and J lie on base BC between B and C, with segments connecting G to F and H to J, forming a trapezoid with inner subdivisions.

Let left single quotation mark k right single quotation mark equals ratio StartFraction upper D upper E Over a EndFraction equals StartFraction upper A upper E Over b EndFraction equals StartFraction upper A upper D Over c EndFraction.

Now upper D upper G equals upper D upper E; therefore upper D upper G equals k a;

therefore upper G upper B equals c minus k a minus k c;

therefore StartFraction upper G upper B Over c EndFraction equals 1 minus k minus k period StartFraction a Over c EndFraction;

therefore upper G upper F left parenthesis which equals upper G upper B period StartFraction b Over c EndFraction right parenthesis equals b minus k b minus k period StartFraction a b Over c EndFraction;

but upper G upper F equals upper D upper E equals k a;

therefore b minus k b minus k period StartFraction a b Over c EndFraction equals k a;

therefore b c equals k period left parenthesis b c plus c a plus a b right parenthesis;

therefore k equals StartFraction b c Over b c plus c a plus a b EndFraction equals StartStartFraction StartFraction 1 Over a EndFraction OverOver StartFraction 1 Over a EndFraction plus StartFraction 1 Over b EndFraction plus StartFraction 1 Over c EndFraction EndEndFraction equals StartStartFraction StartFraction 1 Over a EndFraction OverOver m EndEndFraction (say).

Hence upper A upper D equals StartStartFraction c period StartFraction 1 Over a EndFraction OverOver m EndEndFraction semicolon upper D upper G equals StartFraction 1 Over m EndFraction equals StartStartFraction c period StartFraction 1 Over c EndFraction OverOver m EndEndFraction·

therefore upper G upper B left parenthesis which equals c minus upper A upper D minus upper D upper G right parenthesis equals StartStartFraction c period left parenthesis m minus StartFraction 1 Over a EndFraction minus StartFraction 1 Over c EndFraction right parenthesis OverOver m EndEndFraction equals StartStartFraction c period StartFraction 1 Over b EndFraction OverOver m EndEndFraction;

therefore upper A upper D colon upper D upper G colon upper G upper B colon colon StartFraction 1 Over a EndFraction colon StartFraction 1 Over c EndFraction colon StartFraction 1 Over b EndFraction·

Also upper D upper E equals k a equals StartFraction 1 Over m EndFraction equals StartStartFraction 1 OverOver StartFraction 1 Over a EndFraction plus StartFraction 1 Over b EndFraction plus StartFraction 1 Over c EndFraction EndEndFraction·

Q. E. F.

55. (13)

Three circles of varying sizes meeting at common point P: a large circle (left) with diameter through A and D, a small circle (top) with point C and F, and a medium circle (right, partially dashed) extending to H. Points G and B lie on the horizontal baseline, with multiple chords and radii radiating from P.

Let upper A, upper B, upper C be the centres of the bases of the towers; and a, b, c their radii. Suppose upper P the required Point; and from upper P draw a pair of tangents to each circle, and lines to the centres, which will evidently bisect the angles contained by the pairs of tangents.

Hence angles upper A upper P upper D, upper B upper P upper E, upper C upper P upper F are equal;

therefore sine upper A upper P upper D equals sine upper B upper P upper E equals sine upper C upper P upper F;

i.e. StartFraction a Over upper A upper P EndFraction equals StartFraction b Over upper B upper P EndFraction equals StartFraction c Over upper C upper P EndFraction;

therefore upper A upper P colon upper B upper P colon upper C upper P colon colon a colon b colon c.

Draw a Line through upper A, upper B, and on it take Points upper G, upper H, such that upper A upper G colon upper G upper B colon colon upper A upper H colon upper H upper B colon colon a colon b.

Then the Semicircle, described on upper G upper H, is the locus of all Points whose distances, from upper A and upper B, are proportional to a, b.

Hence, if a Line be drawn through upper B, upper C, and a Semicircle described which shall be the locus of all Points whose distances, from upper B and upper C, are proportional to b, c; the intersection of these two Semicircles will be the Point required.

Q. E. F.

[Note. "The locus of all Points whose distances &c.," if represented algebraically, is evidently a Circle, whose centre is on the Line through upper A, upper B, and which passes through upper G and upper H.]

56. (13, 24)

Draw upper B upper C, upper C upper E, upper B upper D, equal to the given altitudes, so as to form right angles at upper B and upper C: and produce upper D upper B, upper E upper C. Join upper D upper C, and draw upper C upper F perpendicular to it. Join upper E upper B, and draw upper B upper G perpendicular to it. With centre upper B, and distance upper B upper F, describe a circle: with centre upper C, and distance upper C upper G, describe another: let them meet at upper A: and join upper A upper B, upper A upper C.

Call the altitudes of upper A upper B upper C, 'alpha, beta, gamma'.

Now alpha period upper B upper C equals beta period upper C upper A equals gamma period upper A upper B

= twice area of upper A upper B upper C;

also, taking upper B upper C as unit-line,

upper B upper C equals StartFraction 1 Over upper B upper C EndFraction comma upper C upper A equals upper C upper G equals StartFraction 1 Over upper C upper E EndFraction,

upper A upper B equals upper B upper F equals StartFraction 1 Over upper B upper D EndFraction;

therefore StartFraction alpha Over upper B upper C EndFraction equals StartFraction beta Over upper C upper E EndFraction equals StartFraction gamma Over upper B upper D EndFraction;

i.e. alpha, beta, gamma are proportional to given altitudes;

therefore Triangle upper A upper B upper C is similar to required Triangle.

The rest of the construction is obvious.

Q. E. F.

57. (14, 25)

(1) Geometrically.

Let upper A upper B upper C be given Triangle.

(Analysis.)

Suppose the 3 Squares described, and that their upper edges form the Triangle upper A prime upper B prime upper C prime. Join upper A upper A prime, upper B upper B prime, upper C upper C prime.

Now it is evident that, if upper B upper B prime be produced, the perpendiculars dropped, from any Point of it, upon upper A upper B, upper B upper C, will be proportional to upper B prime upper F, upper B prime upper D.

Similarly for upper A upper A prime and upper C upper C prime.

Hence these 3 Lines will meet at the Point from which the perpendiculars, dropped upon the sides of upper A upper B upper C, are proportional to upper B prime upper C prime, upper C prime upper A prime, upper A prime upper B prime.

A triangle with vertices A (top), B (bottom left), C (bottom right). Medial triangle A'B'C' is inscribed with center point O inside. Points F (on AB), G (on AC), D and E (on BC) mark cevian feet. Multiple segments from vertices and medial points cross through O, forming a dense inner network of triangles and quadrilaterals.

Hence, if Squares be described externally on the sides of upper A upper B upper C, and if their outer edges be produced to form a new Triangle upper A double prime upper B double prime upper C double prime: this Triangle, with these 3 Squares, will form a Diagram wholly similar to that formed by the Triangle upper A upper B upper C, with the 3 Squares inside it.

(Synthesis.)

Hence, if Squares be described externally on the sides of the given Triangle; and if their outer edges be produced to form a new Triangle; and if the sides of the given Triangle be divided similarly to those of the new Triangle: their central portions will be the bases of the required Squares.

Q. E. F.

(2) Trigonometrically.

Let a, b, c be the sides of the given Triangle, and m its area; and let x, y, z be the sides of the required Squares.

It is evident that a Circle can be described about the Tetragon upper B upper D upper B prime upper F.

Hence angle upper B prime upper B upper D equals angle upper B prime upper F upper D.

Now, in Triangle upper B prime upper F upper D, we know that

upper B prime upper D sine upper D equals upper B prime upper F sine upper F semicolon

i. e. x sine left parenthesis upper B prime plus upper F right parenthesis equals z sine upper F;

therefore x sine upper B Superscript prime Baseline cosine upper F plus x cosine upper B Superscript prime Baseline sine upper F equals z sine upper F.

Now angle upper B is supplementary to angle upper B prime;

therefore x sine upper B cosine upper F equals left parenthesis z plus x cosine upper B right parenthesis sine upper F;

therefore cotangent upper F equals StartFraction z plus x cosine upper B Over x sine upper B EndFraction equals cotangent upper B prime upper B upper D.

A triangle with vertices A (top), B (bottom left), C (bottom right). Medial triangle A'B'C' is inscribed with center point O inside. Points F (on AB), G (on AC), D and E (on BC) mark cevian feet. Multiple segments from vertices and medial points cross through O, forming a dense inner network of triangles and quadrilaterals.

Now upper B upper D equals x cotangent upper B prime upper B upper D;

therefore upper B upper D equals StartFraction z plus x cosine upper B Over sine upper B EndFraction·

Similarly, upper E upper C equals StartFraction y plus x cosine upper C Over sine upper C EndFraction·

But upper B upper D plus upper E upper C equals a minus x;

therefore StartFraction z plus x cosine upper B Over sine upper B EndFraction plus StartFraction y plus x cosine upper C Over sine upper C EndFraction equals a minus x;

therefore StartFraction x sine left parenthesis upper B plus upper C right parenthesis plus y sine upper B plus z sine upper C Over sine upper B sine upper C EndFraction equals a minus x;

i. e. StartFraction x sine upper A plus y sine upper B plus z sine upper C Over sine upper B sine upper C EndFraction equals a minus x.

Now it is evident that these Triangles are similar; so that

StartFraction a Over x EndFraction equals StartFraction b Over y EndFraction equals StartFraction c Over z EndFraction period

Hence, multiplying the last equation, throughout, by one or other of these equal fractions, we get

StartLayout 1st Row 1st Column Blank 2nd Column StartFraction a sine upper A plus b sine upper B plus c sine upper C Over sine upper B sine upper C EndFraction equals StartFraction a squared Over x EndFraction minus a semicolon 2nd Row 1st Column therefore 2nd Column StartFraction a sine upper A plus b sine upper B plus c sine upper C Over a sine upper B sine upper C EndFraction equals StartFraction a Over x EndFraction minus 1 semicolon 3rd Row 1st Column therefore 2nd Column StartFraction a Over x EndFraction equals StartFraction a sine upper A plus b sine upper B plus c sine upper C Over a sine upper B sine upper C EndFraction plus 1 period EndLayout

Hence, multiplying above and below by one or other of the equal fractions StartFraction a Over sine upper A EndFraction, StartFraction b Over sine upper B EndFraction, StartFraction c Over sine upper C EndFraction,

StartLayout 1st Row 1st Column StartFraction a Over x EndFraction 2nd Column equals StartFraction a squared plus b squared plus c squared Over a b sine upper C EndFraction plus 1 semicolon 2nd Row 1st Column Blank 2nd Column equals StartFraction a squared plus b squared plus c squared Over 2 m EndFraction plus 1 equals StartFraction b Over y EndFraction equals StartFraction c Over z EndFraction period EndLayout

Q. E. F.

58. (14, 25)

It may be assumed that the 3 Points form a Triangle, the chance of their lying in a straight Line being (practically) nil.

A triangle with vertices A (bottom left), B (bottom right), and C (top). Three circular arcs connect the vertices: arc from A to B (bottom, concave up), arc from A to C (left, concave right), and arc from B to C (right, concave left), all meeting at C with tick marks. Points E (left) and D (right) lie on the side arcs, with dashed lines through interior point F.

Take the longest side of the Triangle, and call it 'upper A upper B': and, on that side of it, on which the Triangle lies, draw the semicircle upper A upper F upper B. Also, with centres upper A, upper B, and distances upper A upper B, upper B upper A, draw the arcs upper B upper D upper C, upper A upper E upper C, intersecting at upper C.

Then it is evident that the vertex of the Triangle cannot fall outside the Figure upper A upper B upper D upper C upper E.

Also, if it fall inside the semicircle, the Triangle is obtuse-angled: if outside it, acute-angled. (The chance, of its falling on the semicircle, is practically nil.)

Hence required chance equals StartFraction area of semicircle Over area of fig period upper A upper B upper D upper C upper E EndFraction·

A geometric figure with points A, B (base), C (apex), E, D, and F.

Now let upper A upper B equals 2 a: then area of semicircle equals StartFraction pi a squared Over 2 EndFraction; and area of Fig. upper A upper B upper D upper C upper E equals 2 times sector upper A upper B upper D upper C minus Triangle upper A upper B upper C;

equals 2 period StartFraction 4 pi a squared Over 6 EndFraction minus StartRoot 3 EndRoot period a squared equals a squared period left parenthesis StartFraction 4 pi Over 3 EndFraction minus StartRoot 3 EndRoot right parenthesis semicolon

therefore chance equals StartStartFraction StartFraction pi Over 2 EndFraction OverOver StartFraction 4 pi Over 3 EndFraction minus StartRoot 3 EndRoot EndEndFraction equals StartStartFraction 3 OverOver 8 minus StartFraction 6 StartRoot 3 EndRoot Over pi EndFraction EndEndFraction·

Q. E. F.

59. (14, 25)

Let upper K upper L equals upper M upper N equals a comma

upper K upper N equals upper L upper M equals b comma

upper K upper M equals upper L upper N equals c semicolon

and let angles upper L upper M upper K, upper M upper K upper L, upper K upper L upper M be equal to 'upper A, upper B, upper C'; and similarly for the angles of the other facets.

From upper K draw upper K upper T perpendicular base-facet upper L upper M upper N. Also draw upper K upper R, upper K upper S, perpendicular upper L upper M, upper M upper N. And join upper T upper R, upper T upper M, upper T upper S.

It is easily proved that angles upper T upper R upper M, upper T upper S upper M are right.

The required volume is one third period upper K upper T period upper L upper M upper N. The area of upper L upper M upper N is of course known. All we need is the length of upper K upper T. Now upper K upper T squared equals upper K upper S squared minus upper T upper S squared; and upper K upper S evidently equals c period sine upper B. Hence all we need is the length of upper T upper S.

A diamond-shaped polyhedron with outer vertices K (top), L (left), M (bottom), N (right). Interior points R, S, T form an inner triangle. Edges are labeled a, b, c indicating equal lengths, with multiple triangulating diagonals connecting all vertices, creating a fully triangulated faceted solid.

Now this requires a preliminary Lemma, in itself a very pretty problem, viz.—

LEMMA (1).

Given, in Tetragon upper R upper M upper S upper T, sides upper R upper M, upper M upper S, and angle upper R upper M upper S, and that angles upper T upper R upper M, upper T upper S upper M are right: find upper T upper S.

A small quadrilateral with vertices T (top center), R (left), S (right), and M (bottom center). Diagonal RS connects the two side vertices horizontally, with segments from R and S meeting at both T above and M below, forming two triangles sharing the diagonal RS.

Now StartFraction upper T upper S Over sine upper T upper R upper S EndFraction equals StartFraction upper T upper R Over sine upper T upper S upper R EndFraction;

also upper T upper S cosine upper T upper S upper R plus upper T upper R cosine upper T upper R upper S equals upper R upper S;

StartLayout 1st Row 1st Column therefore StartFraction upper T upper S Over sine upper T upper R upper S EndFraction 2nd Column equals StartFraction upper T upper R Over sine upper T upper S upper R EndFraction comma 2nd Row 1st Column Blank 2nd Column equals StartFraction upper T upper S cosine upper T upper S upper R plus upper T upper R cosine upper T upper R upper S Over sine upper T upper R upper S cosine upper T upper S upper R plus sine upper T upper S upper R cosine upper T upper R upper S EndFraction comma 3rd Row 1st Column Blank 2nd Column equals StartFraction upper R upper S Over sine upper R upper M upper S EndFraction equals StartFraction upper M upper S Over sine upper M upper R upper S EndFraction semicolon 4th Row 1st Column therefore StartFraction upper T upper S Over cosine upper M upper R upper S EndFraction 2nd Column equals StartFraction upper M upper S Over sine upper M upper R upper S EndFraction semicolon i period e period upper T upper S equals upper M upper S cotangent upper M upper R upper S period EndLayout

Q. E. F.

Hence this requires another Lemma, in order to find the value of cotangent upper M upper R upper S (or tangent upper M upper R upper S, which will do as well, and makes a prettier problem).

LEMMA (2).

A plain triangle with vertex R at the upper left, S at the upper right, and M at the bottom center. The triangle is scalene and undecorated, with bold labeled vertices and no interior markings.

Given, in Triangle upper R upper M upper S, sides upper R upper M, upper M upper S, and angle upper R upper M upper S: find tangent upper M upper R upper S.

StartLayout 1st Row 1st Column tangent upper M upper R upper S 2nd Column equals StartFraction sine upper M upper R upper S Over cosine upper M upper R upper S EndFraction equals StartFraction upper R upper S sine upper M upper R upper S Over upper R upper S cosine upper M upper R upper S EndFraction comma 2nd Row 1st Column Blank 2nd Column equals StartFraction upper M upper S sine upper R upper M upper S Over upper R upper M minus upper M upper S cosine upper R upper M upper S EndFraction period EndLayout

Q. E. F.

Hence, in Tetragon upper R upper M upper S upper T, we have by Lemma (1),

upper T upper S equals upper M upper S cotangent upper M upper R upper S semicolon

and, by Lemma (2), cotangent upper M upper R upper S equals StartFraction upper R upper M minus upper M upper S cosine upper R upper M upper S Over upper M upper S sine upper R upper M upper S EndFraction,

equals StartFraction c cosine upper A minus c cosine upper B cosine upper C Over c cosine upper B sine upper C EndFraction equals StartFraction cosine upper A minus cosine upper B cosine upper C Over cosine upper B sine upper C EndFraction semicolon

therefore upper T upper S equals StartFraction c Over sine upper C EndFraction dot left parenthesis cosine upper A minus cosine upper B cosine upper C right parenthesis.

Now upper K upper T squared equals upper K upper S squared minus upper T upper S squared;

therefore it equals left parenthesis c sine upper B right parenthesis squared minus StartFraction c squared Over sine squared upper C EndFraction dot left parenthesis cosine upper A minus cosine upper B cosine upper C right parenthesis squared,

equals StartFraction c squared Over sine squared upper C EndFraction dot left brace left parenthesis sine upper B sine upper C right parenthesis squared minus left parenthesis cosine upper A minus cosine upper B cosine upper C right parenthesis squared right brace;

therefore upper K upper T equals StartFraction c Over sine upper C EndFraction multiplied by StartRoot sine squared upper B sine squared upper C minus cosine squared upper B cosine squared upper C minus cosine squared upper A plus 2 cosine upper A cosine upper B cosine upper C EndRoot,

equals StartFraction c Over sine upper C EndFraction multiplied by

StartRoot left parenthesis 1 minus cosine squared upper B right parenthesis dot left parenthesis 1 minus cosine squared upper C right parenthesis minus cosine squared upper B cosine squared upper C minus cosine squared upper A plus 2 cosine upper A cosine upper B cosine upper C EndRoot,

= StartFraction c Over sine upper C EndFraction dot StartRoot 1 minus left parenthesis cosine squared upper A plus cosine squared upper B plus cosine squared upper C right parenthesis plus 2 cosine upper A cosine upper B cosine upper C EndRoot,

which is symmetrical, as it ought to be.

A diamond-shaped polyhedron with outer vertices K (top), L (left), M (bottom), N (right) and interior triangle R, S, T. All vertices are connected by triangulating segments radiating from apex K down through R, T, S to M. Edges labeled a, b, c indicate equal lengths on corresponding sides.

Now area of upper L upper M upper N equals StartFraction a b sine upper C Over 2 EndFraction;

hence volume of Tetrahedron

equals StartFraction a b c Over 6 EndFraction period StartRoot 1 minus left parenthesis cosine squared upper A plus cosine squared upper B plus cosine squared upper C right parenthesis plus 2 cosine upper A cosine upper B cosine upper C EndRoot period

Q. E. F.

60. (14, 25)

Let angle upper B upper A upper D equals theta, angle upper C upper A upper D equals phi.

A bold triangle with vertices A (top right), B (bottom left), C (bottom right). Cevian AD drops from A to point D on base BC. Interior point B' lies on AB and D' lies near the center, connected by a dashed line from B' to D'.

Now StartFraction sine left parenthesis upper B plus theta right parenthesis Over sine theta EndFraction equals StartStartFraction c OverOver left parenthesis StartFraction m a Over m plus n EndFraction right parenthesis EndEndFraction

equals StartFraction c period left parenthesis m plus n right parenthesis Over m a EndFraction semicolon

therefore sine upper B cotangent theta plus cosine upper B equals StartFraction c period left parenthesis m plus n right parenthesis Over m a EndFraction;

therefore cotangent theta equals StartFraction c period left parenthesis m plus n right parenthesis Over m a period sine upper B EndFraction minus cotangent upper B comma

equals StartFraction left parenthesis m plus n right parenthesis period left parenthesis a cosine upper B plus b cosine upper A right parenthesis minus m a cosine upper B Over m a sine upper B EndFraction comma

equals StartFraction left parenthesis m plus n right parenthesis b cosine upper A plus n a cosine upper B Over m a sine upper B EndFraction semicolon

A geometric diagram of triangle ABC with interior points B', D, D' and dashed lines.

i period e period cotangent theta equals StartStartFraction left parenthesis m plus n right parenthesis dot StartFraction b Over sine upper B EndFraction dot cosine upper A plus n a cotangent upper B OverOver m a EndEndFraction comma

equals StartFraction left parenthesis m plus n right parenthesis a cotangent upper A plus n a cotangent upper B Over m a EndFraction comma

equals StartFraction left parenthesis m plus n right parenthesis cotangent upper A plus n cotangent upper B Over m EndFraction period

Similarly, cotangent phi equals StartFraction left parenthesis m plus n right parenthesis cotangent upper A plus m cotangent upper C Over n EndFraction.

Q. E. F.

COROLLARIES.

(1) m cotangent theta minus n cotangent phi equals n cotangent upper B minus m cotangent upper C.

(2) StartFraction cotangent upper B plus cotangent phi Over cotangent upper C plus cotangent theta EndFraction equals StartFraction m Over n EndFraction.

(3) If Triangle be equilateral,

cotangent theta equals StartFraction m plus 2 n Over m EndFraction period StartFraction 1 Over StartRoot 3 EndRoot EndFraction comma

cotangent phi equals StartFraction n plus 2 m Over n EndFraction dot StartFraction 1 Over StartRoot 3 EndRoot EndFraction semicolon

therefore StartFraction cotangent theta Over cotangent phi EndFraction equals StartFraction m n plus 2 n squared Over m n plus 2 m squared EndFraction semicolon

therefore StartFraction tangent theta Over tangent phi EndFraction equals StartFraction m n plus 2 m squared Over m n plus 2 n squared EndFraction semicolon

i. e., if upper C upper D prime upper B prime be drawn perpendicular to upper A upper D, StartFraction upper B prime upper D prime Over upper D prime upper C EndFraction equals StartFraction m n plus 2 m squared Over m n plus 2 n squared EndFraction;

e. g., if StartFraction m Over n EndFraction equals one half, StartFraction upper B prime upper D prime Over upper D prime upper C EndFraction equals two fifths·

(4) Let tangent upper A equals 1 comma tangent upper B equals 2 comma tangent upper C equals 3;

then cotangent theta equals StartFraction m plus n plus n period one half Over m EndFraction equals StartFraction 2 m plus 3 n Over 2 m EndFraction,

cotangent phi equals StartFraction m plus n plus m period one third Over n EndFraction equals StartFraction 3 n plus 4 m Over 3 n EndFraction;

therefore StartFraction tangent theta Over tangent phi EndFraction equals StartFraction 6 m n plus 8 m squared Over 6 m n plus 9 n squared EndFraction;

from which, if StartFraction tangent theta Over tangent phi EndFraction were given, we could find StartFraction m Over n EndFraction from a Quadratic Equation.

I tried various values, to find one which would give rational values for m and n, and found that two thirds would do, as it leads to the Quadratic

2 left parenthesis 6 m n plus 9 n squared right parenthesis minus 3 left parenthesis 6 m n plus 8 m squared right parenthesis equals 0 comma

in which left parenthesis upper B squared minus 4 upper A upper C right parenthesis becomes, after dividing all through by 6, left parenthesis 1 squared plus 4.4 .3 right parenthesis, i. e. 49.

The Quadratic is 4 m squared plus m n minus 3 n squared equals 0;

whence StartFraction m Over n EndFraction equals StartFraction negative 1 plus or minus 7 Over 8 EndFraction equals three fourths; which solves the Problem 'Given a Triangle upper A upper B upper C, having the tangents of its angles equal to 1, 2, 3: divide upper B upper C at upper D, so that, if upper A upper D be joined, and upper C upper D prime upper B prime drawn perpendicular to it, the ratio StartFraction upper B prime upper D prime Over upper D prime upper C EndFraction may be two thirds'. The answer is 'Divide it so that StartFraction upper B upper D Over upper D upper C EndFraction equals three fourths'.

61. (14)

We know that the equation

'left parenthesis a squared plus 4 b squared plus 4 c squared right parenthesis plus left parenthesis 4 a squared plus b squared plus 4 c squared right parenthesis

plus left parenthesis 4 a squared plus 4 b squared plus c squared right parenthesis equals 9 left parenthesis a squared plus b squared plus c squared right parenthesis'

is identically true.

Hence a squared plus b squared plus c squared

StartLayout 1st Row 1st Column Blank 2nd Column equals one ninth dot left brace left parenthesis a squared plus 4 b squared plus 4 c squared right parenthesis plus left parenthesis 4 a squared plus b squared plus 4 c squared right parenthesis plus left parenthesis 4 a squared plus 4 b squared plus c squared right parenthesis right brace semicolon 2nd Row 1st Column Blank 2nd Column equals one ninth dot left brace left parenthesis a squared plus 4 b squared plus 4 c squared plus 8 b c minus 4 c a minus 4 a b right parenthesis 3rd Row 1st Column Blank 2nd Column plus left parenthesis 4 a squared plus b squared plus 4 c squared minus 4 b c plus 8 c a minus 4 a b right parenthesis 4th Row 1st Column Blank 2nd Column plus left parenthesis 4 a squared plus 4 b squared plus c squared minus 4 b c minus 4 c a plus 8 a b right parenthesis right brace semicolon 5th Row 1st Column Blank 2nd Column equals one ninth dot left brace left parenthesis negative a plus 2 b plus 2 c right parenthesis squared plus left parenthesis 2 a minus b plus 2 c right parenthesis squared plus left parenthesis 2 a plus 2 b minus c right parenthesis squared right brace semicolon 6th Row 1st Column Blank 2nd Column equals left parenthesis StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction right parenthesis squared plus left parenthesis StartFraction 2 a minus b plus 2 c Over 3 EndFraction right parenthesis squared plus left parenthesis StartFraction 2 a plus 2 b minus c Over 3 EndFraction right parenthesis squared period EndLayout

Now left parenthesis negative a plus 2 b plus 2 c right parenthesis equals 3 left parenthesis b plus c right parenthesis minus left parenthesis a plus b plus c right parenthesis;

therefore, if (a plus b plus c) be a multiple of 3, so also is left parenthesis negative a plus 2 b plus 2 c right parenthesis;

therefore StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction is an integer;

and similarly for the other 2 fractions.

Also it may be proved that, if StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction be equal to a, or b, or c, then a, b, c can be arranged in upper A period upper P period

First, let StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction equals a;

then negative a plus 2 b plus 2 c equals 3 a; i. e. b plus c equals 2 a;

secondly, let StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction equals b;

then negative a plus 2 b plus 2 c equals 3 b; i. e. 2 c equals a plus b;

thirdly, let StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction equals c;

then negative a plus 2 b plus 2 c equals 3 c; i. e. 2 b equals c plus a.

And similarly for the other 2 fractions.

Hence, contranominally, if a, b, c can not be arranged in upper A period upper P period, the 2 sets of squares have no common term.

Q. E. D.

Numerical Examples (not thought out).

a squared | b squared | c squared | left parenthesis StartFraction negative a plus 2 b plus 2 c Over 3 EndFraction right parenthesis squared | left parenthesis StartFraction 2 a minus b plus 2 c Over 3 EndFraction right parenthesis squared | left parenthesis StartFraction 2 a plus 2 b minus c Over 3 EndFraction right parenthesis squared
1 squared | 4 squared | 4 squared | 5 squared | 2 squared | 2 squared
3 squared | 4 squared | 8 squared | 7 squared | 6 squared | 2 squared
4 squared | 5 squared | 9 squared | 8 squared | 7 squared | 3 squared

62. (14)

Let upper A upper B, upper A upper C, be the given Lines, and upper P the given Point.

A triangle with vertices A (bottom left), B (bottom right), and an implied apex, with line EC crossing diagonally from upper right down through interior point P. Points D (on AP), G and E (above P), F and H (on base AB) mark intersections, with dashed lines from P to H.

Through upper P draw upper P upper D parallel to upper A upper B; from upper D upper C cut off upper D upper E equal to upper A upper D; join upper E upper P, and produce it to meet upper A upper B at upper F.

Because upper A upper D equals upper D upper E, and that upper D upper P is parallel to upper A upper B,

therefore upper F upper P equals upper P upper E.

Now let upper G upper P upper H be any other line through upper P;

then angle upper P upper F upper H greater than angle upper P upper E upper G.

Because, in Triangles upper P upper F upper H, upper P upper E upper G, upper P upper F equals upper P upper E, and

angle upper F upper P upper H equals angle upper G upper P upper E comma and angle upper P upper F upper H greater than angle upper P upper E upper G comma

therefore upper P upper H greater than upper P upper G, and Triangle upper P upper F upper H greater than Triangle upper P upper G upper E.

To each add Tetragon upper A upper F upper P upper G;

therefore Triangle upper A upper G upper H greater than Triangle upper A upper E upper F.

And so of any other line through upper P.

Hence upper A upper E upper F is the least possible Triangle.

Q. E. F.

63. (15, 26)

A three-dimensional parallelepiped with outer vertices A, B, D (top face), E, F, G (bottom face). An inner rectangular box is inscribed with vertices C, J, L (upper) and O, H, K, M (lower). Multiple diagonal segments connect outer and inner vertices.

Let each side of each Square = 2.

Then upper L upper G equals StartRoot 3 EndRoot, upper M upper G equals left parenthesis StartRoot 2 EndRoot minus 1 right parenthesis;

therefore upper L upper M left parenthesis equals upper J upper K right parenthesis equals NestedStartRoot 3 minus left parenthesis 2 plus 1 minus 2 StartRoot 2 EndRoot right parenthesis NestedEndRoot

equals 2 Superscript three fourths;

therefore upper O upper J equals upper O upper K equals StartFraction 1 Over 2 Superscript one fourth Baseline EndFraction.

Take upper O as origin, the upper X-axis parallel to to upper A upper D, and the upper Y-axis to upper A upper B; and let upper J upper K be part of the upper Z-axis.

Let equation to plane containing Triangle upper C upper D upper G be

x cosine alpha plus y cosine beta plus z cosine gamma minus p equals 0 comma

where p is length of perpendicular dropped, from upper O, upon this plane, and meeting it somewhere in upper L upper G.

Hence we can find p from equation to upper L upper G, in the upper X upper Z-plane, which will be

x cosine alpha plus z cosine gamma minus p equals 0 semicolon

now this line contains upper L, whose co-ordinates are left parenthesis 1 comma StartFraction 1 Over 2 Superscript one fourth Baseline EndFraction right parenthesis, and upper G, whose co-ordinates are left parenthesis StartRoot 2 EndRoot comma minus StartFraction 1 Over 2 Superscript one fourth Baseline EndFraction right parenthesis;

therefore cosine alpha plus StartFraction 1 Over 2 Superscript one fourth Baseline EndFraction dot cosine gamma minus p equals 0,

and StartRoot 2 EndRoot dot cosine alpha minus StartFraction 1 Over 2 Superscript one fourth Baseline EndFraction dot cosine gamma minus p equals 0;

therefore left parenthesis StartRoot 2 EndRoot minus 1 right parenthesis dot cosine alpha equals StartFraction 2 Over 2 Superscript one fourth Baseline EndFraction dot cosine gamma equals 2 Superscript three fourths Baseline dot cosine gamma;

therefore StartFraction cosine alpha Over 2 Superscript three fourths Baseline EndFraction equals StartFraction cosine gamma Over StartRoot 2 EndRoot minus 1 EndFraction equals StartFraction 1 Over StartRoot 2 Superscript three halves Baseline plus 3 minus 2 Superscript three halves Baseline EndRoot EndFraction equals StartFraction 1 Over StartRoot 3 EndRoot EndFraction;

therefore cosine alpha equals StartFraction 2 Superscript three fourths Baseline Over StartRoot 3 EndRoot EndFraction comma cosine gamma equals StartFraction StartRoot 2 EndRoot minus 1 Over StartRoot 3 EndRoot EndFraction;

therefore p equals StartFraction 2 Superscript three fourths Baseline Over StartRoot 3 EndRoot EndFraction plus StartFraction StartRoot 2 EndRoot minus 1 Over 2 Superscript one fourth Baseline dot StartRoot 3 EndRoot EndFraction equals StartFraction StartRoot 2 EndRoot plus 1 Over 2 Superscript one fourth Baseline dot StartRoot 3 EndRoot EndFraction.

Now area of upper C upper D upper G equals StartRoot 3 EndRoot;

therefore volume of pyramid, whose base is upper C upper D upper G and whose vertex is upper O comma equals StartFraction StartRoot 2 EndRoot plus 1 Over 3 dot 2 Superscript one fourth Baseline EndFraction;

and there are eight such pyramids in the solid;

therefore their sum = StartFraction 8 left parenthesis StartRoot 2 EndRoot plus 1 right parenthesis Over 3 dot 2 Superscript one fourth Baseline EndFraction.

Also volume of pyramid, whose base is upper A upper B upper C upper D, and whose vertex is upper O, = StartFraction 4 Over 3 dot 2 Superscript one fourth Baseline EndFraction;

and there are 2 such pyramids in the solid;

therefore their sum = StartFraction 8 Over 3 dot 2 Superscript one fourth Baseline EndFraction;

therefore volume of solid = StartFraction 8 left parenthesis 2 plus StartRoot 2 EndRoot right parenthesis Over 3 dot 2 Superscript one fourth Baseline EndFraction equals StartFraction 8 dot 2 Superscript one fourth Baseline dot left parenthesis StartRoot 2 EndRoot plus 1 right parenthesis Over 3 EndFraction

Q. E. F.

64. (15)

Left: triangle ABC circumscribes a circle centered at O, with contact points D, E, F and additional labeled points R, Q, S, P, M, N on sides and circle. Right: a separate circle contains inscribed quadrilateral GHLK with diagonals drawn, illustrating related circle-polygon theorems.

Let upper A upper B upper C be the given Triangle, and upper O the given Point; and let upper O upper D, its distance from upper B upper C, be less than either upper O upper E or upper O upper F, its distances from upper C upper A, upper A upper B.

Draw a line upper G upper H equal to upper O upper E, and upper G upper K perpendicular it and equal to upper O upper F; and join upper H upper K; and about the Triangle upper G upper H upper K describe a Circle; and place in it a line upper K upper L equal to upper O upper D; and join upper L upper H.

Because sqs of upper K upper L, upper L upper H = sqs of upper K upper G, upper G upper H, and that upper K upper L is less than either upper K upper G or upper G upper H, therefore upper L upper H is greater than either;

therefore a Circle, with centre upper O, and radius equal to upper L upper H, will cut all three Lines, in two Points each. Describe this Circle.

Then sqs of upper M upper D comma upper D upper O equals sqs of upper P upper E comma upper E upper O semicolon

also sq period of upper L upper H equals sqs of upper R upper F comma upper F upper O semicolon

therefore sqs of upper M upper D comma upper D upper O comma upper L upper H equals sqs of upper P upper E comma upper R upper F comma upper E upper O comma upper F upper O semicolon

but sqs of upper D upper O comma upper L upper H equals sqs of upper K upper L comma upper L upper H comma

equals sqs of upper G upper H comma upper G upper K equals sqs of upper E upper O comma upper F upper O semicolon

therefore sq period of upper M upper D equals sqs of upper P upper E comma upper R upper F semicolon

therefore 4 times sq period of upper M upper D equals 4 times sqs of upper P upper E comma upper R upper F semicolon

i period e period sq period of upper M upper N equals sqs of upper P upper Q comma upper R upper S period

Hence upper M upper N, upper P upper Q, upper R upper S, can be sides of a right-angled Triangle.

Q. E. F.

65. (15)

Calling the angles StartFraction 1 Over x EndFraction, StartFraction 1 Over y EndFraction, StartFraction 1 Over z EndFraction, of 360°, we must have

StartFraction 1 Over x EndFraction plus StartFraction 1 Over y EndFraction plus StartFraction 1 Over z EndFraction equals one half semicolon

an Indeterminate Equation with 3 unknowns.

Evidently none of them can be so small as 2.

(1) Let x equals 3; then StartFraction 1 Over y EndFraction plus StartFraction 1 Over z EndFraction equals one sixth.

Now, if StartFraction 1 Over y EndFraction equals StartFraction k Over k plus l EndFraction times one sixth, StartFraction 1 Over z EndFraction will = StartFraction l Over k plus l EndFraction times one sixth:

hence k can only be 1, or 2, or 3, or 6; and the same is true of l.

(N.B. It is assumed that the fractions StartFraction k Over k plus l EndFraction, StartFraction l Over k plus l EndFraction comma are in their lowest terms.)

Let StartFraction 1 Over y EndFraction be not less than StartFraction 1 Over z EndFraction. Then StartFraction k Over k plus l EndFraction not less than one half.

Then its possible values are one half comma so that StartFraction l Over k plus l EndFraction equals one half

two thirds,      | one third
three fourths, | one fourth
three fifths | two fifths
six sevenths, |    one seventh.

This gives 5 sets of values for StartFraction 1 Over x EndFraction, StartFraction 1 Over y EndFraction, StartFraction 1 Over z EndFraction, viz.:

one third comma one twelfth comma one twelfth semicolon one third comma one ninth comma one eighteenth semicolon one third comma one eighth comma one twenty fourth semicolon one third comma one tenth comma one fifteenth semicolon one third comma one seventh comma one forty second period

(2) Let x equals 4. Then StartFraction 1 Over y EndFraction plus StartFraction 1 Over z EndFraction equals one fourth, and, as before, k can only be 1, or 2, or 4, and the same is true of l. Hence the possible values for StartFraction k Over k plus l EndFraction are one half, so that StartFraction l Over k plus l EndFraction equals one half

two thirds,      | one third
four fifths, | one fifth.

This gives 3 more sets of values for StartFraction 1 Over x EndFraction, StartFraction 1 Over y EndFraction, StartFraction 1 Over z EndFraction comma viz.:

one fourth comma one eighth comma one eighth semicolon one fourth comma one sixth comma one twelfth semicolon one half comma one fifth comma one twentieth period

(3) Let x equals 5; then StartFraction 1 Over y EndFraction plus StartFraction 1 Over z EndFraction equals three tenths.

Hence denominator must contain factor "3", and k can be only 1, or 2, or 5, or 10; and the same is true of l.

Hence possible values of StartFraction k Over k plus l EndFraction are one half, so that StartFraction l Over k plus l EndFraction equals one half

two thirds,      | one third
five sixths, | one sixth.

This gives 2 sets of values for StartFraction 1 Over x EndFraction, StartFraction 1 Over y EndFraction, StartFraction 1 Over z EndFraction comma viz.:—

one fifth comma one fifth comma one tenth semicolon one fifth comma one fourth comma one twentieth semicolon

but the latter (a fact overlooked in thinking out) we have had already.

(4) Let x equals 6; then StartFraction 1 Over y EndFraction plus StartFraction 1 Over z EndFraction equals one third.

Hence k can be only 1, or 3, and the same is true of l.

Hence possible values for StartFraction k Over k plus l EndFraction are one half, so that StartFraction l Over k plus l EndFraction equals one half

three fourths comma one fourth period

This gives 2 sets of values, viz.:—

one sixth comma one sixth comma one sixth semicolon one sixth comma one fourth comma one twelfth semicolon

but the latter (a fact overlooked in thinking out) we have had already.

There is no use in giving, to x, any values greater than 6; for these would make StartFraction 1 Over y EndFraction plus StartFraction 1 Over z EndFraction greater than one third; so that one or other must be greater than one sixth; i. e. either y or z must less than 6, and we should get old values over again.

Hence there are 10 different shapes.

Q. E. F.

The 10 sets of angles (I am not certain that they were all thought out) are

(1)      | 120°,      | 30°,      | 30°;     
(2)      | 120°,      | 40°,      | 20°;     
(3)      | 120°,      | 45°,      | 15°;     
(4)      | 120°,      | 36°,      | 24°;     
(5)      | 120°      | 51 and three sevenths°,      | 8 and four sevenths°;     
(6)      | 90°,      | 45°,      | 45°;     
(7)      | 90°,      | 60°,      | 30°;     
(8)      | 90°,      | 72°,      | 18°;     
(9)      | 72°,      | 72°,      | 36°;     
(10)      | 60°,      | 60°,      | 60°.     

66. (15, 26)

Write k for StartFraction alpha Over alpha plus beta EndFraction. Now the counters must be either both white, or one white and one black. Let chance of first condition be x; hence chance of second is left parenthesis 1 minus x right parenthesis. Hence chance of drawing white is x times 1 plus left parenthesis 1 minus x right parenthesis times one half.

StartLayout 1st Row 1st Column therefore x plus StartFraction 1 minus x Over 2 EndFraction 2nd Column equals k semicolon therefore x equals 2 k minus 1 semicolon 2nd Row 1st Column Blank 2nd Column therefore left parenthesis 1 minus x right parenthesis equals 2 minus 2 k period EndLayout

Let a counter now be drawn and prove white; then chance of 'observed event,' in 1st condition, is 1, and, in 2nd condition, one half;

Hence the chances, of the existence of these two conditions, are proportional to left parenthesis 2 k minus 1 right parenthesis times 1, left parenthesis 2 minus 2 k right parenthesis times one half; i. e. are proportional to 2 k minus 1, 1 minus k;

hence these chances actually are StartFraction 2 k minus 1 Over k EndFraction, StartFraction 1 minus k Over k EndFraction;

hence the chance of now drawing white,

StartLayout 1st Row 1st Column is 2nd Column StartFraction 2 k minus 1 Over k EndFraction times 1 plus StartFraction 1 minus k Over k EndFraction times one half semicolon 2nd Row 1st Column i period e period 2nd Column StartFraction 3 k minus 1 Over 2 k EndFraction period EndLayout

Hence the effect of one repetition of the experiment has been to change k into StartFraction 3 k minus 1 Over 2 k EndFraction.

Hence a second repetition of it will change

StartFraction 3 k minus 1 Over 2 k EndFraction into StartStartFraction 3 times StartFraction 3 k minus 1 Over 2 k EndFraction minus 1 OverOver 2 times StartFraction 3 k minus 1 Over 2 k EndFraction EndEndFraction semicolon i period e period into StartFraction 7 k minus 3 Over 6 k minus 2 EndFraction period

We have now to discover the law (if there is one) for the series

k comma StartFraction 3 k minus 1 Over 2 k EndFraction comma StartFraction 7 k minus 3 Over 6 k minus 2 EndFraction comma

regarding these as identical functions of 1, 2, 3.

We can write the 1st and 2nd term in the form of the 3rd, thus:—

StartFraction k minus 0 Over 0 times k minus left parenthesis negative 1 right parenthesis EndFraction comma StartFraction 3 k minus 1 Over 2 k minus 0 EndFraction comma StartFraction 7 k minus 3 Over 6 k minus 2 EndFraction

and, by inspection, we see that each is of the form

StartFraction left parenthesis 2 Superscript n Baseline minus 1 right parenthesis times k minus left parenthesis 2 Superscript n minus 1 Baseline minus 1 right parenthesis Over left parenthesis 2 Superscript n Baseline minus 2 right parenthesis times k minus left parenthesis 2 Superscript n minus 1 Baseline minus 2 right parenthesis EndFraction comma

where n denotes the place of the term.

Suppose this law to hold for n terms, what will be the effect of repeating the experiment once more?

We know that it changes k into StartFraction 3 k minus 1 Over 2 k EndFraction. Hence the new chance will be

StartStartFraction 3 times StartFraction left parenthesis 2 Superscript n Baseline minus 1 right parenthesis times k minus left parenthesis 2 Superscript n minus 1 Baseline minus 1 right parenthesis Over left parenthesis 2 Superscript n Baseline minus 2 right parenthesis times k minus left parenthesis 2 Superscript n minus 1 Baseline minus 2 right parenthesis EndFraction minus 1 OverOver 2 times StartFraction left parenthesis 2 Superscript n Baseline minus 1 right parenthesis times k minus left parenthesis 2 Superscript n minus 1 Baseline minus 1 right parenthesis Over left parenthesis 2 Superscript n Baseline minus 2 right parenthesis times k minus left parenthesis 2 Superscript n minus 1 Baseline minus 2 right parenthesis EndFraction EndEndFraction semicolon

i period e period StartFraction k times left parenthesis 3 dot 2 Superscript n Baseline minus 3 minus 2 Superscript n Baseline plus 2 right parenthesis minus 3 dot 2 Superscript n minus 1 Baseline plus 3 plus 2 Superscript n minus 1 Baseline minus 2 Over left parenthesis 2 Superscript n plus 1 Baseline minus 2 right parenthesis times k minus left parenthesis 2 Superscript n Baseline minus 2 right parenthesis EndFraction semicolon

i period e period StartFraction left parenthesis 2 Superscript n plus 1 Baseline minus 1 right parenthesis times k minus left parenthesis 2 Superscript n Baseline minus 1 right parenthesis Over left parenthesis 2 Superscript n plus 1 Baseline minus 2 right parenthesis times k minus left parenthesis 2 Superscript n Baseline minus 2 right parenthesis EndFraction semicolon

i. e. the ModifyingAbove n plus 1 With quotation dash vertical bar Superscript th Baseline term of the series will follow the same law. But we know that the law holds for the 1 Superscript st, 2 Superscript nd, and 3 Superscript rd terms. Hence it holds universally.

Hence, after m repetitions of the experiment, the chance of drawing white will be the ModifyingAbove m plus 1 With quotation dash vertical bar Superscript th Baseline term of the above series; i. e. it will be

StartFraction left parenthesis 2 Superscript m plus 1 Baseline minus 1 right parenthesis times k minus left parenthesis 2 Superscript m Baseline minus 1 right parenthesis Over left parenthesis 2 Superscript m plus 1 Baseline minus 2 right parenthesis times k minus left parenthesis 2 Superscript m Baseline minus 2 right parenthesis EndFraction period

Now, for k, write StartFraction alpha Over alpha plus beta EndFraction.

StartLayout 1st Row 1st Column Then chance is 2nd Column StartFraction left parenthesis 2 Superscript m plus 1 Baseline minus 1 right parenthesis times alpha minus left parenthesis 2 Superscript m Baseline minus 1 right parenthesis dot left parenthesis alpha plus beta right parenthesis Over left parenthesis 2 Superscript m plus 1 Baseline minus 2 right parenthesis times alpha minus left parenthesis 2 Superscript m Baseline minus 2 right parenthesis dot left parenthesis alpha plus beta right parenthesis EndFraction semicolon 2nd Row 1st Column i period e period 2nd Column StartFraction left parenthesis 2 Superscript m plus 1 Baseline minus 2 Superscript m Baseline right parenthesis alpha minus left parenthesis 2 Superscript m Baseline minus 1 right parenthesis dot beta Over left parenthesis 2 Superscript m plus 1 Baseline minus 2 Superscript m Baseline right parenthesis alpha minus left parenthesis 2 Superscript m Baseline minus 2 right parenthesis dot beta EndFraction semicolon 3rd Row 1st Column i period e period 2nd Column StartFraction 2 Superscript m Baseline dot left parenthesis alpha minus beta right parenthesis plus beta Over 2 Superscript m Baseline dot left parenthesis alpha minus beta right parenthesis plus 2 beta EndFraction period EndLayout

Q. E. F.

Example—Let chance be nine tenths; and then let experiment be repeated 5 times more.

Here alpha equals 9, beta equals 1;

therefore chance becomes StartFraction 32 times 8 plus 1 Over 32 times 8 plus 2 EndFraction, i.e. StartFraction 257 Over 258 EndFraction.

67. (16, 26)

A geometric diagram showing intersecting 3D wireframe shapes (diamond/pyramid forms) centered on perpendicular axes, with a bold diagonal line cutting through.

Let upper A upper B upper C upper D be the socket. Revolve the Tetrahedron until the plane, in it, upper D upper O upper A has taken the new position upper D prime upper Q upper A prime; and let edge upper D upper A, in its new position, meet the socket-rim upper A upper C at upper R. From upper A prime draw upper A prime upper L perpendicular upper X upper Y-plane. Join upper O upper R, and produce it to upper L. And draw upper R upper M, upper L upper N, the y-ordinates of upper R and upper L.

Then co-ordinates of upper A prime are upper O upper N, upper N upper L, upper L upper A prime.

Call upper O upper M, upper M upper R, 'x prime, y prime'; and upper O upper A, upper O upper R, upper O upper D, 'a, a prime, h'; and angle upper X upper O upper R 'theta'.

It is evident that the vertical axis of the Tetrahedron always coincides with the upper Z-axis.

Hence upper A moves on the surface of a cylinder,

i. e. x squared plus y squared equals a squared left parenthesis 1 right parenthesis

Now angle upper X upper A upper C equals 150 degree;

therefore Equation to upper A upper C is y equals minus StartFraction 1 Over StartRoot 3 EndRoot EndFraction dot left parenthesis x minus a right parenthesis;

i period e period x plus StartRoot 3 EndRoot dot y equals a left parenthesis 2 right parenthesis

Also Equation to upper O upper R is StartFraction x Over cosine theta EndFraction equals StartFraction y Over sine theta EndFraction equals a;

therefore comma at upper R comma StartFraction x prime Over cosine theta EndFraction equals StartFraction y prime Over sine theta EndFraction equals a Superscript prime Baseline semicolon left parenthesis 3 right parenthesis

therefore comma by left parenthesis 2 right parenthesis comma a prime dot cosine theta plus StartRoot 3 EndRoot dot a prime dot sine theta equals a;

therefore a prime equals StartFraction a Over cosine theta plus StartRoot 3 EndRoot dot sine theta EndFraction left parenthesis 4 right parenthesis

Also, by similar white up pointing triangles upper D prime upper Q upper A prime, upper D prime upper O upper R, upper Q upper A Superscript prime Baseline colon upper Q upper D Superscript prime Baseline colon colon upper O upper R colon upper O upper D prime

i. e. a colon h colon colon StartFraction a Over cosine theta plus StartRoot 3 EndRoot dot sine theta EndFraction colon h minus z;

therefore h minus z equals StartFraction h Over cosine theta plus StartRoot 3 EndRoot dot sine theta EndFraction;

but cosine theta equals StartFraction x Over a EndFraction, and sine theta equals StartFraction y Over a EndFraction;

therefore h minus z equals StartFraction a h Over x plus StartRoot 3 EndRoot dot y EndFraction;

i. e. left parenthesis x plus StartRoot 3 EndRoot dot y right parenthesis dot left parenthesis h minus z right parenthesis equals a h left parenthesis 5 right parenthesis

Equations (1) and (5) give the required Locus.

Q. E. F.

68. (16,26)

Let the Nos of bottles, taken out of the 3 days, be 'x, y, z'. Let each bottle have cost 10 v pence, and therefore be sold for 11 v pence.

Then the Treasurer's receipts, on the 3 days, were left parenthesis x minus 1 right parenthesis dot 11 v, y dot 11 v minus v, left parenthesis z minus 1 right parenthesis dot 11 v minus v; yielding, as profits (i. e. as remainders after deducting cost-price of bottles taken out), x v minus 11 v, y v minus v, z v minus 12 v. Then these 3 quantities are equal.

Hence y equals x minus 10, and z equals x plus 1;

therefore total No. of bottles, being left parenthesis x plus y plus z right parenthesis comma equals 3 x minus 9.

Now total profits are left parenthesis x plus y plus z right parenthesis period v minus 24 v; i. e. left parenthesis 3 x minus 33 right parenthesis v;

therefore profit, per bottle = StartFraction left parenthesis 3 x minus 33 right parenthesis dot v Over 3 x minus 9 EndFraction; and this must = 6;

therefore left parenthesis x minus 11 right parenthesis period v equals left parenthesis x minus 3 right parenthesis period 6.

Also z period 11 v equals 11 times 240; i. e. left parenthesis x plus 1 right parenthesis period 11 v equals 11 times 240;

therefore StartFraction x minus 11 Over x plus 1 EndFraction equals StartFraction 6 period left parenthesis x minus 3 right parenthesis Over 240 EndFraction;

therefore left parenthesis x plus 1 right parenthesis period left parenthesis x minus 3 right parenthesis equals 40 period left parenthesis x minus 11 right parenthesis;

therefore x squared minus 2 x minus 3 equals 40 x minus 440;

therefore x squared minus 42 x plus 437 equals 0.

Now 42 squared minus 4 times 437 equals 1764 minus 1748 equals 16;

therefore x equals StartFraction 42 plus or minus 4 Over 2 EndFraction equals 23 or 19;

therefore No. of bottles = 60 or 48; but it is a multiple of 5;

therefore it = 60.

Also left parenthesis x plus 1 right parenthesis period 11 v equals 11 times 240; i. e. 24 v equals 240;

therefore v equals 10;

i. e. the wine was bought @ 8/4 a bottle, and sold @ 9/2 a bottle.

Q. E. F.

69. (17, 26)

§ 1. Let angle upper B upper A upper D equals k period upper A comma angle upper C upper B upper E equals l period upper B comma angle upper A upper C upper F equals m period upper C.

Then angle upper A upper B upper E equals left parenthesis 1 minus l right parenthesis period upper B.

Now angle upper B upper C prime upper D equals angle upper C prime upper A upper B plus angle upper C prime upper B upper A.

i.e. k period upper A plus left parenthesis 1 minus l right parenthesis period upper B equals upper C period left parenthesis 1 right parenthesis

Similarly, l period upper B plus left parenthesis 1 minus m right parenthesis period upper C equals upper A semicolon left parenthesis 2 right parenthesis

and m period upper C plus left parenthesis 1 minus k right parenthesis period upper A equals upper B period left parenthesis 3 right parenthesis

A triangle with vertices A (top), B (bottom left), C (bottom right). Medial triangle A'B'C' is inscribed with B' (upper center), C' (middle left), A' (lower center). Points F (on AB), D (on BC), and E (on AC) mark cevian feet, with multiple crossing segments connecting all labeled points, forming a dense inner network.

From equations (1) and (3), l and m may be found in terms of k: but these, taken along with k, will not be similar functions of the single variable k. We must have k a certain function of upper A, upper B, upper C, and theta (say); l a similar function of upper B, upper C, upper A, and theta; and m a similar function of upper C, upper A, upper B, and theta; i. e. we must have

StartLayout 1st Row 1st Column k 2nd Column equals f left parenthesis upper A comma upper B comma upper C comma theta right parenthesis comma 2nd Row 1st Column l 2nd Column equals f left parenthesis upper B comma upper C comma upper A comma theta right parenthesis comma 3rd Row 1st Column m 2nd Column equals f left parenthesis upper C comma upper A comma upper B comma theta right parenthesis period EndLayout

Now we know, by (1), that k upper A minus l upper B equals upper C minus upper B;

i. e. upper A period f left parenthesis upper A comma upper B comma upper C comma theta right parenthesis minus upper B period f left parenthesis upper B comma upper C comma upper A comma theta right parenthesis equals upper C minus upper B.

Now, as an experiment, let

StartLayout 1st Row 1st Column k period upper A 2nd Column equals x upper A plus y upper B plus z upper C plus theta comma 2nd Row 1st Column l period upper B 2nd Column equals x upper B plus y upper C plus z upper A plus theta semicolon EndLayout

then k upper A minus l upper B equals left parenthesis x minus z right parenthesis period upper A plus left parenthesis y minus x right parenthesis period upper B plus left parenthesis z minus y right parenthesis period upper C;

StartLayout 1st Row 1st Column therefore x minus z 2nd Column equals 0 semicolon i period e period x equals z semicolon 2nd Row 1st Column z minus y 2nd Column equals 1 semicolon i period e period z equals y plus 1 period EndLayout

These conditions will be fulfilled, if we make y equals 1, and x equals z equals 2; so that

StartLayout 1st Row 1st Column k upper A 2nd Column equals 2 upper A plus upper B plus 2 upper C plus theta comma 2nd Row 1st Column l upper B 2nd Column equals 2 upper B plus upper C plus 2 upper A plus theta semicolon EndLayout

which would make

f left parenthesis upper A comma upper B comma upper C comma theta right parenthesis mean StartFraction 2 upper A plus upper B plus 2 upper C plus theta Over upper A EndFraction period

Now this may evidently be simplified by omitting left parenthesis upper A plus upper B plus upper C right parenthesis, which is constant; and we then have k equals StartFraction upper A plus upper C plus theta Over upper A EndFraction; or, in a yet simpler form, by again subtracting 180 degree, k equals StartFraction theta minus upper B Over upper A EndFraction.

A geometric diagram of triangle ABC with multiple interior points A', B', C', D, E, F and intersecting cevians.

StartLayout 1st Row 1st Column Similarly l 2nd Column equals StartFraction theta minus upper C Over upper B EndFraction comma 2nd Row 1st Column m 2nd Column equals StartFraction theta minus upper A Over upper C EndFraction period EndLayout

Q. E. F.

§ 2. We see that k upper A equals theta minus upper B, so that angle upper A upper D upper C is evidently equal to theta; and so are angles upper B upper E upper A, upper C upper F upper B.

This gives us a geometrical construction, viz. to draw lines from upper A, upper B, upper C, so that each makes the same angle theta with the opposite side.

§ 3. Let us now ascertain the limits within which the value of theta must lie.

We know that k upper A equals theta minus upper B.

Now k upper A not greater than upper A semicolon therefore theta minus upper B not greater than upper A; i. e. theta not greater than upper A plus upper B;

i. e. theta not greater than the supplement of upper C;

and of course this is true for each of the three angles upper A, upper B, upper C; i. e. if upper A, upper B, upper C, be the order of the angles in a descending order of magnitude, theta not greater than supplement of upper A.

Again k upper A not less than 0.

Hence theta minus upper B not less than 0; i. e. theta not less than upper B;

and of course this is true for each angle.

Hence if upper A, upper B, upper C, be the order in a descending order of magnitude, theta not less than upper A, and not greater than 180 degree negative upper A.

Q. E. F.

§ 4. We have now to ascertain the ratio which upper B prime upper C prime bears to upper B upper C.

In Triangle upper A upper B upper C prime, whose angles are left parenthesis theta minus upper B right parenthesis, left parenthesis 180 degree negative theta minus upper A right parenthesis, left parenthesis 180 degree negative upper C right parenthesis, we have

StartLayout 1st Row 1st Column upper A upper C prime 2nd Column equals StartFraction upper A upper B Over sine upper A upper C prime upper B EndFraction period sine upper A upper B upper C prime equals StartFraction c Over sine upper C EndFraction period sine left parenthesis theta plus upper A right parenthesis equals StartFraction a Over sine upper A EndFraction period sine left parenthesis theta plus upper A right parenthesis semicolon 2nd Row 1st Column upper B upper C prime 2nd Column equals StartFraction upper A upper B Over sine upper A upper C prime upper B EndFraction period sine upper B upper A upper C prime equals StartFraction c Over sine upper C EndFraction period sine left parenthesis theta minus upper B right parenthesis equals StartFraction a Over sine upper A EndFraction period sine left parenthesis theta minus upper B right parenthesis EndLayout

therefore, by symmetry, upper A upper B prime equals StartFraction a Over sine upper A EndFraction period sine left parenthesis theta minus upper A right parenthesis.

Now upper B prime upper C prime equals upper A upper C prime minus upper A upper B prime;

StartLayout 1st Row 1st Column therefore it 2nd Column equals StartFraction a Over sine upper A EndFraction left brace sine left parenthesis theta plus upper A right parenthesis minus sine left parenthesis theta minus upper A right parenthesis right brace comma 2nd Row 1st Column Blank 2nd Column equals StartFraction a Over sine upper A EndFraction period 2 cosine theta sine upper A equals a Baseline 2 cosine theta period EndLayout

Hence StartFraction a prime Over a EndFraction equals StartFraction b prime Over b EndFraction equals StartFraction c prime Over c EndFraction equals 2 cosine theta.

Q. E. F.

70. (17, 27)

Before folding the Plane containing the Triangles, the locus of their vertices is evidently a Line parallel to their common base. Hence, if the base of the Tetrahedron = 1, we may imagine a slip of paper, whose width is StartFraction StartRoot 3 EndRoot Over 2 EndFraction, attached to the front facet of the Tetrahedron, and wrapped round towards the right; and the upper edge of this slip will evidently be the locus of the vertices. This slip may be conveniently regarded as divided into equilateral Triangles, placed base-downwards and base-upwards alternately, and it is evident that these Triangles will successively cover the facets of the Tetrahedron, in the order 'front, right side, base, left side, front, &c.'; and its upper edge, made up of the bases of the inverted constituent Triangles, will evidently run as follows. Calling the successive Triangles, after the first (which occupies the front facet of the Tetrahedron), 'alpha' (base-up), 'beta' (base-down), 'gamma' (base-up), 'delta' (base-down), 'epsilon' (base-up), and so on, the locus consists of the bases of alpha, gamma, &c. Now 'alpha' will occupy the right facet, its base coinciding with the back-edge of the Tetrahedron; 'beta' will occupy the base of the Tetrahedron, its base coinciding with the front-edge; 'gamma' will occupy the left facet, its base coinciding with the back-edge; and so on. Hence the locus runs down the back-edge; up again; and so on. Which answers Question (1).

Q. E. F.

We may therefore, in answering the other three questions, consider the slip before it is folded, and calculate the positions of the vertices along its upper edge: and the problems thus become 'plane' ones.

(2) Gives us a right-angled Triangle, whose left-hand base-angle is 15°, and whose altitude is StartFraction StartRoot 3 EndRoot Over 2 EndFraction. We must calculate its base, and then, deducting half the base of the initial Triangle (i. e. deducting one half), we shall get the distance, measured along the upper edge of the slip, from the vertex of the initial Triangle to the vertex of the given Triangle; and from that we can calculate how many times we must go down and up the back-edge to reach it. Call the base of this right-angled Triangle 'x.' Then StartFraction StartRoot 3 EndRoot Over 2 EndFraction divided by x equals tangent 15 degree.

Now call tangent 15 degree 't'; then StartFraction 2 t Over left parenthesis 1 minus t squared right parenthesis EndFraction equals tangent 30 degree equals StartFraction 1 Over StartRoot 3 EndRoot EndFraction;

therefore 1 minus t squared equals 2 StartRoot 3 EndRoot period t semicolon t squared plus 2 StartRoot 3 EndRoot period t minus 1 equals 0 semicolon

therefore t equals StartFraction minus 2 StartRoot 3 EndRoot plus or minus 4 Over 2 EndFraction equals left parenthesis rejecting negative value right parenthesis 2 minus StartRoot 3 EndRoot period

therefore x equals StartFraction StartRoot 3 EndRoot Over 2 left parenthesis 2 minus StartRoot 3 EndRoot right parenthesis EndFraction equals StartFraction StartRoot 3 EndRoot Over 2 EndFraction left parenthesis 2 plus StartRoot 3 EndRoot right parenthesis equals StartRoot 3 EndRoot plus three halves period

Deducting one half, we get left parenthesis StartRoot 3 EndRoot plus 1 right parenthesis as the required distance.

Now StartRoot 3 EndRoot equals 1 dot 7 &c.; therefore distance equals 2 dot 7 &c.

Hence we must go down back-edge, up again, and then about dot 7 down again. This answers question (2).

(3) We need to go down the back-edge, and up again; i. e. we must use up the upward bases of 'alpha' and 'gamma'. Hence the base of the required right-angled Triangle is 2 and one half. Hence the required left-hand base-angle is

tangent Superscript negative 1 Baseline left parenthesis StartFraction StartRoot 3 EndRoot Over 2 EndFraction divided by five halves right parenthesis semicolon i period e period tangent Superscript negative 1 Baseline StartFraction StartRoot 3 EndRoot Over 5 EndFraction period

Hence, for the required base-angle, we have StartFraction sine Over cosine EndFraction equals StartFraction StartRoot 3 EndRoot Over 5 EndFraction;

StartLayout 1st Row 1st Column Blank 2nd Column therefore StartFraction sine Over StartRoot 3 EndRoot EndFraction equals StartFraction cosine Over 5 EndFraction equals StartFraction 1 Over StartRoot 28 EndRoot EndFraction semicolon therefore sine equals StartRoot three twenty eighths EndRoot equals StartFraction StartRoot 84 EndRoot Over 28 EndFraction comma 2nd Row 1st Column Blank 2nd Column equals StartFraction rather over 9 Over 28 EndFraction semicolon StartStartFraction 7 vertical bar ModifyingBelow 9 dot With quotation dash OverOver StartFraction 4 vertical bar ModifyingBelow 1 dot 28 ampersand c period With quotation dash Over dot 32 ampersand c period EndFraction EndEndFraction EndLayout

StartLayout 1st Row 1st Column Now left parenthesis by mem period tech period right parenthesis 2nd Column sine Superscript negative 1 Baseline dot 3 equals 17 dot 45 ampersand c period degree period 2nd Row 1st Column Blank 2nd Column sine Superscript negative 1 Baseline dot 4 equals 23 dot 57 ampersand c period degree period EndLayout

and the required angle is about one fifth of the way from one to the other. But the difference is almost exactly 6°. Hence we must add, to the lesser, about 1 and one fifth degrees, or 1 dot 20 degree. And the total will be about 18 dot 65 degree.

(4) Here the right-angled Triangle has, for its base, 3 and one half.

therefore the required base-angle has, for its tangent,

left parenthesis StartFraction StartRoot 3 EndRoot Over 2 EndFraction divided by seven halves right parenthesis semicolon i period e period StartFraction StartRoot 3 EndRoot Over 7 EndFraction semicolon

therefore StartFraction sine Over StartRoot 3 EndRoot EndFraction equals StartFraction cosine Over 7 EndFraction equals StartFraction 1 Over StartRoot 52 EndRoot EndFraction semicolon therefore sine equals StartRoot three fifty seconds EndRoot equals nearly StartRoot one seventeenth EndRoot,

= nearly StartFraction StartRoot 17 EndRoot Over 17 EndFraction. Now StartRoot 17 EndRoot equals 4 dot 12 ampersand c period therefore sine equals dot 24 ampersand c period

Now sine Superscript negative 1 Baseline dot 2 equals 11 dot 53 &c°.; and we must go about half-way to the next angle, viz. 17 dot 45 &c°. The difference is about 6°; therefore we must add about 3°. Hence the answer is about 14 dot 53°.

71. (18)

Let upper A upper B upper C be the given Triangle, and upper P the given Point.

Large triangle with vertices A (top left), B (bottom left), C (bottom right). Two horizontal transversals cross the interior: upper one through H, G and lower through F, E, N. Interior point P marks central intersections, with base points K, L, D, M on BC.

Bisect the sides of upper A upper B upper C at upper D, upper E, upper F; and join these Points.

First, let upper P be within the Triangle upper D upper E upper F.

Draw upper H upper G parallel to upper B upper C, so that its distance from upper B upper C may be double the distance of upper P from upper B upper C; join upper G upper P, upper H upper P, and produce them to meet upper B upper C in upper L, upper M. From upper L draw upper L upper K parallel to upper A upper C; join upper K upper P, and produce it to meet upper A upper C at upper N colon join upper M upper N.

Because upper H upper G is parallel to upper L upper M,

therefore upper G upper P equals upper P upper L comma and upper H upper P equals upper P upper M;

because upper K upper L is parallel to upper G upper N comma and that upper L upper P equals upper P upper G,

therefore upper K upper P equals upper P upper N semicolon therefore upper M upper N is parallel to upper H upper K.

Now the Triangles upper P upper G upper H comma upper P upper L upper M, are equal in all respects;

therefore upper G upper H equals upper L upper M period Similarly upper K upper L equals upper G upper N comma and upper M upper N equals upper H upper K.

If upper P lies on upper F upper E, upper H upper G and upper L upper M vanish, and the Hexagon becomes a Parallelogram.

If upper P lies at upper D, the Hexagon becomes the line upper B upper C.

If upper P lies outside the Triangle upper D upper E upper F, the Problem is insoluble.

Q. E. F.

72. (18, 27)

We know that, if a bag contained 3 counters, 2 being black and one white, the chance of drawing a black one would be two thirds; and that any other state of things would not give this chance.

Now the chances, that the given bag contains left parenthesis alpha right parenthesis upper B upper B, left parenthesis beta right parenthesis upper B upper W comma left parenthesis gamma right parenthesis upper W upper W comma are respectively one fourth comma one half comma one fourth.

Add a black counter.

Then the chances, that it contains left parenthesis alpha right parenthesis upper B upper B upper B comma left parenthesis beta right parenthesis upper B upper W upper B, left parenthesis gamma right parenthesis upper W upper W upper B, are, as before, one fourth, one half, one fourth.

Hence the chance, of now drawing a black one,

equals one fourth dot 1 plus one half dot two thirds plus one fourth dot one third equals two thirds period

Hence the bag now contains upper B upper B upper W (since any other state of things would not give this chance).

Hence, before the black counter was added, it contained upper B upper W, i. e. one black counter and one white.

Q. E. F.

THE END.

WORKS BY C. L. DODGSON.

PUBLISHED BY

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THE FORMULÆ OF PLANE TRIGONOMETRY, printed with symbols, instead of words, to express the goniometrical ratios. (First published in 1861.) Crown 4to, sewed. Price 1s.

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CURIOSA MATHEMATICA.

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TRANSCRIBER’S NOTES

For the sake of clarity, the notation used by the author for element of opening downwards and segment has been replaced with modern terminology.

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