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

Oxford
HORACE HART, PRINTER TO THE UNIVERSITY

[Pg vii]

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 ${\Large\unicode{0x2ad9}}$ and ${\Large\unicode{0x2313}}$ to represent the words ' ${\text{sine}}$ ' and ' ${\text{cosine}}$ '.

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

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

In this diagram, ${OP}$ being taken as the unit of length, ${PN}$ is the sine of the angle ${NOP}$ , and ${ON}$ its cosine.

In each of my two symbols I have retained the semicircle in the symbol ${\sin}$ , I have merely moved ${PN}$ to the middle; and, in the symbol ${\cos}$ , I have lengthened ${ON}$ , taking it a little beyond the curve, in order to avoid confusion with the existing symbol for ' ${\text{semicircle}}$ '.

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. ${b^{2} + c^{2} - a^{2}}$ 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 ${2 + 3 \gt 4}$ , they fail to fulfil the second, as ${2^{2} + 3^{2}}$ is not ${\gt 4^{2}}$ .

[Pg ix]

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,[Pg x] 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[Pg xi] of the Educational Times, for Problem 61, viz. "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."

"Let ${3m}$ , ${21m}$ , ${30m}$ be the three Numbers; then $3m + 21m + 30m = 3 \times 18m.$ Also ${\quad(3m)^{2} + (21m)^{2} + (30m)^{2} = (6m)^{2} + (15m)^{2} + (33m)^{2}}$

${\quad = (5m)^{2} + (13m)^{2} + (34m)^{2} = (10m)^{2} + (17m)^{2} + (31m)^{2}}$

Now, if we denote, by ' ${\alpha}$ ', the property "which cannot be arranged in ${A. P.}$ , 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 ' ${\alpha}$ '."

[Pg xii]

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 ${\cdot 7}$ 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[Pg xiii] 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[Pg xiv] 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[Pg xv] 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. ${\tfrac{6}{7}}$ .[Pg xvi] 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 ${A}$ , ${B}$ , ${C}$ , and ${P}$ : I merely thought of them by their positions (e. g. 'the foot of the perpendicular from ${P}$ on ${BC}$ ')."

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. ${\tfrac{3}{4}}$ , 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 ${2^\tfrac{3}{4}}$ . 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[Pg xviii] 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. 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 ' ${2bc . ca}$ ', instead of ' ${4bc . ca}$ '.

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

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

In No. 44, I said that the denominator would be of the form ${(10^{n} - 1) . 10^{m}}$ . This last factor is superfluous: i. e. ${m = 0}$ .

In No. 50, I made a mistake near the end, bringing out ${\tfrac{41}{108}}$ , instead of ${\tfrac{50}{108}}$ .

In No. 57, in the last paragraph, I replaced the denominator ' ${a \sin B \sin C}$ ' by (what I imagined to be its equivalent) ' ${2m}$ '. Apparently I was under the delusion that ' ${a \sin B \sin C}$ ' was the same thing as ' ${\sin A \,. bc}$ '!

In No. 70, section (3), I forgot to add in the ${\cdot 45}$ , thus making the answer half a degree wrong. And, in section (4), I forgot to add in the ${\cdot 53}$ , 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. ${\dfrac{abc}{3}\cdot\sqrt{\cos A \cos B \cos C}}$ .

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!

FOOTNOTES:

[Pg xx]

[1] In the trigonometrical Problems, I have used the symbols ${\Large\unicode{0x2ad9}}$ and ${\Large\unicode{0x2313}}$ , to represent the words 'sine' and 'cosine'.

[Pg xxi]

SUBJECTS CLASSIFIED.

CHAPTER I.

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.

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.

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.

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?

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

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.

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?

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[Pg 3] forming with the given Lines and the line joining their intersection to the given Point, two equal Triangles.

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 ${\dfrac{2}{6}}$ . What are the coins?

A Triangle ${ABC}$ has another ${A'B'C'}$ inscribed in it, so that ${\angle BA′C′ = \angle CB′A′ = \angle AC′B′ = \theta}$ ; thus making it similar to the first Triangle. Find ratio between homologous sides. And solve for " ${\theta = 90°}$ ".

The Triangles can be proved similar thus:— $\begin{array}{c} \angle C′A′B′ + \angle B′A′C = \text{supp. of}~ \theta,\\ \angle B′A′C + \angle A′CB′ = \text{supp. of}~ \theta;\\ \therefore \text{these pairs are equal};~ \therefore \angle C′A′B′ = C. \end{array}$

Let ${C′A′ = ka;~\therefore A′B′ = kb,~ \text{and}~ B′C′ = kc}$ . We have to find ${k}$ .

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.

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.

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

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?

In a given Triangle place a line parallel to the base, such that if, from its ends, lines be drawn, parallel to the[Pg 5] sides and terminated by the base, they shall be together equal to the first line.

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.

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?

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.

Sum, (1) to ${n}$ terms, (2) to 100 terms, the series $1\,.\,3\,.\,5 + 2\,.\,4\,.\,6 + \&c.$

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

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[Pg 6] 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?

If, from the vertices of a triangle ${ABC}$ , the lines ${AD}$ , ${BE}$ , ${CF}$ be drawn, intersecting at ${O}$ : find the ratio ${\dfrac{DO}{DA}}$ in terms of the two ratios ${\dfrac{EO}{EB}}$ , ${\dfrac{FO}{FC}}$ .

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

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.

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[Pg 7] 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?

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.

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.

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.

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[Pg 8] uniformly during any one day: the clock is always going, and goes uniformly.

Sum the Series ${1\,.\,5 + 2\,.\,6 +\&c.}$ (1) to ${n}$ terms; (2) to 100 terms.

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

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.

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.

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.

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?

There are 3 bags, ' ${A}$ ', ' ${B}$ ', and ' ${C}$ '. ' ${A}$ ' contains 3 red counters, ' ${B}$ ' 2 red and one white, ' ${C}$ ' 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?

${A}$ and ${B}$ begin, at 6 a. m. on the same day, to walk along a road in the same direction, ${B}$ having a start of 14 miles, and each walking from 6 a. m. to 6 p. m. daily. ${A}$ walks 10 miles, at a uniform pace, the first day, 9 the second, 8 the third, and so on: ${B}$ 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?

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

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[Pg 10] "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."

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.

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.

If ${a}$ , ${b}$ be two numbers prime to each other, a value may be found for ${n}$ which will make ${(a^{n} - 1)}$ a multiple of ${b}$ .

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

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.

$\left. \begin{array}{c} \frac{x}{y} = x - z;\\ \frac{x}{z} = x - y;\\ \end{array} \right\} \left. \begin{array}{c} \qquad \text{(1)}\\ \qquad \text{(2)}\\ \end{array} \right.$ and find the limits, if any, between which the real values lie.

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 $(\frac{\beta\gamma}{\alpha} + \frac{\gamma\alpha}{\beta} + \frac{\alpha\beta}{\gamma})$ is equal to the semiperimeter of the Triangle.

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.

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

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.

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!

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.

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.

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

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;

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

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.

Given a Triangle ${ABC}$ , and that its base ${BC}$ is divided at ${D}$ in the ratio ${m}$ to ${n}$ : find the angles ${BAD}$ , ${CAD}$ .

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.

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.

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.

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.

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

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 ${\dfrac{\alpha}{\alpha + \beta}}$ . 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?

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.

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[Pg 17] the wine taken out during the day, had come to the same amount every time); but of course only the Salesman knew this.

If, from each of the angles of a given Triangle ${ABC}$ , 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 ${B}$ and ${C}$ , being equal to ${A}$ , 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.

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.

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.

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.

FOOTNOTES:

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

[Pg 19]

CHAPTER II.

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

${\quad \cos A= \dfrac{5\alpha^{2}-\beta^{2}-\gamma^{2}}{2\sqrt{2\alpha^{2}-\beta^{2}+2\gamma^{2}}.{\sqrt{2\alpha^{2}+2\beta^{2}-\gamma^{2}}}}}$ .

Let ${AB}$ , ${AD}$ be given sides, and ${B}$ , ${D}$ the right ${\angle}$ s; and let ${AB = b}$ , ${AD = d}$ .

(1) ${\quad BC = \dfrac{d-b\cos A}{\sin A};\quad CD = \dfrac{b-d\cos A}{\sin A};}$

The required ratio is equal to $\frac{\sin A \sin B \sin C}{\sin \theta (1 + \cos A \cos B \cos C) + \cos \theta \sin A \sin B \sin C}.$

If ${\theta = 90°}$ , this = ${\dfrac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C}.}$

If ${s}$ = semi-perimeter, ${m}$ = area, ${v}$ = volume; then $a^{2} + b^{2} + c^{2} = 2.(s^{2} - \frac{v}{s} - \frac{m^{2}}{s^{2}}).$

If ' ${2M}$ ' = 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 $= \frac{32M^{3}}{(b^{2} + c^{2} - a^{2}).(c^{2} + a^{2} - b^{2})}.$

The first course gives chance = ${\tfrac{1}{2}}$ ; the second, ${\tfrac{5}{12}}$ . Hence the first is best.

(1) Divide base ${BC}$ , at ${E}$ , so that ${\dfrac{BE}{EC} = \dfrac{\sin 2C}{\sin 2B}}$ .

(2) At ${B}$ , ${C}$ , make right angles ${ABD}$ , ${ACD}$ ; and join ${AD}$ cutting ${BC}$ at ${E}$ , which is the required Point.

(1) ${\dfrac{n . \overline{n + 1} . \overline{n + 4} . \overline{n + 5}}{4}}$ ; ${\quad\text{(2)}\quad 27,573,000}$ .

Calling the given altitudes ' ${\alpha}$ , ${\beta}$ , ${\gamma}$ '; and the fraction $\frac{2\alpha^{2} \beta^{2} \gamma^{2} . (\alpha^{2} + \beta^{2} + \gamma^{2}) - (\beta^{4} \gamma^{4} + \gamma^{4} \alpha^{4} + \alpha^{4} \beta^{4})}{4\alpha^{4} \beta^{4} \gamma^{4}}$ \unicode{x2018} $k^2$ \unicode{x2019},

${\dfrac{DO}{DA} + \dfrac{EO}{EB} + \dfrac{FO}{FC} = 1}$ ; whence any one can be found in terms of the other two.

$\begin{align} (1)\quad &\dfrac{n.(n+1).(2n+13)}{6};\\ (2)\quad &358550.\\ \end{align}$

$\dfrac{4 + \sqrt{3}}{12} - \dfrac{1 + \sqrt{3}}{2\pi};\quad \text{i.e.}\,\, \text{about}\,\, \cdot 044$ .

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

And the 4th set is ${x = \dfrac{k^{2}}{k - 1}}$ , ${y = z = k}$ ; where ${k}$ has any value.

The portion, cut off from the second side, is equal to $\frac{(\alpha\sin C+\gamma\sin A)(2\gamma\cos A+\beta)}{\alpha\cos C+\gamma\cos A+\beta}+\frac{\beta\cos A+\gamma\cos 2A}{\sin A}.$

Side ${AB}$ must be divided at ${D}$ , ${G}$ , so that $AD :\, DG :\, GB\, \colon\colon\, \frac{1}{a} :\, \frac{1}{c} :\, \frac{1}{b};$ and similarly for the other sides. Also each side of the Hexagon = ${\dfrac{1}{\tfrac{1}{a} + \tfrac{1}{b} + \tfrac{1}{c}}}$ .

Draw ${BC}$ , ${CE}$ , ${BD}$ equal to the given altitudes, so as to form right ${\angle}$ s at ${B}$ and ${C}$ : and produce ${DB}$ , ${EC}$ . Join ${DC}$ , and draw ${CF \perp}$ to it. Join ${EB}$ , and draw ${BG \perp}$ to it. With centre ${B}$ , and distance ${BF}$ , describe a circle: with centre ${C}$ , and distance ${CG}$ describe another: let them meet at ${A}$ : and join ${AB}$ , ${AC}$ . Triangle ${ABC}$ may be proved to be similar to required Triangle. The rest of the construction is obvious.

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.

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 $\frac{a}{x} = \frac{b}{y} = \frac{c}{z} = \frac{a^{2}+b^{2}+c^{2}}{2m} + 1.$

Calling lengths of the 3 pairs of edges ' ${a}$ , ${b}$ , ${c}$ ', and the corresponding ${\angle}$ s, in each facet, ' ${A}$ , ${B}$ , ${C}$ '; volume = $\frac{abc}{6}.\sqrt{1 - (\cos^{2}A + \cos^{2}B + \cos^{2}C) + 2\cos A\cos B\cos C.}$

If each side of each Square = 2, the volume = $\frac{8.2^{\tfrac{1}{4}}.(\sqrt{2}+1)}{3}.$

If the centre of the horizontal facet be taken as the Origin, and if the ${X}$ -axis pass through one of the vertices of that facet, and the ${Y}$ -axis be parallel to the opposite edge of that facet, and the ${Z}$ -axis be perpendicular to that facet: and if the altitude (measured downwards) of the Tetrahedron be called ' ${h}$ ', and the intercept on the ${X}$ -axis be called ' ${a}$ ': the Equations to the Locus are $\begin{aligned} (x+\sqrt{3}.y).(h-z) &= ah; \\ x^{2} + y^{2} &= a^{2}. \end{aligned}$

(1) ${k = \dfrac{\theta-B}{A}}$ ; ${l = \dfrac{\theta-C}{B}}$ ; ${m = \dfrac{\theta-A}{C}}$ .

(2) Calling new Triangle ' ${A'B'C'}$ ', $\frac{a'}{a} = \frac{b'}{b} = \frac{c'}{c} = 2\cos{\theta}.$

(1) Down the back-edge; up again; and so on. (2) about ${\cdot 7}$ of the way down the back-edge. (3) About ${18 \cdot 65°}$ . (4) About ${14 \cdot 53°}$ .

[Pg 28]

CHAPTER III.

Also, if we write ' ${2m^{2}}$ ' for '2' (which will not interfere with the problem, as we can divide by ${m^{2}}$ , and get ${\dfrac{u^{2}}{m^{2}} + \dfrac{v^{2}}{m^{2}} = 2}$ ), the above form becomes ' ${(m^{2}+k), (m^{2}-k)}$ '.

Now, as these are squares, their resemblance to $\unicode{x2018}(a^{2} + b^{2} + 2ab), (a^{2} + b^{2} - 2ab)\unicode{x2019}$ at once suggests itself; so that the problem depends on the known one of finding ${a}$ , ${b}$ , such that ${(a^{2} + b^{2})}$ is a square; and we can then take ${2ab}$ as ${k}$ .

A general form for this is $\begin{array}{c} \begin{aligned} a &= x^{2}-y^{2},\\ b &= 2xy;\\ \therefore a^{2}+b^{2} &= (x^{2}+y^{2})^{2};\\ \end{aligned}\\ \therefore \text{the formula}\,\unicode{x2018}u^{2} + v^{2} = 2 m^{2}\unicode{x2019}\,\text{becomes}\\ (x^{2}-y^{2}+2xy)^{2} + (x^{2}-y^{2}-2xy)^{2} = 2(x^{2}+y^{2})^{2};\\ \text{i. e. } \large(\dfrac{x^{2}-y^{2}+2xy}{x^{2}+y^{2}})^{2} + \large(\dfrac{x^{2}-y^{2}-2xy}{x^{2}+y^{2}})^{2}=2. \end{array}$

Let ${ABC}$ be the Triangle, and ${DE}$ the required line, so that ${BD + CE = BC}$ .

${\therefore \angle \text{s} \quad BDE, \quad CED}$ , are bisected by ${DF}$ , ${EF}$ , and ${F}$ is centre of ${\odot}$ escribed to ${\triangle ADE}$ .

Drop, from ${F}$ , ${\perp}$ s on ${BD}$ , ${DE}$ , ${EC}$ ; then these ${\perp}$ s are equal.

Bisect ${\angle A}$ by ${AF}$ : from ${F}$ draw ${FB'}$ , ${FC'}$ , ${\perp AC}$ , ${AB}$ : also draw ${FA' \perp BC}$ and equal to ${FB'}$ : and through ${A'}$ draw ${DE \perp FA'}$ , i. e. ${\parallel BC}$ . Then ${DE}$ shall be line required.

${\because \angle}$ s at ${A'}$ , ${B'}$ , ${C'}$ , are right, and ${FA' = FB'= FC'}$ ,

Now ${\angle BFD = \angle FDA'}$ ; ${\therefore}$ it ${= \angle BDF}$ ; ${\therefore BF = BD}$ ;

Let ${ABCD}$ be the Tetragon; and let the 3 sides, ${AB}$ , ${BC}$ , ${CD}$ , be bisected by vertices of the Parallelogram ${EFGH}$ .

${\because}$ , in Triangle ${BCD}$ , sides ${BC}$ , ${CD}$ are bisected at ${F}$ and ${G}$ ,

Let ${ABC}$ be the given Triangle, and ${A'B'C'}$ the required Triangle, so that ${\angle BA'C' = \angle CA'B'}$ , &c.

Evidently ${A'C'}$ , ${A'B'}$ are equally inclined to a line drawn,[Pg 31] from ${A'}$ , ${\perp BC}$ ; and so of the others: i. e. these ${\perp}$ s bisect the ${\angle}$ s at ${A'}$ , ${B'}$ , ${C'}$ ;

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

${\therefore}$ Triangle ${BC'A'}$ is similar to Triangle ${BCA}$ ; and so of the others; $\begin{aligned} \therefore BA' & = \frac{c}{a} \cdot BC' = \frac{c}{a} \cdot (c - AC'),\\ & = \frac{c}{a} \cdot (c - \frac{b}{c} \cdot AB'),\\ & = \frac{c^{2}}{a} - \frac{b}{a} \cdot (b - CB'),\\ & = \frac{c^{2}}{a} - \frac{b^{2}}{a} + \frac{b}{a} \cdot \frac{a}{b} \cdot CA',\\ & = \frac{c^{2}}{a} - \frac{b^{2}}{a} + a - BA';\\ \therefore 2BA' & = \frac{c^{2}+a^{2}-b^{2}}{a} = \frac{2ca\cos {B}}{a};\\ \therefore BA' & = c\cos {B}; \end{aligned}$

${\therefore A'}$ is foot of ${\perp}$ drawn, from ${A}$ , to ${BC}$ . Hence the construction is obvious.

At first sight, it would appear that, as the state of the bag, after the operation, is necessarily identical with its state[Pg 32] before it, the chance is just what it then was, viz. ${\tfrac{1}{2}}$ . This, however, is an error.

The chances, before the addition, that the bag contains (a) 1 white (b) 1 black, are (a) ${\tfrac{1}{2}}$ (b) ${\tfrac{1}{2}}$ . Hence the chances, after the addition, that it contains (a) 2 white (b) 1 white, 1 black, are the same, viz. (a) ${\tfrac{1}{2}}$ (b) ${\tfrac{1}{2}}$ . Now the probabilities, which these 2 states give to the observed event, of drawing a white counter, are (a) certainty (b) ${\tfrac{1}{2}}$ . 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) ${\tfrac{1}{2} \cdot 1}$ (b) ${\tfrac{1}{2} \cdot \tfrac{1}{2}}$ ; i. e. (a) ${\tfrac{1}{2}}$ (b) ${\tfrac{1}{4}}$ ; i. e. (a) 2 (b) 1. Hence the chances are (a) ${\tfrac{2}{3}}$ (b) ${\tfrac{1}{3}}$ . Hence, after the removal of a white counter, the chances, that the bag now contains (a) 1 white (b) 1 black, are for (a) ${\tfrac{2}{3}}$ and for (b) ${\tfrac{1}{3}}$ .

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

$\begin{aligned} \therefore &\frac{\alpha^{2} + a^{2} - 4c^{2}}{2\alpha a} + \frac{\alpha^{2} + a^{2} - 4b^{2}}{2\alpha a} = 0;\\ \therefore 2\alpha^{2} &+ 2a^{2} - 4b^{2} - 4c^{2} = 0;\\ \therefore \alpha^{2}& = -a^{2} + 2b^{2} + 2c^{2}.\\ \text{Similarly}, \quad\beta^{2} &= 2a^{2} - b^{2} + 2c^{2};\\ \gamma^{2}& = 2a^{2} + 2b^{2} - c^{2}. \end{aligned}$

To eliminate ${b}$ , ${c}$ , let us multiply by ${k}$ , ${l}$ , ${m}$ , so taken that $\begin{aligned} & 2k - l + 2m = 0,\\ \text{and}\quad & 2k + 2l - m = 0;\\ \therefore \quad & 3(l - m) = 0;~ \text{i. e.}\, l = m;\\ \therefore \quad & 2k = -l = -m; \end{aligned}$ hence we may make ${k = -1, \quad l = 2, \quad m = 2}$ ; $\begin{align} & \quad \therefore \quad - \alpha^{2} + 2\beta^{2} + 2\gamma^{2} = 9a^{2};\\ & \quad \text{i. e.}\quad a^{2} = \frac{- \alpha^{2} + 2\beta^{2} + 2\gamma^{2}}{9};\\ & \quad \therefore \quad BC ~(\text{which} = 2a) = \tfrac{2}{3}\sqrt{- \alpha^{2} + 2\beta^{2} + 2\gamma^{2}}, \&c., \end{align}$ which gives lengths of sides.

$\begin{array}{l} \text{Also} \quad \cos A & = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}\\ & = \dfrac{2\alpha^{2} - \beta^{2} + 2\gamma^{2} + 2\alpha^{2} + 2\beta^{2} - \gamma^{2} + \alpha^{2} - 2\beta^{2} - 2\gamma^{2}}{2 . \sqrt{2\alpha^{2} - \beta^{2} + 2\gamma^{2}} . \sqrt{2\alpha^{2} + 2\beta^{2} - \gamma^{2}}}\\ & = \dfrac{5\alpha^{2} - \beta^{2} - \gamma^{2}}{\text{den.}};\quad \text{and so for other angles.} \end{array}$

Let ${AB}$ , ${AD}$ be given sides, and ${B}$ , ${D}$ the right ${\angle}$ s; and let ${AB = b}$ , ${AD = d}$ .

$\begin{aligned} \text{Now}\quad AE &= AD.\sec A = d\sec A;\\ \therefore \quad BE &= d\sec A - b.\\ \text{Also} \quad BC & = BE.\tan E = (d\sec A - b).\cot A,\\ & = \dfrac{d - b\cos A}{\sin A}; \end{aligned}$ similarly, ${CD = \dfrac{b-d\cos {A}}{\sin {A}}}$ ; which answers (1).

$\begin{aligned} \text{Also area} &= \tfrac{1}{2} \cdot (AB.BC + AD.DC),\\ &= \tfrac{1}{2} \cdot \frac{b.(d - b\cos {A}) + d.(b-d\cos {A})}{\sin {A}},\\ &= \frac{2bd - (b^{2} + d^{2})\cos {A}}{2\sin {A}}; \quad \text{which answers (2)}. \end{aligned}$

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 ${mk}$ shillings. Hence the thing ends when the last man is again called on to hand on the heap, which then contains ${(mk + m - 1)}$ shillings, the penultimate man now having nothing, and the first man having ${(m - 2)}$ 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 $\begin{align} mk + m - 1 &= 4(m-2),\\ \text{or else}\quad 4(mk + m - 1) &= m - 2. \end{align}$

The first equation gives ${mk = 3m - 7}$ , i. e. ${k = 3 - \dfrac{7}{m}}$ , which evidently gives no integral values other than ${m = 7}$ , ${k = 2}$ .

The second gives ${4mk = 2 - 3m}$ , which evidently gives no positive integral values.

Let ${AB}$ , ${AC}$ , be the given Lines, and ${P}$ the given Point; and join ${AP}$ .

Through ${A}$ draw ${EAF}$ , ${\perp AP}$ , and bisected at ${A}$ ; from ${E}$ , ${F}$ , draw ${EG}$ , ${FH}$ , parallel to ${AP}$ , and meeting ${AB}$ , ${AC}$ , at ${G}$ , ${H}$ ; join ${GH}$ , and on it describe a semicircle cutting ${AP}$ at ${K}$ ; and join ${KG}$ , ${KH}$ . Then ${\angle GKH}$ is a right angle. From ${P}$ draw ${PL}$ , ${PM}$ , parallel to ${KG}$ , ${KH}$ .

Now Triangle ${APL}$ has, to Triangle ${AKG}$ , the duplicate ratio of ${AP}$ to ${AK}$ ;

also Triangles ${AKG}$ , ${AKH}$ , are equal, being on the same base ${AK}$ , and having equal altitudes ${AE}$ , ${AF}$ ;

${\therefore}$ Triangles ${APL}$ , ${APM}$ are equal: and ${\angle LPM}$ is evidently equal to ${\angle GKH}$ ; ${\therefore}$ it is a right angle.

The chance, that the pocket contains 2 balls, is ${\tfrac{2}{3}}$ ; and, if it does, the 'expectation' is the average value of $(y + z), (z + x), (x + y); \text{ i.e. it is } \frac{2s}{3}.$

Also the chance, that it contains only one, is ${\dfrac{1}{3}}$ ; and, if it does, the 'expectation' is ${\dfrac{s}{3}}$ .

Hence total 'expectation' ${= \dfrac{4s}{9} + \dfrac{s}{9} = \dfrac{5s}{9}}$ . $\therefore~\frac{5s}{9} = 30d.;\quad\therefore~s = 54d.= 4/6.$

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

Now ${\quad\dfrac{BA'}{A'C'} = \dfrac{\sin (B + \theta)}{\sin B};\quad\text{and}~\dfrac{A'C}{A'B'} = \dfrac{\sin \theta}{\sin C}.}$

${\therefore~BA = \dfrac{\sin (B + \theta)}{\sin B} . ka;\quad\text{and}~A'C = \dfrac{\sin \theta}{\sin C} . kb}$

${= \dfrac{a}{\dfrac{a . \sin (B + \theta)}{\sin B} + \dfrac{b\sin \theta}{\sin C}} = \dfrac{\sin A}{\dfrac{\sin A\sin (B + \theta)}{\sin B} + \dfrac{\sin B\sin \theta}{\sin C}}}$

$\begin{aligned} = &\frac{\sin A\sin B\sin C}{\sin A\sin (B+\theta)\sin C+\sin ^{2} B\sin \theta}\\ = &\frac{\sin A\sin B\sin C}{\sin A\sin C(\sin B\cos \theta+\cos B\sin \theta) + \left(1-\cos ^{2} B\right)\sin \theta}\\ = &{\sin A\sin B\sin C}\over\displaystyle{\left. \begin{array}{r}{\sin \theta + \sin \theta(\sin A\sin C\cos B - \cos {^{2}}B) \atop\quad {}+ \cos \theta\sin A\sin B\sin C} \end{array}\right\}}\\ = &{\sin A\sin B\sin C}\over\displaystyle{\left. \begin{array}{r}{\sin \theta + \sin \theta\cos B(\sin A\sin C + \cos (A+C)) \atop\quad{}+ \cos \theta\sin A\sin B\sin C} \end{array}\right\}}\\ = &\frac{\sin A\sin B\sin C}{\sin \theta(1+\cos A\cos B\cos C)+\cos \theta\sin A\sin B\sin C}. \end{aligned}$

Cor. Let ${\theta = 90°}$ ; then ${k = \dfrac{\sin A\sin B\sin C}{1+\cos A\cos B\cos C}}$ .

We know that ${m = \sqrt{s . (s-a) . (s-b) . (s-c)}}$ ; $\begin{array}{l} \therefore m^{2} &= s . (s-a) . (s-b) . (s-c);\\ \therefore \frac{m^{2}}{s} &= s^{3} - s^{2} . (a + b + c) + s . (bc + ca + ab) - abc;\\ &= s^{3} - 2s^{3} + s . (bc + ca + ab) - v;\\ \therefore &\frac{m^{2}}{s^{2}} + \frac{v}{s} + s^{2} = bc + ca + ab;\\ \therefore &2.\left(\frac{m^{2}}{s^{2}} + \frac{v}{s} + s^{2}\right) = (a + b + c)^{2} - (a^{2} + b^{2} + c^{2});\\ &=4s^{2} - (a^{2} + b^{2} + c^{2});\\ \therefore\quad &a^{2} + b^{2} + c^{2} = 2.(s^{2} -\frac{v}{s} - \frac{m^{2}}{s^{2}}). \end{array}$

Let ${A}$ , ${B}$ , be the centres of the Circles; ${C}$ , ${D}$ , their points of intersection; and ${CFDE}$ the Tetragon whose area is required.

Let the sides of the Triangle ${ABC}$ be ${a}$ , ${b}$ , ${c}$ ; and its ${\angle}$ s ${\alpha}$ , ${\beta}$ , ${\gamma}$ .

Hence area of Triangle ${FCE =\tfrac{1}{2}.ab.\tan{\alpha}.\tan{\beta}.\sin{\gamma}}$ ;

${\therefore}$ area of Tetragon = ${\dfrac{ab\sin{\alpha}\sin{\beta}\sin{\gamma}}{\cos{\alpha}\cos{\beta}}}$ .

$\begin{aligned} \therefore \text{area of Tetragon} &= ab \cdot \frac{8M^{3}}{a^{2} b^{2} c^{2}} \cdot \frac{4bc.ca}{(b^{2} + c^{2} - a^{2})(c^{2} + a^{2} - b^{2})};\\ &=\frac{32M^{3}}{(b^{2}+c^{2}-a^{2}).(c^{2}+a^{2}-b^{2})}. \end{aligned}$

This simply expresses the identity $\begin{array}{l} & 3(a^{2}+b^{2}+c^{2})\\ & = (a+b+c)^{2}+(b^{2}-2bc+c^{2})+(c^{2}-2ca+a^{2})+(a^{2}-2ab+b^{2});\\ & = (a+b+c)^{2}+(b-c)^{2}+(c-a)^{2}+(a-b)^{2}.\\ \end{array}$

$\begin{align} 3(1^{2}+2^{2}+3^{2})&= 6^{2}+1^{2}+2^{2}+1^{2}.\\ 3(1^{2}+3^{2}+7^{2})&= 11^{2}+4^{2}+6^{2}+2^{2}. \end{align}$

Let ${ABCD}$ be an inscribed Tetragon. Join ${AC}$ : and about Triangle ${ACD}$ describe a Circle.

Now, if this Circle does not pass through ${B}$ , let it cut ${CB}$ , or ${CB}$ produced, in ${B′}$ or ${B″}$ . Join ${AB′}$ , ${AB″}$ .

Then ${\angle AB′C}$ , or ${\angle AB″C}$ , is supplementary to ${\angle ADC}$ ;

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 ${AC}$ as the Point ${D}$ .

The 'a priori' chances of possible states of first bag are ' ${W}$ , ${\tfrac{1}{2}}$ ; ${B}$ , ${\tfrac{1}{2}}$ '. Hence chances, after putting ${W}$ in, are ' ${WW}$ , ${\tfrac{1}{2}}$ ; ${WB}$ , ${\tfrac{1}{2}}$ '. The chances, which these give to the 'observed event', are 1, ${\tfrac{1}{2}}$ . Hence chances of possible states ' ${W}$ , ${B}$ ', after the event, are proportional to 1, ${\tfrac{1}{2}}$ ; i. e. to 2, 1; i. e. their actual values are ${\tfrac{2}{3}}$ , ${\tfrac{1}{3}}$ .

Now, in first course, chance of drawing ${W}$ is ${\tfrac{1}{2} \cdot \tfrac{2}{3} + \tfrac{1}{2} \cdot \tfrac{1}{3}}$ ; i. e. ${\tfrac{1}{2}}$ .

And, in second course, chances of possible states ' ${WWBB}$ , ${WBBB}$ ' are ${\tfrac{2}{3}}$ , ${\tfrac{1}{3}}$ : hence chance of drawing ${W}$ is ${\tfrac{2}{3} \cdot \tfrac{1}{2} + \tfrac{1}{3} \cdot \tfrac{1}{4}}$ ; i. e. ${\tfrac{5}{12}}$ .

From ${D}$ , ${E,}$ draw ${DF}$ , ${EG}$ , parallel to the sides. Then ${DF + EG = DE}$ .

Bisect ${\angle}$ s ${B}$ , ${C}$ , by ${BH}$ , ${CH}$ , meeting at ${H}$ ; through ${H}$ draw ${DE}$ parallel to ${BC}$ ; and from ${D}$ , ${E}$ , draw ${DF}$ , ${EG}$ , parallel to ${AC}$ , ${AB}$ .

(1) Call required Point ${E}$ . From ${E}$ draw ${EF}$ , ${FG \perp}$ sides. Join ${FG}$ . From ${F}$ , ${G}$ , draw ${FH}$ , ${GK \perp BC}$ . Call ${BE}$ ' ${x}$ ', and ${EC}$ ' ${y}$ '.

$\begin{array}{l} \text{Now}\, FH \text{ must } = GK.\\ \text{Also } EF = x\sin B; \text{ and } FH & = EF\sin FEH,\\ & = EF\cos B,\\ & = x\sin B\cos B;\\ \qquad\qquad\qquad\text{similarly, } GK & = y\sin C\cos C. \end{array}$ $\begin{aligned} \text{But } FH = GK; & \therefore~ x\sin B\cos B = y\sin C\cos C;\\ & \therefore~ \frac{x}{y} = \frac{\sin 2C}{\sin 2B}. \end{aligned}$

(2) At ${B}$ , ${C}$ , make right angles ${ABD}$ , ${ACD}$ ; and join ${AD}$ , cutting ${BC}$ at ${E}$ . From ${E}$ draw ${EF}$ , ${EG\, \perp}$ sides; and join ${FG}$ .

${\because BD}$ , ${FE}$ are ${\perp AB, \quad \therefore}$ they are ${\parallel}$ ; ${\therefore AF : FB \colon\colon AE : ED}$ ;

${\because CD}$ , ${GE}$ are ${\perp AC, \quad \therefore}$ they are ${\parallel}$ ; ${\therefore AG : GC \colon\colon AE : ED}$ ;

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

The chances of the orders ${ABC}$ , ${ACB}$ , ${BAC}$ , ${BCA}$ , ${CAB}$ , ${CBA}$ , are, a priori, ${\tfrac{1}{6}}$ 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:— $\begin{aligned} \text{for} \quad & ABC, \quad \tfrac{1}{2} \times \tfrac{1}{3}; \quad \text{i. e.} \quad \tfrac{1}{6}.\\ & ACB, \quad \tfrac{1}{2} \times \tfrac{1}{4}; \quad \text{i. e.} \quad \tfrac{1}{8}.\\ & BAC, \quad \tfrac{2}{3} \times \tfrac{1}{2}; \quad \text{i. e.} \quad \tfrac{1}{3}.\\ & BCA, \quad \tfrac{2}{3} \times \tfrac{1}{4}; \quad \text{i. e.} \quad \tfrac{1}{6}.\\ & CAB, \quad \tfrac{3}{4} \times \tfrac{1}{2}; \quad \text{i. e.} \quad \tfrac{3}{8}.\\ & CBA, \quad \tfrac{3}{4} \times \tfrac{1}{3}; \quad \text{i. e.} \quad \tfrac{1}{4}. \end{aligned}$

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 $\begin{aligned} & \tfrac{1}{34} · \{4× \tfrac{3}{4} +3× \tfrac{2}{3} +8× \tfrac{3}{4} +4× \tfrac{1}{2} +9× \tfrac{2}{3} +6× \tfrac{1}{2}\};\\ \text{i. e.}\quad & \tfrac{1}{34} ×\{3+2+6+2+6+3\};\quad \text{i. e.}\, \tfrac{22}{34};\quad \text{i. e.}\, \tfrac{11}{17}. \end{aligned}$

Let ${ABC}$ be the given Triangle, and ${P}$ the required Point. Draw ${PQ\perp BC}$ , and ${PR\perp AB}$ . Then ${PQ = PR}$ .

$\begin{array}{l} \colon\colon {\Large\frac{AD}{AB}} : {\Large\frac{AD}{DC}},\\ \colon\colon DC:AB. \end{array}$

From ${A}$ draw ${AD\perp BC}$ . Produce ${BA}$ to ${E}$ , making ${AE}$ equal to ${DC}$ . Join ${EC}$ . From ${A}$ draw ${AP}$ parallel to ${EC}$ ; and from ${P}$ draw ${PQ\perp BC}$ , and ${PR\perp AB}$ . $\begin{aligned} \text{Then}~ \frac{PQ}{PC} = \frac{AD}{DC} & = \frac{AD}{AB} · \frac{AB}{DC},\\ & = \frac{PR}{PB} · \frac{AB}{AE},\\ & = \frac{PR}{PB} · \frac{PB}{PC} = \frac{PR}{PC};\\ \therefore PQ & = PR. \end{aligned}$

${\therefore \quad}$ the ${(n+ 1)}$ th term is ${\overline{n+1}.\overline{n+3}.\overline{n+5}}$ ;

${= \overline{n+1}.\overline{n+2}.\overline{n+5} + \overline{n+1}.\overline{n+5}}$

${= \overline{n+1}.\overline{n+2}.(\overline{n+3} + 2)+\overline{n+1}.(\overline{n+2} + 3);}$

${= \overline{n+1}.\overline{n+2}.\overline{n+3} + 2.\overline{n+1}.\overline{n+2} + \overline{n+1}.\overline{n+2} + 3.\overline{n+1};}$

${= \overline{n+1}.\overline{n+2}.\overline{n+3} + 3.\overline{n+1}.\overline{n+2} + 3.\overline{n+1}.}$

${\therefore S = \dfrac{n.\overline{n+1}.\overline{n+2}.\overline{n+3}}{4} + n.\overline{n+1}.\overline{n+2} + \dfrac{3}{2}.n.\overline{n+1}+C;}$

${\therefore S = n.\overline{n+1}.\left(\dfrac{n^{2}+5n+6}{4} + n + 2 + \tfrac{3}{2}\right);}$

${\quad\quad= n.\overline{n+1}.\dfrac{n^{2}+9n+20}{4} = \dfrac{n.\overline{n+1}.\overline{n+4}.\overline{n+5}}{4}.}$

${\therefore \dfrac{\sin A}{\beta\gamma} = \dfrac{\sin B}{\gamma\alpha} = \dfrac{\sin C}{\alpha\beta} = k}$ (say);

${\therefore \sin A = k\beta\gamma,\quad \sin B = k\gamma\alpha,\quad \sin C = k\alpha\beta}$ .

${\therefore \sin ^{2}A (1-\sin ^{2}B) = \sin ^{2}C + \sin ^{2}B (1-\sin ^{2}A) -2\sin C\cos A\sin B}$ ;

${\therefore \sin ^{2}A -\sin ^{2}A\sin ^{2}B = \sin ^{2}C+ \sin ^{2}B - \sin ^{2}A\sin ^{2}B -2\sin B\sin C\cos A}$ ;

${\therefore, \text{squaring},\quad (\sin ^{4}A + {\&c.}) -2\sin ^{2}A\sin ^{2}B -2\sin ^{2}A\sin ^{2}C +2\sin ^{2}B\sin ^{2}C = 4\sin ^{2}B\sin ^{2}C(1-\sin ^{2}A)}$ ;

${\therefore (\sin ^{4}A + {\&c.}) - 2(\sin ^{2}B\sin ^{2}C + {\&c.}) +4\sin ^{2}A\sin ^{2}B\sin ^{2}C =0}$ ;

${\therefore}$ , substituting for ${\sin A}$ , &c., and dividing by ${k^{4}}$ , [Pg 46] ${(\beta^{4}\gamma^{4} + {\&c.}) -2\alpha^{2}\beta^{2}\gamma^{2}.(\alpha^{2} + {\&c.}) +4k^{2}\alpha^{4}\beta^{4}\gamma^{4} =0}$ ;

${\therefore \quad k^{2} = \dfrac{2\alpha^{2}\beta^{2}\gamma^{2}(\alpha^{2} + \&c.) - (\beta^{4}\gamma^{4} + {\&c.})}{4\alpha^{4}\beta^{4}\gamma^{4}}}$ .

${\therefore \quad a = \dfrac{\gamma}{\sin {B}} = \dfrac{\gamma}{k\gamma\alpha} = \dfrac{1}{k\alpha}}$ , \text{&c.}; which answers (1).

Also area ${= \dfrac{bc\sin {A}}{2} = \dfrac{1}{2} \cdot \dfrac{1}{k\beta} \cdot \dfrac{1}{k\gamma} \cdot k\beta\gamma = \dfrac{1}{2k}}$ ;

The original chances, as to states of bag, are $\begin{align} \text{for}\quad &2 W &\tfrac{1}{4};\\ &1 W, 1 B &\tfrac{1}{2};\\ &2 B &\tfrac{1}{4}. \end{align}$ ${\therefore}$ the chances, after adding 2 W and 1 B, are $\begin{align} \text{for}\quad &4 W, 1 B &\tfrac{1}{4};\\ &3 W, 2 B &\tfrac{1}{2};\\ &2 W, 3 B &\tfrac{1}{4}. \end{align}$

Now the chances, which these give to the observed event, drawing 2 W and 1 B, are ${\tfrac{3}{5}}$ , ${\tfrac{3}{5}}$ , ${\tfrac{3}{10}}$ .

${\therefore}$ the chances, after this event, are proportional to ${\tfrac{3}{20}}$ , ${\tfrac{3}{10}}$ , ${\tfrac{3}{40}}$ ; i. e. to 2, 4, 1. Hence they are ${\tfrac{2}{7}}$ , ${\tfrac{4}{7}}$ , ${\tfrac{1}{7}}$ .

Hence the chances, as to states, now are $\begin{align} \text{for}\, & 2\, W &\tfrac{2}{7};\\ & 1\,W, 1\,B &\tfrac{4}{7};\\ & 2 \,B &\tfrac{1}{7}. \end{align}$

${\therefore}$ the chances, after adding ${1 W}$ , are $\begin{align} \text{for}\, & 3 \,W &\tfrac{2}{7};\\ & 2 \,W, 1\, B &\tfrac{4}{7};\\ & 1 \,W, 2\, B &\tfrac{1}{7}. \end{align}$

Now the chances, which these give to the observed event, of drawing ${1\,W}$ , are 1, ${\tfrac{2}{3}}$ , ${\tfrac{1}{3}}$ .

${\therefore}$ the chances, after this event, are proportional to ${\tfrac{2}{7}}$ , ${\tfrac{8}{21}}$ , ${\tfrac{1}{21}}$ ;

i. e. to 6, 8, 1. Hence they are ${\tfrac{6}{15}}$ , ${\tfrac{8}{15}}$ , ${\tfrac{1}{15}}$ .

Hence the chance, that the bag now contains 2 white, is ${\tfrac{6}{15}}$ ; i. e. ${\tfrac{2}{5}}$ .

$\begin{aligned} \text{Because} &\frac{DO}{OA} = \frac{\triangle DOC}{\triangle OAC} = \frac{\triangle DOB}{\triangle OAB} = \frac{\triangle OBC}{\triangle OCA + \triangle OAB};\\ \qquad \quad\therefore &\frac{DO}{DA} = \frac{\triangle OBC}{\triangle ABC}.\\ \text{Similarly,} &\frac{EO}{EB} = \frac{\triangle OCA}{\triangle ABC}, \quad \text{and} \quad \frac{FO}{FC} = \frac{\triangle OAB}{\triangle ABC}.\\ \text{Hence} &\frac{DO}{DA} + \frac{EO}{EB} + \frac{FO}{FC} = 1. \end{aligned}$

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

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

This may be done by re-arranging the patients, so that the ${E}$ -class may begin from one end of the row, and the ${A}$ -class from the other: and the least possible number for the ${EA}$ -class is the common portion, i. e. ${(\epsilon-\overline{1-\alpha})}$ , i. e. ${(\epsilon+\alpha-1)}$ .

Then, as already shown, the least possible number for the ${EAL}$ -class is the common portion, i. e. ${(\epsilon+\alpha-1-\overline{1-\lambda})}$ , i. e. ${(\epsilon+\alpha+\lambda-2)}$ .

Let ${ABC}$ be the given Triangle, and ${A'B'C'}$ the required one; and let the ratio, which ${B'C'}$ has to ${BC}$ , be ' ${k}$ '; so that ${k}$ is less than 1.

Since ${BB'=CC'}$ , and that ${BC}$ , ${B'C'}$ , are parallel, it may easily be proved, by dropping perpendiculars from ${B'}$ , ${C'}$ , upon ${BC}$ , which must necessarily be equal, that ${\angle\text{s}}$ ${B'BC}$ , ${C'CB}$ , are equal.

Similarly, ${\angle\text{s}}$ ${A'AC}$ , ${C'CA}$ , are equal; and so are ${\angle\text{s}}$ ${A'AB}$ , ${B'BA}$ .

Hence, if ${BB'}$ , ${CC'}$ , be produced to meet at ${D}$ , Triangle ${DBC}$ will be isosceles, with a vertical ${\angle}$ equal to ${2A}$ .

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

Bisect the sides, and draw perpendiculars, meeting at ${D}$ . Join ${D}$ to the vertices ${B}$ , ${C}$ . From ${DB}$ cut off ${DB'= k.DB}$ . From ${B'}$ draw ${B'C'}$ parallel to ${BC}$ .

And if, from ${B'}$ , ${C'}$ , parallels to ${AB}$ , ${AC}$ , be drawn, it may easily be proved that they meet on ${DA}$ , and that they are respectively equal to ${k.AB}$ , ${k.AC}$ .

If remaining bag be ${A}$ , chance of observed event = ${\tfrac{1}{2}}$ chance of drawing white from ${B}$ and black from ${C}$ + ${\tfrac{1}{2}}$ chance of drawing black from ${B}$ and white from ${C}$ :

i. e. it ${=\tfrac{1}{2}.\{\tfrac{2}{3}\times\tfrac{1}{2}+\tfrac{1}{3}\times\tfrac{1}{2}\} = \tfrac{1}{4}}$ .

Similarly, if remaining bag be ${B}$ , it is ${\tfrac{1}{2}.\{\tfrac{5}{6}.\tfrac{1}{2}+\tfrac{1}{6}.\tfrac{1}{2}\}=\tfrac{1}{4}}$ ; and, if it be ${C}$ , it is ${\tfrac{1}{2}.\{\tfrac{5}{6}.\tfrac{1}{3}+\tfrac{1}{6}.\tfrac{2}{3}\}=\tfrac{7}{36}}$ .

${\therefore}$ chances of remaining bag being ${A}$ , ${B}$ , or ${C}$ , are as ${\tfrac{1}{4}}$ to ${\tfrac{1}{4}}$ to ${\tfrac{7}{36}}$ ; i. e. as 9 to 9 to 7. ${\therefore}$ they are, in value, ${\tfrac{9,\,9,\,7}{25}}$ .

Now, if remaining bag be ${A}$ , chance of drawing white from it is ${\tfrac{5}{6}}$ ; ${\therefore}$ chance, on this issue, is ${\tfrac{5}{6}.\tfrac{9}{25}=\tfrac{3}{10}}$ ; similarly, for ${B}$ , it is ${\tfrac{2}{3}.\tfrac{9}{25} = \tfrac{6}{25}}$ ; and, for ${C}$ , ${\tfrac{1}{2}.\tfrac{7}{25}=\tfrac{7}{50}}$ . And entire chance of drawing white from the remaining bag is the sum of these; i. e. ${\tfrac{15+12+7}{50}=\tfrac{34}{50} = \tfrac{17}{25}}$ .

Let ${ABC}$ be the given Triangle; and let its sides be divided internally at ${A'}$ , ${B'}$ , ${C'}$ , in extreme and mean ratio.

${\therefore x=\dfrac{-a\pm a\sqrt{5}}{2} = \dfrac{a}{2}.(\sqrt{5}-1)}$ , the other sign being excluded by the terms of the question.

Then area of Triangle ${AB'C'}$ $\begin{aligned} & = \frac{1}{2}.\frac{c}{2}.(\sqrt{5} -1).\{b- \frac{b}{2}.(\sqrt{5} -1)\}.\sin A,\\ & = \frac{1}{8}.(\sqrt{5}-1)(3-\sqrt{5})bc.\sin A,\\ & = \frac{1}{4}.(4\sqrt{5}-8).M=(\sqrt{5}-2).M. \end{aligned}$

Hence the sum of these 3 Triangles ${= 3.(\sqrt{5} -2).M}$ , and area of Triangle ${A'B'C'=(7-3\sqrt{5}).M}$ .

$(a^{2}+b^{2}).(c^{2}+d^{2})=a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}.$ $\begin{array}{rl} (a^{2}+b^{2}).(c^{2}+d^{2}) &= a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2};\\ \begin{array}{r}~\\ \text{or else}\qquad\end{array} & \left.\begin{array}{l} = a^{2}c^{2}+b^{2}d^{2}+2acbd+a^{2}d^{2}+b^{2}c^{2}-2adbc,\\ = a^{2}c^{2}+b^{2}d^{2}-2acbd+a^{2}d^{2}+b^{2}c^{2}+2adbc; \end{array}\right\}\\ \begin{array}{r}\text{i. e.}\qquad\\ \text{or else}\qquad\end{array} & \left.\begin{array}{l} = (ac+bd)^{2}+(ad-bc)^{2},\\ = (ac-bd)^{2}+(ad+bc)^{2}. \end{array}\right\} \end{array}$

Now, if these last 2 sets are identical, ${(ac + bd)}$ must = ${(ad+bc)}$ ; for it cannot = ${(ac-bd)}$ ;

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.

Let ${ABC}$ be the Triangle: and suppose ${B'C'}$ so placed that ${B'D}$ , ${C'E}$ , drawn parallel to the sides, shall together = ${2B'C'}$ .

By Euc. I. 34, ${B'D = C'B}$ , and ${C'E = B'C}$ : $\therefore~ B'C+C'B = 2B'C'.$

Hence, if ${B'L}$ be cut off equal to half ${B'C}$ , ${C'L = \text{half}\,\, C'B.}$

In ${BC'}$ take any point ${F}$ : draw ${FG}$ , ${\parallel BC}$ , and equal to half ${BF}$ : and join ${BG}$ .

Similarly, in ${CB'}$ take any point ${H}$ : draw ${HK}$ , ${\parallel BC}$ , and equal to half ${HC}$ : and join ${CK}$ .

Produce ${BG}$ , ${CK}$ , to meet at ${L}$ : and through ${L}$ draw ${C'B' \parallel BC}$ : and from ${B'}$ , ${C'}$ , draw ${B'D}$ , ${C'E}$ , ${\parallel}$ the sides.

${\because FG = \text{half}\,\, FB;}$ ${\therefore}$ , by similar Triangles, ${C'L = \text{half}\,\, C'B}$ ;

${\therefore ~ C'B' = ~ \text{half sum of} ~ C'B, B'C}$ ; i. e. ${C'B+B'C = 2B'C'}$ .

On July 1, watch gained on clock 5m. in 10h.; i. e. ${\tfrac{1}{2}}$ m. 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. ${\dfrac{11}{24 \times 12}}$ sec. in 5m.

Now, when true time is 12h. 5m. on July 31, clock is (2m. 19sec. + 11sec.) fast of it; i. e. says '12h. ${7\tfrac{1}{2}}$ m.' Hence, if true time be put 5m. back, clock must be put 5m. ${\tfrac{11}{288}}$ sec. back; i. e. must be put back to 12h. 2m. ${29\dfrac{277}{288}}$ sec.

$\begin{aligned} &\therefore \text{the (n+1)th term is}\, (n+1).(n+5) = (n+1).\{(n+2)+3\}, \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=(n+1).(n+2) + 3(n+1);\\ &\therefore S_{n} = \dfrac{n.(n+1).(n+2)}{3} + 3 \cdot \dfrac{n.(n+1)}{2} + C; \text{and}\, C = 0;\\ &\therefore S_{n} = n.(n+1).\left(\dfrac{n+2}{3} + \dfrac{3}{2}\right) = \dfrac{n.(n+1).(2n+13)}{6}. \end{aligned}$

Also ${S_{100} = \dfrac{100.101.213}{6} = \dfrac{100.101.71}{2} = \dfrac{100.7171}{2} = \dfrac{717100}{2} = 358550}$ .

$\begin{array}{l} \therefore \frac{dv}{dx} = \sqrt{r^{2}-x^{2}} + \sqrt{r^{2}-4x^{2}}-x^{2}.\left(\frac{1}{\sqrt{r^{2}-x^{2}}} + \frac{4}{\sqrt{r^{2}-4x^{2}}}\right) = 0;\\ \therefore (r^{2}-x^{2}).\sqrt{r^{2}-4x^{2}} + (r^{2}-4x^{2}).\sqrt{r^{2}-x^{2}} = x^{2}.(4\sqrt{r^{2}-x^{2}} + \sqrt{r^{2}-4x^{2}});\\ \therefore (r^{2}-2x^{2}).\sqrt{r^{2}-4x^{2}} = -(r^{2}-8x^{2}).\sqrt{r^{2}-x^{2}};\\ \therefore r^{4}-4(r^{2}x^{2}+4x^{4}).(r^{2}-4x^{2}) = (r^{4}-16r^{2}x^{2}+64x^{4}).(r^{2}-x^{2});\\ \therefore r^{6}-8r^{4}x^{2}+20r^{2}x^{4}-16x^{6} = r^{6}-17r^{4}x^{2}+80r^{2}x^{4}-64x^{6}; \end{array}$

${\therefore}$ , omitting ${r^{6}}$ , and dividing by ${x^{2}}$ , $\begin{aligned} 48x^{4}-60r^{2}x^{2}+9r^{4} & =0;\\ \text{i. e.}~ 16x^{4}-20r^{2}x^{2}+3r^{4} & =0; \end{aligned}$ ${\therefore {\dfrac{x^{2}}{r^{2}} = \dfrac{20\pm\sqrt{208}}{32} = \dfrac{5-\sqrt{13}}{8}}}$ (upper sign being inadmissible, though this was not thought out.)

Let ${A}$ be the given Point, and ${C}$ the centre of the given Circle. Join ${AC}$ , and let ${ABD}$ be the required Line. From ${B}$ draw the Chord ${BE}$ parallel to ${AC}$ . Then ${\angle\, DBE = \angle\, A}$ . Hence Arc ${DE}$ = Arc ${BD}$ ; i. e. Arc ${BE}$ is bisected by ${D}$ ; i. e. ${D}$ is on perpendicular from ${C}$ .

Join ${AC}$ . From ${C}$ draw ${CD}$ perpendicular to ${AC}$ . Join ${AD}$ cutting Circle at ${B}$ . From ${B}$ draw ${BE}$ parallel to ${AC}$ .

It is easily proved that Arc ${BD}$ = Arc ${DE}$ . Hence Arc ${BD}$ subtends, in the Circle, an angle = ${\angle\, DBE}$ = ${\angle\, A}$ .

Let ${ABC}$ be the given Triangle; and let the portions of the radii, outside the Triangle, have to the radius the given ratios ${k:1}$ , ${l:1}$ , ${m:1}$ . (N.B. ${k}$ , ${l}$ , ${m}$ , are supposed to be proper fractions.)

From ${B}$ draw ${BD\,\perp\,BA}$ , and ${BE\,\perp\,BC}$ ; and make ${BD}$ have, to ${BE}$ , the ratio ${1-m:1-k}$ . Through ${D}$ draw ${DF}$ parallel to ${BA}$ , and ${EF}$ parallel to ${BC}$ ; and join ${BF}$ . From ${F}$ draw ${FG\, \perp\, BA}$ , and ${FH\,\perp\,BC}$ .

Similarly, draw ${CO}$ so that the ${\perp}$ s, drawn from any Point of it to ${CA}$ and ${CB}$ , are in the ratio ${1-l:1-k}$ ; and produce ${BF}$ to meet it at ${O}$ .

With centre ${O}$ , and distance ${OK}$ , describe a Circle; and produce ${OB'}$ , ${OC'}$ , to meet it at ${L}$ , ${M}$ .

$\begin{aligned} \text{Now}\quad & OK:OA'\colon\colon \quad 1\quad:1-k;\\ & OA':OB'\colon\colon 1-k\,:1-l;\\ \therefore\quad & OK:OB'\colon\colon \quad 1 \quad:1-l;\\ \end{aligned}$

Similarly ${B'L:\text{radius}\colon\colon l:1}$ , and ${C'M:\text{radius}\colon\colon m:1}$ .

Join ${DB}$ , ${DC}$ : then ${\angle DBC'= 45°}$ , and ${\angle B'DC = \angle B'CD.}$

Then ${\angle B'DC = \angle DCE}$ ; ${\therefore}$ ${\angle B'CD = \angle DCE}$ .

Hence construction. Draw ${CE \perp AB}$ : bisect ${\angle ACE}$ : at ${B}$ make ${\angle ABD = 45°}$ . Let these lines meet at ${D}$ . Through ${D}$ draw ${B'DC'\perp AB}$ .

$\begin{aligned} & \text{Also}~ \angle B'DC = \angle DCE = \angle DCB';\\ & \therefore~ DB'= B'C;\\ & \therefore~ C'B'= ~\text{sum of}~ BC',~ CB'. \end{aligned}$

${\therefore \dfrac{\sqrt{3}}{2} = 1 - \dfrac{1}{2x^{2}}}$ ; ${\therefore \dfrac{1}{2x^{2}} = \dfrac{2-\sqrt{3}}{2}}$ ;

${\therefore}$ areas of Sectors are ${\pi.\dfrac{2+\sqrt{3}}{12}}$ and ${\dfrac{\pi}{6}}$ ;

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

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

${= \dfrac{5\cdot 732}{12} - \dfrac{2\cdot 732}{(\tfrac{44}{7})} = \cdot 478 - \dfrac{\cdot 248}{(\tfrac{4}{7})}}$ ,

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 ' ${A}$ ', ' ${B}$ ', and ' ${C}$ ': viz.—

Now the chance of the observed event is, in case (1), ${1\times\tfrac{4}{9} = \tfrac{4}{9}}$ ; in case (2), ${1\times\tfrac{1}{9} = \tfrac{1}{9}}$ ; in case (3), ${\tfrac{2}{3}\times 1 = \tfrac{2}{3}}$ ; in case (4), ${\tfrac{2}{3}\times \tfrac{1}{9} = \tfrac{2}{27}}$ ; in case (5), ${\tfrac{1}{3}\times 1 = \tfrac{1}{3}}$ ; and in case (6), ${\tfrac{1}{3}\times \tfrac{4}{9} = \tfrac{4}{27}}$ .

Hence the chances of existence, for these 6 states, are proportional to '12, 3, 18, 2, 9, 4'. Hence their actual values are ' ${\tfrac{1}{4}, \tfrac{1}{16}, \tfrac{3}{8}, \tfrac{1}{24}, \tfrac{3}{16}, \tfrac{1}{12}}$ '.

Hence the chance of the unknown counter being red is the sum of ${\tfrac{1}{4}\times 1, \tfrac{1}{16}\times 1, \tfrac{3}{8}\times\tfrac{2}{3}, \tfrac{1}{24}\times\tfrac{2}{3}, \tfrac{3}{16}\times\tfrac{1}{3}, \tfrac{1}{12}\times\tfrac{1}{3}}$ ;

i. e. it is ${\dfrac{36+9+36+4+9+4}{9\times 16}}$ ; which ${= \dfrac{98}{9\times 16} = \tfrac{49}{72}}$ .

Then ${(2\times 10-\overline{x-1}).\dfrac{x}{2} = 14 + \{2\times 2 + \overline{x-1}\,.\,2\} \cdot \dfrac{x}{2}}$ ;

${\therefore 3x^{2}-19x+28=0}$ ; ${\therefore x = \dfrac{19\pm 5}{6} = 4}$ or ${\dfrac{7}{3}}$ .

Now the above solution has taken no account of the discontinuity of increase, or decrease of pace, and is the true solution[Pg 61] 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 ' ${\dfrac{7}{3}}$ ' only indicates that a meeting occurs during the third day. To find the hour of this, let ${y =}$ no. of hours.

Now in 2 days ${A}$ has got to the end of 19 miles, ${B}$ to the end of ${(14 + 6)}$ , i. e. 20.

i. e. ${y \,. \dfrac{2}{3} = 1 + y \,. \dfrac{1}{2};\quad \therefore\, y = 6}$ .

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

From ${D}$ draw ${DE}$ , ${DF}$ , parallel to the sides; and from ${E}$ and ${F}$ draw ${EG}$ , ${FH}$ , ${\perp BC}$ .

Also, ${\because}$ Triangles ${AED}$ , ${AFD}$ , are equal and on the same base ${AD}$ ,

Make ${CE=BD}$ ; from ${E}$ draw ${EF}$ , ${EG}$ , parallel to the sides; and from ${F}$ , ${G}$ , draw ${FH}$ , ${GK}$ , ${\perp BC}$ .

(1) As there was certainly at least one ${W}$ in the bag at first, the 'a priori' chances for the various states of the bag, ' ${WWWW}$ , ${WWWB}$ , ${WWBB}$ , ${WBBB,}$ ' were ' ${\tfrac{1}{8}}$ , ${\tfrac{3}{8}}$ , ${\tfrac{3}{8}}$ , ${\tfrac{1}{8}}$ '.

These would have given, to the observed event, the chances '1, ${\tfrac{1}{2}}$ , ${\tfrac{1}{6}}$ , 0'.

Hence the chances, after the event, for the various states, are proportional to ' ${\tfrac{1}{8}\cdot 1}$ , ${\tfrac{3}{8}\cdot\tfrac{1}{2}}$ , ${\tfrac{3}{8}\cdot\tfrac{1}{6}}$ '; i. e. to ' ${\tfrac{1}{8}}$ , ${\tfrac{3}{16}}$ , ${\tfrac{1}{16}}$ '; i.e. to '2, 3, 1'. Hence their actual values are ' ${\tfrac{1}{3}}$ , ${\tfrac{1}{2}}$ , ${\tfrac{1}{6}}$ '.

Hence the chance, of now drawing ${W}$ , is ' ${\tfrac{1}{3}\cdot 1 + \tfrac{1}{2}\cdot\tfrac{1}{2}}$ '; i. e. it is ${\tfrac{7}{12}}$ .

(2) If he had not spoken, the 'a priori' chances for the states ' ${WWWW}$ , ${WWWB}$ , ${WWBB}$ , ${WBBB}$ , ${BBBB}$ ', would have been ' ${\dfrac{1,\,4,\,6,\,4,\,1}{16}}$ '.

These would have given, to the observed event, the chances '1, ${\tfrac{1}{2}}$ , ${\tfrac{1}{6}}$ , 0, 0'.

Hence the chances, after the event, for the various states, are proportional to ' ${\tfrac{1}{16} \cdot 1}$ , ${\tfrac{1}{4} \cdot \tfrac{1}{2}}$ , ${\tfrac{1}{6} \cdot \tfrac{3}{8}}$ '; i. e. to '1, 2, 1'. Hence their actual values are ' ${\tfrac{1}{4}}$ , ${\tfrac{1}{2}}$ , ${\tfrac{1}{4}}$ '.

Hence the chance, of now drawing ${W}$ , is ' ${\tfrac{1}{4} \cdot 1 + \tfrac{1}{2} \cdot \tfrac{1}{2}}$ '; i. e. it is ${\tfrac{1}{2}}$ .

Let ${ABC}$ be the given Triangle. Bisect its angles, and draw ${\perp}$ s to them, forming the Triangle ${A′B′C′}$ .

${\therefore \,A′= 180° - (CBA′ + BCA′) = \dfrac{B+C}{2} = 90°- \dfrac{A}{2};}$ ${\therefore \,BA′= a\, . \dfrac{\cos \dfrac{C}{2}}{\cos \dfrac{A}{2}}.}$

Similarly, ${BC'= c. \dfrac{\cos \tfrac{A}{2}}{\cos \tfrac{C}{2}}}$ ; $\begin{aligned} \therefore A'C' & =\frac{ a\cos ^{2}\frac{C}{2} +c\cos ^{2}\frac{A}{2}}{\cos \frac{A}{2}\cos \frac{C}{2}} = \frac{a.\frac{s.(s-c)}{ab} + c.\frac{s.(s-a)}{bc}}{\frac{s}{b}·\sqrt{\frac{(s-a).(s-c)}{ac}}},\\ & =\frac{s-c+s-a}{\sin \frac{B}{2}} = \frac{b}{\sin \frac{B}{2}}. \end{aligned}$

${\therefore}$ area of ${A'B'C' = \dfrac{bc\cos \dfrac{A}{2}}{2\sin \dfrac{B}{2}\sin \dfrac{C}{2}}}$ ;

$\begin{aligned} \therefore \frac{\text{area of } A'B'C'}{\text{area of }ABC} &= \frac{bc\cos \frac{A}{2}}{2\sin \frac{B}{2}\sin \frac{C}{2}}·\frac{2}{bc\sin A},\\ & = \frac{\cos \frac{A}{2}}{\sin \frac{B}{2}\sin \frac{C}{2}.2\sin \frac{A}{2}\cos \frac{A}{2}},\\ & = \frac{1}{2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}},\\ & = \frac{abc}{2(s-a).(s-b).(s-c)}. \end{aligned}$

Let ${ABC}$ be the given Triangle; and let ${BFD}$ , ${CFE}$ , be the required lines, so that ${FB = FC}$ , and Tetragon ${AEFD}$ = Triangle ${FBC}$ . And call the angle ${FBC}$ ' ${\theta}$ '. It will suffice to calculate this angle.

${\therefore~ \text{Triangles } DBC, EBC, \text{ together } = \text{Triangle }ABC;}$

${\therefore~ \dfrac{1}{2} \cdot \dfrac{a^{2}}{\cot~\theta+\cot~ C} + \dfrac{1}{2} \cdot \dfrac{a^{2}}{\cot~\theta+\cot B} = \dfrac{1}{2} \cdot \dfrac{a^{2}}{\cot~ B+\cot~C};}$

${\therefore~ \dfrac{1}{\cot~\theta+\cot~ C} +\dfrac{1}{\cot~\theta+\cot~ B} = \dfrac{1}{\cot~ B+\cot~ C};}$

${\therefore~ \dfrac{2\cot~\theta+(\cot~ B+\cot~ C)}{\cot^{2}~ \theta+\cot~\theta \cdot(\cot~ B+\cot~ C)+\cot B \cot~C}= \text{do.};}$

${\therefore~ \cot^{2}~ \theta+\cot~\theta \cdot(\cot~ B+\cot~ C)+\cot~ B~\cot~ C = 2\cot~\theta\cdot(\cot~ B+\cot~ C)+(\cot~ B+\cot~ C)^{2};}$

${\therefore~ \cot^{2}~ \theta-\cot~\theta.(\cot~ B+\cot~ C) -(\cot^{2}~ B+\cot~ B~\cot~ C+\cot^{2}~ C)=0;}$

${\therefore~ \cot~ \theta = \dfrac{1}{2}. \{\cot~B+\cot~C \pm\surd(5\cot^{2}~B+6\cot~ B~\cot~ C+5\cot^{2}~C)\}}$ .

Let ${k}$ be a No. not containing 2 or 5 as a factor, i. e. let it be prime to 10. Then, if ${\dfrac{1}{k}}$ 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 ${(10^{n}-1)}$ .

And since this fraction ${=\dfrac{1}{k}}$ , and that ${k}$ is prime to 10, and so prime to ${10^{m}}$ , the factor ${(10^{n}-1)}$ 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 ${(a^{n}-1)}$ a multiple of ${b}$ .

(1) With radix 10, find a value, for ${n}$ , which will make ${(10^{n}-1)}$ a multiple of 7.

Taking 8 as radix, we get ${\tfrac{1}{9} = \cdot\dot{0}\dot{7} = \dfrac{7}{8^{2}-1}}$ .

Taking 7 as radix, we get $\tfrac{1}{13} = \cdot\dot{0}3524563142\dot{1} = \frac{35245631421}{7^{12}-1}.$

Divide each rod into ${(n + 1)}$ 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.

$\begin{array}{l} \text{The chance of one failure}&\text{is}\, \frac{n - 1}{n};\\ \therefore\quad \unicode{x201d}\qquad \unicode{x201d}\qquad n \quad\text{failures}&\text{is} \,\left(\frac{n - 1}{n}\right)^{n}\\ & = \left(1 - \frac{1}{n}\right)^{n}. \end{array}$

${\therefore}$ the chance that no rod is broke in the middle ${= (1 - m)^\tfrac{1}{m}}$ , when ${m = 0}$ ;

Now ${(1 + 0)^{\tfrac{1}{0}} = e}$ . Hence if, in the series for ${e}$ , we call the sum of the odd terms ' ${a}$ ', and of the even terms ' ${b}$ '; then ${e = a + b}$ ; and ${(1 - 0)^{\tfrac{1}{0}} = a - b = 2a - e}$ .

$\begin{aligned} \text{Now}\quad a &= 1 + \frac{1}{\mid\underline{2}} + \frac{1}{\mid\underline{4}} + \&c.\\ 1 & = 1\\ \frac{1}{\mid\underline{2}} & = \cdot 5\\ \frac{1}{\mid\underline{4}} & = \cdot 04166666\,\, \&c.\\ \frac{1}{\mid\underline{6}} & = \cdot 00138888 \,\,\&c.\\ \frac{1}{\mid\underline{8}} & = \cdot 00002480 \,\,\&c.\\ \frac{1}{\mid\underline{10}} & = \cdot 00000027 \,\,\\&c.\\ & \quad \, \text{——————————}\\ \therefore \, a & = 1\cdot 5430806 \,\, \&c.\\ \therefore\, 2a & = 3\cdot 0861612 \,\, \&c.\\ e & = 2\cdot 7182818 \, {\&c.}\\ & \quad\, \text{——————————}\\ \therefore (1 - 0)^{\tfrac{1}{0}} & = \cdot 3678793 \,\, \&c.\\ \therefore \text{Ans.} & = 1 - (1 - 0)^{\tfrac{1}{0}} = \cdot 6321207 \,\, \&c. \end{aligned}$

If we make a Triangle ${D′E′F′}$ , having its angles equal to the given angles, and having ${D′}$ as its assigned vertex, the Problem may be solved, if we can circumscribe, about the Triangle ${D′E′F′}$ , a Triangle similar to ${ABC}$ .

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

This Lemma may be solved as follows. Let ${G}$ , ${H}$ , be the centres of the Circles. Join ${GH}$ , and divide it, at ${K}$ , proportionally to ${BDC}$ .

Join ${KD'}$ ; through ${D'}$ draw ${B′D′C′ \perp KD'}$ ; and from ${G}$ , ${H}$ , draw ${GL}$ , ${HM}$ , ${\perp B′C′}$ .

Now it may be easily proved that $LD'\, \colon D′M \colon\colon GK\, \colon KH \colon\colon BD\, \colon DC.$

But ${B′D′}$ , ${D′C′}$ , are doubles of ${LD'}$ , ${D′M}$ ; $\therefore B′D′\, \colon D′C′ \colon\colon BD\, \colon DC.$

[The construction is now obvious, viz. to join ${B′F′}$ , ${C′E′}$ , and produce them to meet, on the third Circle (as they may be easily proved to do), at ${A′}$ ; then to divide ${AB}$ , ${AC}$ , at ${F}$ and ${E}$ , proportionally to ${A′F′B′}$ , ${A′E′C′}$ ; and then to join ${DE}$ , ${DF}$ .]

Subtracting, we get ${x \,.\, \left(\dfrac{1}{y} - \dfrac{1}{z}\right) = y - z}$ ;

${\therefore\, x = yz \,. \,\dfrac{y - z}{z - y} = -yz}$ , unless ${y = z}$ , in which case ${x = \dfrac{0}{0}}$ .

Hence, if ${x}$ be finite, and if ${y\neq z}$ , either ${x}$ or ${y = 0}$ , and also either ${x}$ or ${z = 0}$ ; i. e. either ${x = 0}$ , or else ${y = z = 0}$ . But the latter is excluded by our hypothesis. Hence ${x = 0}$ . Hence ${yz = 0}$ ; i. e. either ${y}$ or ${z = 0}$ , and the other may take any value.

This gives us 2 more sets of values, viz. $\begin{aligned} x & = y = 0;\, z \text{ has any value};\\ x & = z = 0;\, y \text{ has any value}. \end{aligned}$

This gives us a 4th set of values, viz. ${x = \dfrac{k^{2}}{k - 1} ,\, y = z = k}$ ; where ${k}$ has any value.

Now ${y}$ and ${z}$ may evidently have any real values, but ${x}$ is restricted by the equation $y^{2} - xy + x = 0,$ in which y cannot be real, unless ${(x^{2} - 4x) \gt 0}$ . 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.

Let ${ABC}$ be the given Triangle, ${A′}$ , ${B′}$ , ${C′}$ , the centres of the semicircles, and ${DE}$ , ${FG}$ , ${HJ}$ , the common tangents; so that ${DE = \alpha}$ , ${FG = \beta}$ , and ${HJ = \gamma}$ .

Join ${B′D}$ , ${C′E}$ ; and from ${C′}$ draw ${C′K \perp B′D}$ . Hence ${CK = \alpha}$ .

${\qquad \,\, \text{and}\quad \gamma = \surd\left\{(-a+b+c)\,.(a-b+c)\right\};}$

If the Tetrahedron be cut by a vertical Plane containing one of the slant edges, the section is a Triangle whose base is ${\dfrac{\sqrt{3}}{2}}$ , and whose sides are ${\dfrac{\sqrt{3}}{2}, 1}$ ;

hence cosine of smaller base-angle $= \left(\tfrac{3}{4} + 1 - \tfrac{3}{4}\right)\cdot\dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}};$

${\therefore}$ volume of Tetrahedron ${= \dfrac{1}{3}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{4} = \dfrac{\sqrt{2}}{12}}$ .

Also altitude of Pyramid = altitude of Triangle whose base is ${\sqrt{2}}$ , and whose sides are 1, 1;

${\therefore}$ volume of Pyramid = ${\dfrac{1}{3}\cdot\dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{6}}$ .

At first, the chance that bag ${H}$ shall contain $\begin{array}{ll} 2\, W\, \text{counters}, &\text{is}\, \tfrac{1}{4}.\\ 1\, W\, \text{and}~1~ B, & \text{is}\, \tfrac{1}{2}.\\ 2\, B, & \text{is}\,\, \tfrac{1}{4}. \end{array}$

${\therefore}$ , after adding a ${W}$ , the chance that it shall contain $\begin{aligned} &3~W, &\text{is }\,\tfrac{1}{4}.\\ &2~W, 1~B, &\text{is }\,\tfrac{1}{2}.\\ &1~W, 2~B, &\text{is }\,\tfrac{1}{4}. \end{aligned}$ hence the chance of drawing a W from it is $\tfrac{1}{4}\times 1 + \tfrac{1}{2}\times \tfrac{2}{3} + \tfrac{1}{4}\times \tfrac{1}{3}: \text{ i. e. }\tfrac{2}{3}.$ ${\therefore}$ the chance of drawing a ${B}$ is ${\tfrac{1}{3}}$ .

After transferring this (unseen) counter to bag ${K}$ , the chance that it shall contain $\begin{array}{llll} 3 W, & & \text{is }\frac{2}{3}\times\frac{1}{4}; & \text{ i. e. }\,\frac{1}{6}.\\ 2 W, \text{ and} & 1 B, & \text{is }\frac{2}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{4}; & \text{ i. e. }\,\frac{5}{12}.\\ 1 W, & 2 B, & \text{is }\frac{2}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{1}{2}; & \text{ i. e. }\,\frac{1}{3}.\\ 3 B, & & \text{is }\frac{1}{3}\times \frac{1}{4}; & \text{ i. e. }\frac{1}{12}; \end{array}$ ${\therefore}$ the chance of drawing a ${W}$ from it is $\tfrac{1}{6}\times 1 + \tfrac{5}{12}\times \tfrac{2}{3} + \tfrac{1}{3}\times \tfrac{1}{3}; \text{ i. e. }\,\tfrac{5}{9}.$ ${\therefore}$ the chance of drawing a ${B}$ is ${\tfrac{4}{9}}$ .

Before transferring this to bag ${H}$ , the chance that bag ${H}$ shall contain $\begin{array}{llll} 2 W, & \text{is} & \tfrac{1}{4}\times 1 + \tfrac{1}{2}\times\tfrac{1}{3}; & \text{ i. e. }\,\tfrac{5}{12}.\\ 1 W, 1 B, & & \tfrac{1}{2}\times\tfrac{2}{3} + \tfrac{1}{4}\times\tfrac{2}{3}; & \text{ i. e. }\,\tfrac{1}{2}.\\ 2 B, & & \tfrac{1}{4}\times\tfrac{1}{3}; & \text{ i. e. }\tfrac{1}{12}. \end{array}$ ${\therefore}$ , after transferring it, the chance that bag ${H}$ shall contain $\begin{array}{llll} 3 W, & \text{is} & \tfrac{5}{12}\times\tfrac{5}{9}; & \text{ i. e. }\tfrac{25}{108}.\\ 2 W, 1 B, & & \tfrac{5}{12}\times\tfrac{4}{9} + \tfrac{1}{2}\times\tfrac{5}{9}; & \text{ i. e. }\tfrac{50}{108}.\\ 1 W, 2 B, & & \tfrac{1}{2}\times\tfrac{4}{9} + \tfrac{1}{12}\times\tfrac{5}{9}; & \text{ i. e. }\tfrac{29}{108}.\\ 3 B, & & \tfrac{1}{12}\times\tfrac{4}{9}; & \text{ i. e. }\tfrac{4}{108}. \end{array}$

Hence the chance of drawing a ${W}$ is $\tfrac{1}{108}\times\{25\times 1 + 50\times\tfrac{2}{3} + 29\times\tfrac{1}{3}\}; \text{ i. e. } \tfrac{17}{27}.$ i. e. the odds are 17 to 10 on its happening.

Let ${DE}$ be the line required. Draw ${DF}$ , ${EG}$ , ${\perp}$ the base. Then their sum is equal to ${DE}$ .

Bisect ${DE}$ at ${H}$ , and draw ${HK}$ ${\perp}$ the base: then it is evident that ${HK}$ is the ${A. M.}$ of ${DF}$ , ${EG}$ , and is equal to half their sum; i. e. it is equal to half of ${DE}$ . Hence a Circle, drawn with centre ${H}$ and at distance ${HD}$ , will pass through ${E}$ and ${K}$ , and will touch the base at ${K}$ .

Through ${H}$ draw ${LHM}$ parallel to ${AC}$ . Then ${DA}$ is evidently bisected at ${L}$ . Also ${LM}$ passes through the centre of the Circle. Hence, if ${DN}$ be drawn ${\perp}$ ${LM}$ (or ${CA}$ ), it is a chord of the Circle, and is bisected at ${R}$ . Produce ${ND}$ to meet the base produced at ${S}$ . Hence ${SDN}$ cuts the Circle, and ${SK}$ touches it at ${K}$ . But ${S}$ can be found, and ${SK}$ can then be taken, so that sq. of ${SK}$ may be equal to rect. of ${SD}$ , ${SN}$ .

From ${D}$ draw ${DN \perp AC}$ , and produce it to meet the base produced at ${S}$ . Take ${SK}$ , so that its square may be equal to rect. of ${SD}$ , ${SN}$ .

Bisect ${DA}$ at ${L}$ , and from ${L}$ draw ${LM}$ parallel to ${AC}$ ; and from ${K}$ draw ${KH}$ ${\perp}$ the base, to meet ${LM}$ at ${H}$ . Join ${DH}$ ,[Pg 75] and produce it to meet ${AC}$ at ${E}$ , and draw ${DF}$ , ${EG}$ , ${\perp}$ the base.

[N. B. This proof is incomplete. I have assumed, without proving it, that ${DH = HK}$ . It may be proved thus. Because sq. of ${SK}$ = rect. of ${SD}$ , ${SN}$ , ${\therefore DN}$ is a chord of a Circle which touches the base at ${K}$ ; ${\therefore LM}$ , which bisects it at right angles, passes through the centre. But ${KH}$ also passes through the centre; ${\therefore H}$ is the centre; ${\therefore HD = HK}$ .]

No. (3) received ${x}$ , took out (2 + 4), and put in ${\dfrac{x}{2}}$ ; so that the sack then contained ${(x.\tfrac{3}{2} - 6)}$ . Let us write ' ${a}$ ' for ' ${\tfrac{3}{2}}$ .'

No. (5) received ${(xa - 6)}$ , took out (4 + 1), and put in enough to multiply, by ${a}$ , its contents when he received it. The sack now contained ${(xa^{2} - 6a - 5)}$ .

${\qquad\quad = \dfrac{(6.3^{3} + 5.3^{2}.2 + 4.3.2^{2} + 6.2^{3}).2}{3^{4} - 5.2^{4}};}$

${\qquad\quad = \dfrac{(162 + 90 + 48 + 48).2}{81 - 80} = 696 = 2l.\,18s.\,0d.}$

Let ${ABC}$ be the given Triangle, and ${P}$ the given Point; and call its trilinear co-ordinates ' ${\alpha}$ , ${\beta}$ , ${\gamma}$ '.

From ${P}$ draw ${PA′}$ , ${PB′}$ , ${PC′}$ , ${\perp}$ the sides, and therefore equal to ${\alpha}$ , ${\beta}$ , ${\gamma}$ . Produce ${PA′}$ and ${PC′}$ to ${A''}$ and ${C''}$ , making ${A′A'' = PA′}$ , and ${C′C″= PC′}$ . From ${C″}$ draw ${C″D \perp AC}$ , and produce it to ${E}$ , making ${DE = C''D}$ . Join ${EA''}$ , cutting ${AC}$ in ${R}$ , and ${BC}$ in ${S}$ . Join ${C''R}$ , cutting ${AB}$ in ${Q}$ . Join ${PQ}$ , ${PS}$ .

The path of the ball is evidently ${PQRSP}$ ; and we have to calculate the length of ${AR}$ .

From ${P}$ draw ${PU}$ , ${PV}$ , parallel to ${AB}$ , ${AC}$ ; from ${C'}$ draw ${C'W \perp PV}$ ; and from ${A''}$ draw ${A''F \perp AC}$ .

By similar Triangles, ${DR : RF \colon\colon DE : A''F \colon\colon C''D : A''F}$ ;

${\therefore~ C″D,~ \text{which}~ = 2 C'W + PB',~ = 2\gamma \cos A + \beta}$ .

${\therefore~ DR = \dfrac{(\alpha \sin C + \gamma \sin A) \cdot (2 \gamma \cos A + \beta)}{\alpha \cos C + \gamma \cos A + \beta}}$ .

${\qquad\DeclareMathOperator\cosec{cosec}\qquad\qquad=\beta \cot A + \gamma \cosec A = \dfrac{\beta \cos A + \gamma}{\sin A}}$ ;

${\therefore~ AB' - DB' = \dfrac{\beta \cos A + \gamma}{\sin A} - 2 \gamma \sin A}$ ,

${\qquad\quad\qquad\qquad= \dfrac{\beta \cos A + \gamma (1 - 2 \sin^{2} A)}{\sin A}}$

${\therefore~ AR = \dfrac{(\alpha \sin C + \gamma \sin A) \cdot (2 \gamma \cos A + \beta)}{\alpha \cos C + \gamma \cos A + \beta}}$ ${+ \dfrac{\beta \cos A + \gamma \cos 2 A}{\sin A}}$ .

Let ${\unicode{x2018}k\unicode{x2019} = \text{ratio}\, \dfrac{DE}{a} = \dfrac{AE}{b} = \dfrac{AD}{c}}$ .

${\therefore~~ GF~(\text{which } = GB . \dfrac{b}{c}) = b - kb - k . \dfrac{ab}{c}}$ ;

${\therefore~~ k = \dfrac{bc}{bc + ca + ab} = \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} = \dfrac{\tfrac{1}{a}}{m}}$ (say).

Hence ${AD = \dfrac{c . \dfrac{1}{a}}{m}; DG = \dfrac{1}{m} = \dfrac{c . \dfrac{1}{c}}{m}}$ ·

${\therefore~~ GB~(\text{which }= c - AD - DG) = \dfrac{c . (m - \dfrac{1}{a} - \dfrac{1}{c})}{m} = \dfrac{c . \tfrac{1}{b}}{m}}$ ;

${\therefore~~ AD : DG : GB \colon\colon \dfrac{1}{a} : \dfrac{1}{c} : \dfrac{1}{b}}$ ·

Also ${DE = ka = \dfrac{1}{m} = \dfrac{1}{\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}}}$ ·

Let ${A}$ , ${B}$ , ${C}$ be the centres of the bases of the towers; and ${a}$ , ${b}$ , ${c}$ their radii. Suppose ${P}$ the required Point; and from ${P}$ 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.

Draw a Line through ${A}$ , ${B}$ , and on it take Points ${G}$ , ${H}$ , such that ${AG : GB \colon\colon AH : HB \colon\colon a : b}$ .

Then the Semicircle, described on ${GH}$ , is the locus of all Points whose distances, from ${A}$ and ${B}$ , are proportional to ${a}$ , ${b}$ .

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

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

Draw ${BC}$ , ${CE}$ , ${BD}$ , equal to the given altitudes, so as to form right ${\angle}$ s at ${B}$ and ${C}$ : and produce ${DB}$ , ${EC}$ . Join ${DC}$ , and draw ${CF \, \perp}$ to it. Join ${EB}$ , and draw ${BG \, \perp}$ to it. With centre ${B}$ , and distance ${BF}$ , describe a circle: with centre ${C}$ , and distance ${CG}$ , describe another: let them meet at ${A}$ : and join ${AB}$ , ${AC}$ .

Suppose the 3 Squares described, and that their upper edges form the Triangle ${A'B'C'}$ . Join ${AA'}$ , ${BB'}$ , ${CC'}$ .

Now it is evident that, if ${BB'}$ be produced, the perpendiculars dropped, from any Point of it, upon ${AB}$ , ${BC}$ , will be proportional to ${B'F}$ , ${B'D}$ .

Hence these 3 Lines will meet at the Point from which the perpendiculars, dropped upon the sides of ${ABC}$ , are proportional to ${B'C'}$ , ${C'A'}$ , ${A'B'}$ .

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

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.

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.

${\DeclareMathOperator\cot{cot}\therefore \cot F = \dfrac{z+x\cos B}{x\sin B} = \cot B'BD}$ .

Now it is evident that these Triangles are similar; so that $\frac{a}{x} = \frac{b}{y} = \frac{c}{z}.$

Hence, multiplying the last equation, throughout, by one or other of these equal fractions, we get $\begin{aligned} & \frac{a\sin A+b\sin B+c\sin C}{\sin B\sin C} = \frac{a^{2}}{x} -a;\\ \therefore~ & \frac{a\sin A+b\sin B+c\sin C}{a\sin B\sin C} = \frac{a}{x} -1;\\ \therefore~ & \frac{a}{x} = \frac{a\sin A+b\sin B+c\sin C}{a\sin B\sin C} +1. \end{aligned}$

Hence, multiplying above and below by one or other of the equal fractions ${\dfrac{a}{\sin A}}$ , ${\dfrac{b}{\sin B}}$ , ${\dfrac{c}{\sin C}}$ , $\begin{aligned} \frac{a}{x} & = \frac{a^{2}+b^{2}+c^{2}}{ab\sin C} +1;\\ & = \frac{a^{2}+b^{2}+c^{2}}{2m} +1 = \frac{b}{y} = \frac{c}{z}. \end{aligned}$

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

Take the longest side of the Triangle, and call it ' ${AB}$ ': and, on that side of it, on which the Triangle lies, draw the semicircle ${AFB}$ . Also, with centres ${A}$ , ${B}$ , and distances ${AB}$ , ${BA}$ , draw the arcs ${BDC}$ , ${AEC}$ , intersecting at ${C}$ .

Then it is evident that the vertex of the Triangle cannot fall outside the Figure ${ABDCE}$ .

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 ${= \dfrac{\text{area of semicircle}}{\text{area of fig. }ABDCE}}$ ·

Now let ${AB = 2a}$ : then area of semicircle ${= \dfrac{\pi a^{2}}{2}}$ ; and area of Fig. ${ABDCE = 2 \times \text{sector } ABDC - \text{Triangle } ABC}$ ; $= 2.\frac{4\pi a^{2}}{6} - \sqrt{3}.a^{2} = a^{2} . (\frac{4\pi}{3} - \sqrt{3});$

${\therefore \text{chance }= \dfrac{\dfrac{\pi}{2}}{\dfrac{4\pi}{3}-\sqrt{3}} = \dfrac{3}{8- \dfrac{6\sqrt{3}}{\pi}}}$ ·

and let ${\angle}$ s ${LMK}$ , ${MKL}$ , ${KLM}$ be equal to ' ${A}$ , ${B}$ , ${C}$ '; and similarly for the ${\angle}$ s of the other facets.

From ${K}$ draw ${KT}$ ${\perp}$ base-facet ${LMN}$ . Also draw ${KR}$ , ${KS}$ , ${\perp}$ ${LM}$ , ${MN}$ . And join ${TR}$ , ${TM}$ , ${TS}$ .

The required volume is ${\tfrac{1}{3} . KT . LMN}$ . The area of ${LMN}$ is of course known. All we need is the length of ${KT}$ . Now ${KT^{2} = KS^{2} - TS^{2}}$ ; and ${KS}$ evidently ${= c.\sin B}$ . Hence all we need is the length of ${TS}$ .

Given, in Tetragon ${RMST}$ , sides ${RM}$ , ${MS}$ , and ${\angle RMS}$ , and that ${\angle}$ s ${TRM}$ , ${TSM}$ are right: find ${TS}$ .

also ${TS \cos TSR + TR \cos TRS = RS}$ ; $\begin{aligned} \therefore~ \frac{TS}{\sin TRS} & = \frac{TR}{\sin TSR},\\ & = \frac{TS\cos TSR + TR \cos TRS}{\sin TRS \cos TSR + \sin TSR \cos TRS},\\ & = \frac{RS}{\sin RMS} = \frac{MS}{\sin MRS};\\ \therefore~ \frac{TS}{\cos MRS} & = \frac{MS}{\sin MRS};~ \text{i.e. } TS = MS \cot MRS. \end{aligned}$

Hence this requires another Lemma, in order to find the value of ${\cot MRS}$ (or ${\tan MRS}$ , which will do as well, and makes a prettier problem).

Given, in Triangle ${RMS}$ , sides ${RM}$ , ${MS}$ , and ${\angle RMS}$ : find ${\tan MRS}$ .

$\begin{aligned} \tan MRS & = \frac{\sin MRS} {\cos MRS} = \frac{RS \sin MRS} {RS \cos MRS},\\ & = \frac{MS \sin RMS} {RM-MS \cos RMS}. \end{aligned}$

Hence, in Tetragon ${RMST}$ , we have by Lemma (1), $TS = MS \cot MRS;$ and, by Lemma (2), ${\cot MRS = \dfrac{RM-MS \cos RMS} {MS \sin RMS}}$ , $= \frac{c\cos A-c\cos B\cos C} {c\cos B\sin C} = \frac{\cos A-\cos B\cos C} {\cos B\sin C};$

${\therefore \text{it} = (c\sin B)^{2}-\dfrac{c^{2}} {\sin ^{2}C}\cdot(\cos A-\cos B\cos C)^{2}}$ ,

${=\dfrac{c^{2}}{\sin ^{2}C}\cdot\{(\sin B\sin C)^{2}-(\cos A-\cos B\cos C)^{2}\}}$ ;

therefore ${\quad KT = \dfrac{c} {\sin C}}$ multiplied by ${\sqrt{\sin ^{2}B\sin ^{2}C-\cos ^{2}B\cos ^{2}C-\cos ^{2}A+2\cos A\cos B\cos C}}$ , $= \frac{c}{\sin C}\,\, \text{multiplied by}$ ${\sqrt{(1-\cos ^{2}B)\cdot(1-\cos ^{2}C)-\cos ^{2}B\cos ^{2}C-\cos ^{2}A+2\cos A\cos B\cos C}}$ ,

= ${\dfrac{c} {\sin C} \cdot \sqrt{1-(\cos ^{2}A+\cos ^{2}B+\cos ^{2}C)+2\cos A\cos B\cos C}}$ ,

${= \dfrac{abc}{6}.\sqrt{1-(\cos ^{2}A+\cos ^{2}B+\cos ^{2}C)+2\cos A\cos B\cos C}.}$

${\text{i. e.}\cot\theta = \dfrac{(m+n)\cdot\dfrac{b}{\sin B}\cdot\cos A+na\cot B}{ma},}$

${\qquad\therefore~ \dfrac{\cot\theta}{\cot\phi} = \dfrac{mn+2n^{2}}{mn+2m^{2}};}$

${\qquad\therefore~ \dfrac{\tan\theta}{\tan\phi} = \dfrac{mn+2m^{2}}{mn+2n^{2}};}$

i. e., if ${CD'B'}$ be drawn ${\perp}$ to ${AD}$ , ${\dfrac{B'D'}{D'C} = \dfrac{mn+2m^{2}}{mn+2n^{2}}}$ ;

e. g., if ${\dfrac{m}{n} = \dfrac{1}{2}}$ , ${\dfrac{B'D'}{D'C}=\dfrac{2}{5}}$ ·

from which, if ${\dfrac{\tan\theta}{\tan\phi}}$ were given, we could find ${\dfrac{m}{n}}$ from a Quadratic Equation.

I tried various values, to find one which would give rational values for ${m}$ and ${n}$ , and found that ${\tfrac{2}{3}}$ would do, as it leads to the Quadratic $2(6mn+9n^{2})-3(6mn+8m^{2}) = 0,$ in which ${(B^{2}-4AC)}$ becomes, after dividing all through by 6, ${(1^{2}+4.4.3)}$ , i. e. 49.

whence ${\dfrac{m}{n}=\dfrac{-1\pm 7}{8}=\dfrac{3}{4}}$ ; which solves the Problem 'Given a Triangle ${ABC}$ , having the tangents of its angles equal to 1, 2, 3: divide ${BC}$ at ${D}$ , so that, if ${AD}$ be joined, and ${CD'B'}$ drawn ${\perp}$ to it, the ratio ${\dfrac{B'D'}{D'C}}$ may be ${\dfrac{2}{3}}$ '. The answer is 'Divide it so that ${\dfrac{BD}{DC}=\dfrac{3}{4}}$ '.

${\qquad\qquad\qquad\qquad\qquad+ (4a^{2} + 4b^{2} + c^{2}) = 9(a^{2} + b^{2} + c^{2})}$ '

$\begin{aligned} &= \dfrac{1}{9}·\{(a^{2}+4b^{2}+4c^{2})+(4a^{2}+b^{2}+4c^{2})+(4a^{2}+4b^{2}+c^{2})\};\\ &= \dfrac{1}{9}·\{(a^{2}+4b^{2}+4c^{2}+8bc-4ca-4ab)\\ &\quad\qquad+ (4a^{2}+b^{2}+4c^{2}-4bc+8ca-4ab)\\ &\quad\qquad+ (4a^{2}+4b^{2}+c^{2}-4bc-4ca+8ab)\};\\ &= \dfrac{1}{9}·\{(-a+2b+2c)^{2} + (2a-b+2c)^{2} + (2a+2b-c)^{2}\};\\ &= \left(\dfrac{-a+2b+2c}{3}\right)^{2} + \left(\dfrac{2a-b+2c}{3}\right)^{2} + \left(\dfrac{2a+2b-c}{3}\right)^{2}. \end{aligned}$

${\therefore}$ , if ( ${a+b+c}$ ) be a multiple of 3, so also is ${(-a+2b+2c)}$ ;

Also it may be proved that, if ${\dfrac{-a+2b+2c}{3}}$ be equal to ${a}$ , or ${b}$ , or ${c}$ , then ${a}$ , ${b}$ , ${c}$ can be arranged in ${A. P.}$

Hence, contranominally, if ${a}$ , ${b}$ , ${c}$ can not be arranged in ${A. P.}$ , the 2 sets of squares have no common term.

${a^{2}}$	${b^{2}}$	${c^{2}}$	${\left(\dfrac{-a+2b+2c}{3}\right)^{2}}$	${\left(\dfrac{2a-b+2c}{3}\right)^{2}}$	${\left(\dfrac{2a+2b-c}{3}\right)^{2}}$
${1^{2}}$	${4^{2}}$	${4^{2}}$	${5^{2}}$	${2^{2}}$	${2^{2}}$
${3^{2}}$	${4^{2}}$	${8^{2}}$	${7^{2}}$	${6^{2}}$	${2^{2}}$
${4^{2}}$	${5^{2}}$	${9^{2}}$	${8^{2}}$	${7^{2}}$	${3^{2}}$

Through ${P}$ draw ${PD}$ parallel to ${AB}$ ; from ${DC}$ cut off ${DE}$ equal to ${AD}$ ; join ${EP}$ , and produce it to meet ${AB}$ at ${F}$ .

Because, in Triangles ${PFH}$ , ${PEG}$ , ${PF=PE}$ , and $\angle FPH = \angle GPE,\quad \text{and}\quad \angle PFH \gt \angle PEG,$

${\therefore PH \gt PG}$ , and ${\text{Triangle} ~ PFH \gt \text{Triangle} ~ PGE}$ .

Take ${O}$ as origin, the ${X}$ -axis ${\parallel}$ to ${AD}$ , and the ${Y}$ -axis to ${AB}$ ; and let ${JK}$ be part of the ${Z}$ -axis.

Let equation to plane containing Triangle ${CDG}$ be $x\cos \alpha +y\cos \beta +z\cos \gamma -p=0,$ where ${p}$ is length of perpendicular dropped, from ${O}$ , upon this plane, and meeting it somewhere in ${LG}$ .

Hence we can find ${p}$ from equation to ${LG}$ , in the ${XZ}$ -plane, which will be $x\cos \alpha +z\cos \gamma -p=0;$ now this line contains ${L}$ , whose co-ordinates are [Pg 93] ${\left(1, \dfrac{1}{2^\tfrac{1}{4}}\right)}$ , and ${G}$ , whose co-ordinates are ${\left(\sqrt{2}, -\dfrac{1}{2^\tfrac{1}{4}}\right)}$ ;

and ${\quad\sqrt{2} \cdot \cos \alpha - \dfrac{1}{2^\tfrac{1}{4}} \cdot \cos \gamma-p=0}$ ;

${\therefore (\sqrt{2}-1) \cdot \cos \alpha = \dfrac{2}{2^\tfrac{1}{4}} \cdot \cos \gamma = 2^\tfrac{3}{4} \cdot \cos \gamma}$ ;

${\therefore \dfrac{\cos \alpha}{2^\tfrac{3}{4}}=\dfrac{\cos \gamma}{\sqrt{2}-1}=\dfrac{1}{\sqrt{2^\tfrac{3}{2}+3-2^\tfrac{3}{2}}}=\dfrac{1}{\sqrt{3}}}$ ;

${\therefore \cos \alpha = \dfrac{2^\tfrac{3}{4}}{\sqrt{3}},\, \cos \gamma = \dfrac{\sqrt{2}-1}{\sqrt{3}}}$ ;

${\therefore p = \dfrac{2^\tfrac{3}{4}}{\sqrt{3}} + \dfrac{\sqrt{2}-1}{2^\tfrac{1}{4}\cdot\sqrt{3}} = \dfrac{\sqrt{2}+1}{2^\tfrac{1}{4}\cdot\sqrt{3}}}$ .

${\therefore}$ volume of pyramid, whose base is ${CDG}$ and whose vertex is ${O, = \dfrac{\sqrt{2}+1}{3\cdot 2^\tfrac{1}{4}}}$ ;

Also volume of pyramid, whose base is ${ABCD}$ , and whose vertex is ${O}$ , = ${\dfrac{4}{3 \cdot 2^\tfrac{1}{4}}}$ ;

${\therefore}$ volume of solid = ${\dfrac{8(2+\sqrt{2})}{3\cdot 2^\tfrac{1}{4}} = \dfrac{8 \cdot 2^\tfrac{1}{4} \cdot(\sqrt{2}+1)}{3}}$

Let ${ABC}$ be the given Triangle, and ${O}$ the given Point; and let ${OD}$ , its distance from ${BC}$ , be less than either ${OE}$ or ${OF}$ , its distances from ${CA}$ , ${AB}$ .

Draw a line ${GH}$ equal to ${OE}$ , and ${GK \perp}$ it and equal to ${OF}$ ; and join ${HK}$ ; and about the Triangle ${GHK}$ describe a Circle; and place in it a line ${KL}$ equal to ${OD}$ ; and join ${LH}$ .

Because sqs of ${KL}$ , ${LH}$ = sqs of ${KG}$ , ${GH}$ , and that ${KL}$ is less than either ${KG}$ or ${GH}$ , ${\therefore LH}$ is greater than either;

${\therefore}$ a Circle, with centre ${O}$ , and radius equal to ${LH}$ , will cut all three Lines, in two Points each. Describe this Circle.

Calling the angles ${\dfrac{1}{x}}$ , ${\dfrac{1}{y}}$ , ${\dfrac{1}{z}}$ , of 360°, we must have $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \tfrac{1}{2};$ an Indeterminate Equation with 3 unknowns.

Now, if ${\dfrac{1}{y} = \dfrac{k}{k+l} \times \tfrac{1}{6}}$ , ${\dfrac{1}{z}}$ will = ${\dfrac{l}{k+l} \times \tfrac{1}{6}}$ :

(N.B. It is assumed that the fractions ${\dfrac{k}{k+l}}$ , ${\dfrac{l}{k+l},}$ are in their lowest terms.)

Let ${\dfrac{1}{y}~ \text{be}\, \nless\, \dfrac{1}{z}}$ . Then ${\dfrac{k}{k+l}\, \nless\, \tfrac{1}{2}}$ .

Then its possible values are ${\tfrac{1}{2},~ \text{so that}~ \dfrac{l}{k+l} = \tfrac{1}{2}}$

This gives 5 sets of values for ${\tfrac{1}{x}}$ , ${\tfrac{1}{y}}$ , ${\tfrac{1}{z}}$ , viz.: $\tfrac{1}{3}, \tfrac{1}{12}, \tfrac{1}{12};\quad \tfrac{1}{3}, \tfrac{1}{9}, \tfrac{1}{18};\quad \tfrac{1}{3}, \tfrac{1}{8}, \tfrac{1}{24}; \quad\tfrac{1}{3}, \tfrac{1}{10}, \tfrac{1}{15};\quad \tfrac{1}{3}, \tfrac{1}{7}, \tfrac{1}{42}.$

(2) Let ${x = 4}$ . Then ${\dfrac{1}{y} + \dfrac{1}{z} = \tfrac{1}{4}}$ , and, as before, ${k}$ can only be 1, or 2, or 4, and the same is true of ${l}$ . Hence the[Pg 96] possible values for ${\dfrac{k}{k+l}}$ are ${\tfrac{1}{2}}$ , so that ${\dfrac{l}{k+l} = \tfrac{1}{2}}$

This gives 3 more sets of values for ${\tfrac{1}{x}}$ , ${\tfrac{1}{y}}$ , ${\dfrac{1}{z},}$ viz.:

$\frac{1}{4}, \frac{1}{8}, \frac{1}{8}; \quad\frac{1}{4}, \frac{1}{6}, \frac{1}{12};\quad \frac{1}{2}, \frac{1}{5}, \frac{1}{20}.$

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 ${\dfrac{k}{k+l}}$ are ${\tfrac{1}{2}}$ , so that ${\dfrac{l}{k+l} = \tfrac{1}{2}}$

This gives 2 sets of values for ${\dfrac{1}{x}}$ , ${\dfrac{1}{y}}$ , ${\dfrac{1}{z},}$ viz.:— $\tfrac{1}{5}, \tfrac{1}{5}, \tfrac{1}{10};\quad \tfrac{1}{5}, \tfrac{1}{4}, \tfrac{1}{20};$ but the latter (a fact overlooked in thinking out) we have had already.

Hence possible values for ${\dfrac{k}{k+l}}$ are ${\tfrac{1}{2}}$ , so that ${\dfrac{l}{k+l} = \tfrac{1}{2}}$ $\tfrac{3}{4}, \qquad\qquad\tfrac{1}{4}.$

This gives 2 sets of values, viz.:— $\tfrac{1}{6}, \tfrac{1}{6}, \tfrac{1}{6};\quad\tfrac{1}{6}, \tfrac{1}{4}, \tfrac{1}{12};$

There is no use in giving, to ${x}$ , any values greater than 6; for these would make ${\dfrac{1}{y} + \dfrac{1}{z} \gt \tfrac{1}{3}}$ ; so that one or other must be ${\gt \tfrac{1}{6}}$ ; i. e. either ${y}$ or ${z}$ must ${\lt 6}$ , and we should get old values over again.

Write ${k}$ for ${\dfrac{\alpha}{\alpha + \beta}}$ . 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 ${(1-x)}$ . Hence chance of drawing white is ${x \times 1 + (1-x) \times \tfrac{1}{2}}$ . $\begin{aligned} \therefore x + \frac{1-x}{2} &= k;\quad \therefore x = 2k-1;\\ &\therefore (1-x) = 2-2k. \end{aligned}$

Let a counter now be drawn and prove white; then chance of 'observed event,' in 1st condition, is 1, and, in 2nd condition, ${\tfrac{1}{2}}$ ;

Hence the chances, of the existence of these two conditions, are proportional to ${(2k-1) \times 1}$ , ${(2-2k) \times \tfrac{1}{2}}$ ; i. e. are proportional to ${2k-1}$ , ${1-k}$ ;

hence the chance of now drawing white, $\begin{aligned} \text{is} \quad &\frac{2k-1}{k} \times 1 + \frac{1-k}{k} \times \tfrac{1}{2};\\ \text{i. e.} \quad &\frac{3k-1}{2k}. \end{aligned}$

Hence the effect of one repetition of the experiment has been to change ${k}$ into ${\dfrac{3k-1}{2k}}$ .

Hence a second repetition of it will change $\frac{3k-1}{2k}\quad\text{into}\quad \frac{3 \times \frac{3k-1}{2k} - 1}{2 \times \frac{3k-1}{2k}};\quad \text{i. e. into}\quad \frac{7k-3}{6k-2}.$

We have now to discover the law (if there is one) for the series $k,\quad \frac{3k-1}{2k},\quad \frac{7k-3}{6k-2},$ regarding these as identical functions of 1, 2, 3.

We can write the 1st and 2nd term in the form of the 3rd, thus:— $\frac{k-0}{0\times k-(-1)},\quad \frac{3k-1}{2k-0},\quad\frac{7k-3}{6k-2}$ and, by inspection, we see that each is of the form $\frac{(2^{n}-1)\times k-(2^{n-1}-1)}{(2^{n}-2)\times k-(2^{n-1}-2)},$ 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 ${\dfrac{3k - 1}{2k}}$ . Hence the new chance will be $\frac{3\times\frac{(2^{n} - 1)\times k - (2^{n - 1} - 1)}{(2^{n} - 2)\times k - (2^{n - 1} - 2)} - 1}{2\times\frac{(2^{n} - 1)\times k - (2^{n - 1} - 1)}{(2^{n} - 2) \times k - (2^{n - 1} - 2)}};$

$\text{i.e.}\quad\frac{k\times (3\cdot 2^{n} - 3 - 2^{n} + 2) - 3\cdot 2^{n - 1} + 3 + 2^{n - 1} - 2}{(2^{n + 1} - 2)\times k - (2^{n} - 2)};$

$\text{i.e.}\quad\frac{(2^{n + 1} - 1)\times k - (2^{n} - 1)}{(2^{n + 1} - 2)\times k - (2^{n} - 2)};$

i. e. the ${\overline{n + 1}\mid^{\text{th}}}$ term of the series will follow the same law. But we know that the law holds for the ${1^{\text{st}}}$ , ${2^{\text{nd}}}$ , and ${3^{\text{rd}}}$ terms. Hence it holds universally.

Hence, after ${m}$ repetitions of the experiment, the chance of drawing white will be the ${\overline{m + 1}\mid^{\text{th}}}$ term of the above series; i. e. it will be $\frac{(2^{m + 1} - 1)\times k - (2^{m} - 1)}{(2^{m + 1} - 2)\times k - (2^{m} - 2)}.$

$\begin{aligned} \text{Then chance is} \quad &\frac{(2^{m + 1} - 1)\times \alpha - (2^{m} - 1)\cdot (\alpha + \beta)}{(2^{m + 1} - 2)\times \alpha - (2^{m} - 2)\cdot (\alpha + \beta)};\\ \text{i.e.}\quad &\frac{(2^{m + 1} - 2^{m})\alpha - (2^{m} - 1)\cdot \beta}{(2^{m + 1} - 2^{m})\alpha - (2^{m} - 2)\cdot \beta};\\ \text{i.e.}\quad &\frac{2^{m} \cdot (\alpha - \beta) + \beta}{2^{m} \cdot (\alpha - \beta) + 2\beta}. \end{aligned}$

Example—Let chance be ${\tfrac{9}{10}}$ ; and then let experiment be repeated 5 times more.

${\therefore}$ chance becomes ${\dfrac{32\times 8 + 1}{32\times 8 + 2}}$ , i.e. ${\tfrac{257}{258}}$ .

Let ${ABCD}$ be the socket. Revolve the Tetrahedron until the plane, in it, ${DOA}$ has taken the new position ${D'QA'}$ ; and let edge ${DA}$ , in its new position, meet the socket-rim ${AC}$ at ${R}$ . From ${A'}$ draw ${A'L \perp XY}$ -plane. Join ${OR}$ , and produce it to ${L}$ . And draw ${RM}$ , ${LN}$ , the ${y}$ -ordinates of ${R}$ and ${L}$ .

Call ${OM}$ , ${MR}$ , ' ${x'}$ , ${y'}$ '; and ${OA}$ , ${OR}$ , ${OD}$ , ' ${\alpha}$ , ${\alpha'}$ , ${h}$ '; and ${\angle XOR}$ ' ${\theta}$ '.

It is evident that the vertical axis of the Tetrahedron always coincides with the ${Z}$ -axis.

${\therefore \text{Equation to} ~ AC ~ \text{is} ~ y = - \dfrac{1}{\sqrt{3}}\cdot (x - \alpha)}$ ; $\text{i. e.}\quad x + \sqrt{3} \cdot y = a \qquad\text{(2)}$ Also Equation to ${OR}$ is ${\dfrac{x}{\cos \theta} = \dfrac{y}{\sin \theta} = a}$ ; $\therefore, ~ \text{at} ~ R, ~ \frac{x'}{\cos \theta} = \frac{y'}{\sin \theta} = a';\qquad\text{(3)}$ ${\therefore,~ \text{by} ~ (2),~ \a' \cdot \cos \theta + \sqrt{3} \cdot \alpha' \cdot \sin \theta = \alpha}$ ; $\therefore ~ \a' = \frac{a}{\cos \theta + \sqrt{3} \cdot \sin \theta}\qquad\text{(4)}$

Also, by similar ${\triangle}$ s ${D'QA'}$ , ${D'OR}$ , ${QA'\colon QD' \colon\colon OR \colon OD'}$

i. e. ${a \colon h \colon\colon \dfrac{a}{\cos \theta + \sqrt{3}\cdot \sin \theta} \colon h - z}$ ;

Let the Nos of bottles, taken out of the 3 days, be ' ${x}$ , ${y}$ , ${z}$ '. Let each bottle have cost ${10v}$ pence, and therefore be sold for ${11v}$ pence.

Then the Treasurer's receipts, on the 3 days, were ${(x - 1)\cdot 11v}$ , ${y \cdot 11~v - v}$ , ${(z - 1)\cdot 11~v - v}$ ; yielding, as profits (i. e. as remainders after deducting cost-price of bottles taken out), ${xv - 11~v}$ , ${yv - v}$ , ${zv - 12~v}$ . Then these 3 quantities are equal.

${\therefore}$ profit, per bottle = ${\dfrac{(3x - 33) \cdot v}{3x - 9}}$ ; and this must = 6;

§ 1. Let ${\angle BAD = k . A,\quad \angle CBE = l . B,\quad \angle ACF= m . C}$ .

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 ${A}$ , ${B}$ , ${C}$ , and ${\theta}$ (say); ${l}$ a similar function of ${B}$ , ${C}$ , ${A}$ , and ${\theta}$ ; and ${m}$ a similar function of ${C}$ , ${A}$ , ${B}$ , and ${\theta}$ ; i. e. we must have $\begin{aligned} k &= f(A, B, C, \theta),\\ l &= f(B, C, A, \theta),\\ m &= f(C, A, B, \theta). \end{aligned}$

Now, as an experiment, let $\begin{aligned} k . A &= xA + yB + zC + \theta,\\ l . B &= xB + yC + z A + \theta; \end{aligned}$ then ${kA - lB = (x - z) . A + (y - x) . B + (z - y) . C}$ ;

These conditions will be fulfilled, if we make ${y = 1}$ , and ${x = z = 2}$ ; so that $\begin{aligned} kA &= 2A + B + 2C + \theta,\\ lB &= 2B + C + 2A + \theta; \end{aligned}$ which would make $f(A, B, C, \theta) ~\text{mean} ~\frac{2A + B + 2C + \theta}{A}.$

Now this may evidently be simplified by omitting ${(A + B + C)}$ , which is constant; and we then have ${k = \dfrac{A + C + \theta}{A}}$ ; or, in a yet simpler form, by again subtracting ${180°}$ , ${k = \dfrac{\theta - B}{A}}$ .

$\begin{aligned} \text{Similarly}\quad l &= \frac{\theta - C}{B},\\ m &= \frac{\theta - A}{C}. \end{aligned}$

§ 2. We see that ${kA = \theta - B}$ , so that ${\angle}$ ${ADC}$ is evidently equal to ${\theta}$ ; and so are ${\angle}$ s ${BEA}$ , ${CFB}$ .

This gives us a geometrical construction, viz. to draw lines from ${A}$ , ${B}$ , ${C}$ , 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.

Now ${kA \ngtr A;\quad \therefore \theta - B \ngtr A}$ ; i. e. ${\theta \ngtr A + B}$ ;

and of course this is true for each of the three angles ${A}$ , ${B}$ , ${C}$ ; i. e. if ${A}$ , ${B}$ , ${C}$ , be the order of the angles in a descending order of magnitude, ${\theta \ngtr}$ supplement of ${A}$ .

Hence if ${A}$ , ${B}$ , ${C}$ , be the order in a descending order of magnitude, ${\theta \nless A}$ , and ${\ngtr 180° - A}$ .

In Triangle ${ABC′}$ , whose ${\angle}$ s are ${(\theta - B)}$ , ${(180° - \theta - A)}$ , ${(180° - C)}$ , we have $\begin{aligned} AC′ &= \frac{AB}{\sin AC′B} . \sin ABC′ = \frac{c}{\sin C} . \sin (\theta + A) = \frac{a}{\sin A} . \sin (\theta + A);\\ BC′ &= \frac{AB}{\sin AC′B} . \sin BAC′ = \frac{c}{\sin C} . \sin (\theta - B) = \frac{a}{\sin A} . \sin (\theta - B)\\ \end{aligned}$

${\therefore}$ , by symmetry, ${AB′ =\dfrac{a}{\sin A} . \sin (\theta - A)}$ .

Now ${B′C′ = AC′ - AB′}$ ; $\begin{aligned} \therefore \text{it} & = \frac{a}{\sin A}\{\sin (\theta + A) - \sin (\theta - A)\},\\ &= \frac{a}{\sin A} . 2 \cos \theta \sin A = a 2 \cos \theta. \end{aligned}$

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 ${\dfrac{\sqrt{3}}{2}}$ , 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[Pg 106] 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).

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 ${\dfrac{\sqrt{3}}{2}}$ . We must calculate its base, and then, deducting half the base of the initial Triangle (i. e. deducting ${\tfrac{1}{2}}$ ), 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 ${\dfrac{\sqrt{3}}{2} \div x = \tan 15°}$ .

Now call ${\tan 15°}$ 't'; then ${\dfrac{2t}{(1 - t^{2})} = \tan 30° = \dfrac{1}{\sqrt{3}}}$ ;

${\therefore t = \dfrac{-2\sqrt{3} \pm 4}{2} = \text{(rejecting negative value)} ~2 - \sqrt{3}.}$

${\therefore x = \dfrac{\sqrt{3}}{2 (2 - \sqrt{3})} = \dfrac{\sqrt{3}}{2} (2 + \sqrt{3}) = \sqrt{3} + \dfrac{3}{2}.}$

Deducting ${\tfrac{1}{2}}$ , we get ${(\sqrt{3} + 1)}$ as the required distance.

Hence we must go down back-edge, up again, and then about ${\cdot 7}$ 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 ${2 \tfrac{1}{2}}$ . Hence the required left-hand base-angle is $\tan^{-1}(\frac{\sqrt{3}}{2} \div \frac{5}{2});~ \text{i. e.}~ \tan^{-1}\frac{\sqrt{3}}{5}.$

Hence, for the required base-angle, we have ${\dfrac{\sin }{\cos } = \dfrac{\sqrt{3}}{5}}$ ; $\begin{aligned} &\therefore \frac{\sin }{\sqrt{3}} = \frac{\cos }{5} = \frac{1}{\sqrt{28}};\quad \therefore \sin = \sqrt{\frac{3}{28}} = \frac{\sqrt{84}}{28},\\ &= \frac{\text{rather over 9}}{28};\quad \dfrac{7\mid\underline{9\cdot}}{\tfrac{\qquad\qquad 4\mid \underline{1 \cdot 28~{\&c.}}}{\qquad\qquad\qquad\cdot 32~{\&c.}}}\\ \end{aligned}$

$\begin{aligned} \text{Now (by mem. tech.)}\quad &\sin ^{-1} \cdot 3 = 17 \cdot 45 ~\&c.°.\\ &\sin ^{-1} \cdot 4 = 23 \cdot 57 ~\&c.°. \end{aligned}$ and the required angle is about ${\dfrac{1}{5}}$ of the way from one to the other. But the difference is almost exactly 6°. Hence we must add, to the lesser, about ${1 \tfrac{1}{5}}$ degrees, or ${1\cdot 20°}$ . And the total will be about ${18\cdot 65°}$ .

${\therefore}$ the required base-angle has, for its tangent, $\left(\frac{\sqrt{3}}{2} \div \frac{7}{2}\right);~ \text{i. e.} ~\frac{\sqrt{3}}{7};$ ${\therefore \dfrac{\sin}{\sqrt{3}} = \dfrac{\cos}{7} = \dfrac{1}{\sqrt{52}};~ \therefore \sin = \sqrt{\dfrac{3}{52}} =~ \text{nearly} \sqrt{\dfrac{1}{17}}}$ ,

= nearly ${\frac{\sqrt{17}}{17}}$ . Now ${\sqrt{17} = 4\cdot 12 ~\&c.~ \therefore \sin = \cdot 24 ~\&c.}$

Now ${\sin^{-1}\cdot 2 = 11\cdot 53}$ &c°.; and we must go about half-way to the next angle, viz. ${17\cdot 45}$ &c°. The difference is about 6°; ${\therefore}$ we must add about 3°. Hence the answer is about ${14\cdot 53}$ °.

Draw ${HG}$ parallel to ${BC}$ , so that its distance from ${BC}$ may be double the distance of ${P}$ from ${BC}$ ; join ${GP}$ , ${HP}$ , and produce them to meet ${BC}$ in ${L}$ , ${M}$ . From ${L}$ draw ${LK}$ parallel to ${AC}$ ; join ${KP}$ , and produce it to meet ${AC}$ at ${N\colon}$ join ${MN}$ .

If ${P}$ lies on ${FE}$ , ${HG}$ and ${LM}$ vanish, and the Hexagon becomes a Parallelogram.

We know that, if a bag contained 3 counters, 2 being black and one white, the chance of drawing a black one would be ${\tfrac{2}{3}}$ ; and that any other state of things would not give this chance.

Now the chances, that the given bag contains ${(\alpha)~ BB}$ , ${(\beta)~ BW,~ (\gamma)~ WW, \text{are respectively} \dfrac{1}{4}, \dfrac{1}{2}, \dfrac{1}{4}}$ .

Then the chances, that it contains ${(\alpha)~ BBB, ~(\beta) ~BWB}$ , ${(\gamma)~ WWB}$ , are, as before, ${\dfrac{1}{4}}$ , ${\dfrac{1}{2}}$ , ${\dfrac{1}{4}}$ .

Hence the chance, of now drawing a black one, $= \frac{1}{4} \cdot 1 + \frac{1}{2} \cdot \frac{2}{3} + \frac{1}{4} \cdot \frac{1}{3} = \frac{2}{3}.$

Hence the bag now contains ${BBW}$ (since any other state of things would not give this chance).

Hence, before the black counter was added, it contained ${BW}$ , i. e. one black counter and one white.

WORKS BY C. L. DODGSON.

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.

EUCLID AND HIS MODERN RIVALS. (First published in 1879.) Second Edition, 1885. Crown 8vo, cloth. Price 6s.

SUPPLEMENT TO FIRST EDITION OF "EUCLID AND HIS MODERN RIVALS," containing a Notice of Henrici's Geometry, together with Selections from the Reviews. (First published in 1885.) Crown 8vo, sewed. Price 1s.

EUCLID, BOOKS I, II, with Notes. (First published in 1882.) Sixth Edition, 1888. Crown 8vo, cloth. Price 2s.

Part I. A New Theory of Parallels. (First published in 1888.) Fourth Edition, 1895. Crown 8vo, cloth. Price 2s.

Part II. Pillow-Problems, thought out during Wakeful Hours. (First published in 1893.) Fourth Edition, 1895. Crown 8vo, cloth. Price 2s.

TRANSCRIBER’S NOTES

For the sake of clarity, the notation used by the author for ${\Large\unicode{0x2ad9}}$ and ${\Large\unicode{0x2313}}$ has been replaced with modern terminology.

Due to a lack of distinction between 'a' and ' ${\alpha}$ ' in the printed version, certain expressions have been corrected to avoid potential mathematical errors.

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

(1) ${ABC}$ ,	(4) ${BCA}$ ,
(2) ${ACB}$ ,	(5) ${CAB}$ ,
(3) ${BAC}$ ,	(6) ${CBA}$ .

(1)	120°,	30°,	30°;
(2)	120°,	40°,	20°;
(3)	120°,	45°,	15°;
(4)	120°,	36°,	24°;
(5)	120°	${51\tfrac{3}{7}}$ °,	${8\tfrac{4}{7}}$ °;
(6)	90°,	45°,	45°;
(7)	90°,	60°,	30°;
(8)	90°,	72°,	18°;
(9)	72°,	72°,	36°;
(10)	60°,	60°,	60°.

The Project Gutenberg eBook of Curiosa mathematica, part II

Curiosa Mathematica

PREFACE TO FOURTH EDITION.

PREFACE TO SECOND EDITION.

INTRODUCTION.

FOOTNOTES:

CONTENTS.

SUBJECTS CLASSIFIED.

PILLOW-PROBLEMS.

CHAPTER I.

FOOTNOTES:

CHAPTER II.

CHAPTER III.

WORKS BY C. L. DODGSON.

${\tfrac{2}{3}}$ ,	${\tfrac{1}{3}}$
${\tfrac{3}{4}}$ ,	${\tfrac{1}{4}}$
${\tfrac{3}{5}}$	${\tfrac{2}{5}}$
${\tfrac{6}{7}}$ ,	${\tfrac{1}{7}}$ .

${\tfrac{2}{3}}$ ,	${\tfrac{1}{3}}$
${\tfrac{4}{5}}$ ,	${\tfrac{1}{5}}$ .

${\tfrac{2}{3}}$ ,	${\tfrac{1}{3}}$
${\tfrac{5}{6}}$ ,	${\tfrac{1}{6}}$ .