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Now, if ru, r, be the roots of (1), we have r; +r2 = 2d cos 8 ; and since (r, e) satisfies (2), we have

(2d cos 0–77) (a 0 + b sin e) +n = 0; or, in rectangular coordinates,

(2x2 + y2 – 2dx) (ax +by) +c= 0, a circular cubic, having a double point at the origin, and its double focus at the point (d, 0), the centre of the circles.

(2). Calling the fixed point N, draw a normal PNG cutting the axis in G; then, if S be the focus, a circle centre S and radius SP will pass through G; and, considering the series of concentric circles formed by taking successive positions of PNG, we see that the points G range along a fixed straight line, and that therefore, by (1), the locus of P is a circular cubic having a double point at N and the double focus at S.

5502. (By W. S. B. WOOLHOUSE, F.R.A.S.)–(1) Two chords are drawn in a circle; all that is known, appertaining to them, being that they intersect within the circle; determine the respective probabilities that a third random chord shall intersect neither, only one, or both of thern within the circle, (2) Two chords are drawn in a circle; all that is known, appertaining to them, being that they do not intersect within the circle; determine the respective probabilities before stated.

I. Solution by Professor HENRY STANLEY Monck, M.A. 1. Let the ends of the two chords be numbered in order (i.e., in the same direction) 1, 2, 3, 4. Since they intersect, the chords must be formed by joining 1, 3 and 2, 4. We now draw a third chord. Let one of its ends fall between 1 and 2 (by properly selecting one starting point we can always provide that this shall be so). Then the other extremity being taken altogether at random, and the points 1, 2, 3, 4 being random points, the chances are equal that it lies between 1 and 2, between 2 and 3, between 3 and 4, and between 4 and 1. In the first of these cases the third chord will not intersect either of the former, in the second and fourth it will intersect one of them, in the third both. Hence the chances are that it will intersect both ả, that it will intersect neither \, that it will intersect one only 1.

This differs from Miss BLACKWOOD's result (in Quest. 5461, see Reprint, Vol. XXVIII., p. 109), as, taking the chance of two chords drawn at random intersecting as ķ, the chance of three intersecting each other would be $.x = 1; instead of th: For brevity, I speak of intersection within the circle as intersection simply.

2. Since the two chords do not intersect, must be connected with 2, and 3 with 4, or else 1 with 4, and 2 with 3. It will be enough to consider the former, as the points are arbitrary. Let one extremity of the new chord lie between 1 and 2, as before. The chances are again equal that the other will lie between 1 and 2, 2 and 3, 3 and 4, and 4 and 1, and the chances will be the same as in the former case. But it is now as probable that the first extremity of the new chord will lie between 2 and 3 as between 1 and 2; and in the latter case if the second extremity lies between 1 and 2 or 3 and 4, there will be one intersection, while, if it lies between 2 and 3 or 1 and 4, there will be none. Taking the whole eight cases together, therefore, the new chord will only intersect both the former chords in one, it will intersect one only in four, and none in three. The chances therefore in this instance are still į for a single intersection, ģ for intersecting both, and for intersecting neither.

[Mr. WOOLHOUSE remarks that “this solution appears plausible, but is not correct. It would be true if 1, 2, 3, 4 were fixed and equidistant. But when 1, 2, 3, 4 vary, it is fallacious.” On communicating this remark to Professor Monck, he sent thereon the following comments :-“I certainly do not assume the four points to be fixed points, and cannot see where the error in my solution lies. The whole dispute seems to me to turn on whether certain chords ought to be counted twice over or once only: Something may be said for Mr. Woolhouse's way of counting (which is really counting the chord once when both extremities are in the same arc, and twice when they are in different arcs), but he has not said it. Suppose for example the question were, what is the chance that a random chord will intersect a fixed diameter, is it not clearly ?? Yet this can only be reached, I apprehend, by counting the chords which intersect the diameter twice, and those which do not intersect it once only.” See also a further Note (135) on p. 62 of this volume.]

II. Solution by the PROPOSER. Let a, b, y, o denote the four arcs into which the circumference of the circle is divided by the two chords, as shown for the two cases of the problem in the the annexed diagrams. Also let P, P' designate the random points connected by the new chord. On each arc the number of possible positions of a random point may be estimated by its length. This being premised, the total number of positions of P, P are each 27, and the total number of chords, reckoning two, viz., PP', P'P, for each pair of points, is (25)2 = (a +8+9+8)2 a’ +82 +0° +8* + 2aB + 2ay + 2ad

+ 2By +238 + 298......... (1). This development, in its several terms, exhibits an analysis of the corresponding classes of chords. Thus, az expresses the number of chords having P, P' both on the arc a; 2aB expresses the number of chords having the points in the arcs a and B, viz., P in a, P' in B, and P' in a, Pin B. The magnitudes of the arcs a, b, y, 8 being unknown, it becomes requisite to effect a summation of the number of positions for every possible set of values, the same being known to be equally probable. For this purpose the limits of integration must be such as to exclude negative values, a, b, y, or 8. Let

L= a +B
M = a +B+g
21 = a +++.

7

Then, taking a, L, M, successively, as variables, we shall have the fol. lowing limits, viz., a from 0 to L, L from 0 to M, and M from 0 to 27. Hence, for the total number of chords, we obtain, remembering the limits just stated,

LALIM

ModM

(27)

:(2). 2

2.3 For the number (am) of chords contained by the arc a,

1 aodadLIM = LLIM

Mчам

(27) 3.4

3.4.5 Also, for the number (2ab) of chords connecting the arcs a and B,

2a (L-a) dadLdM

SSSda

SSS

(3).

SSS 2aBdadLAM = SSS

L
1

(275

.(4). 3.4

3.4.5 The probability of the chord PP' falling wholly on the arc a is therefore (3)

ty; and the probability of its falling upon the arcs a and B is (2) (4)

to. Similarly the probability of PP' falling upon any one, or upon (2) any specified two, of the arcs a, b, y, 8 is '; and thus the combinations severally represented by the ten terms of (1) have an equal probability. Now, referring to the diagram showing the first case of the problem, we perceive by inspection that the combinations are

for no intersection, a”, B2, 79, 82;
for one intersection, 2aB, 2By, 298, 2að;

for two intersections, 2ay, 288; the respective probabilities of which are therefore to, no, no, that is $, 5, š, Again, turning to the second diagram, the combinations are

for no intersection, a, B2, 72, 82, 288;
for one intersection, 2aß, 287, 298, 2að;
for two intersections,

2ar ; the respective probabilities of which are therefore to, io, to, that is, }, }, to

Note. Estimating the total intersections of the three chords in the two cases of the problem, the results obtained show that the respective probabilities are as stated hereunder:

INTERSECTIONS.

1 2 3

Case (1).

승 Case (2). In the problem of three random chords, before any chord is drawn, it is known that the chance of Case (1) occurring is , and that of Case (2) is s. Therefore, multiplying the two sets of values just stated respectively by

and ş, and adding the products, we find that for three random chords the probabilities of 0, 1, 2, 3 intersections within the circle are }, }, }, 1's, respectively, the same as determined in an example contained in my general investigation of those problems for n chords. See my solution to Quest. 1894, Reprint, Vol. V., p. 110. It will be seen that the probability of

25

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14, 25

15, 26

three intersections agrees with Miss BLACKWOOD's solution to Quest. 5461, (Reprint, Vol. XXVIII., p. 109), the notes appended to which gave rise to the recent discussion on the subject of "random chords.”

The problem may be solved without the aid of the integral calculus in the following manner :

Let any six points taken in order round the circumference of the circle

CASE (1).

CASE (2).
be numbered 1, 2, 3, 4, 5,
6. Then, beginning from
the first point, it is easy,

Chiords
Third
New Chords

New inter

Third

inter- not interfrom an inspection of

chord.

inter

chord,
secting.
sections secting.

sections the diagram, to make up the annexed tables for

56 13, 24

0 the two cases.

56 In that

12, 34 46 1

35 46 relating to Case (1), the

26 45 0

36 45 first column contains

36 2

45 36 every pair of chords that

26 35 1

46 35 intersect; the second

35 26 1

56 34 column contains the

36 25

13,45

26 1 third chord, or that con

34

46 25 2 necting the two remain

36 24 1

56 24 ing points; and the last

46 23

56 0 column shows the num

56 23 ber of intersections made

15, 23

46 by the third chord. The

24 36 table appertaining to Case (2) is made up in

34 26 like manner, only the first column here contains

16, 23 every pair of chords that do not intersect. For

24 35 the system of six points, as regards Case (1) we

25 34 0 have ten combinations, viz., four with no inter

34 26 section, four with one intersection, and two with

35 24 1 two intersections, the respective probabilities of

45 23 0 which are therefore mo to, ko, or 3, 5, ž. In Case (2) there are twenty combinations, viz. ten with no intersection, eight with one intersection, and two with two intersections, the respective probabilities of which are therefore zo, do, or }, , to, as before found.

As Case (1) exhibits ten combinations, and Case (2) exhibits twenty combinations, the respective probabilities of their promiscuous occurrence are } and .

The argument is completed by conceiving the six points to vary so as to severally occupy every position on the circumference. It is evident that every possible formation of diagram must then result for the two cases ;

and as the combinations arising from each separate system of six points give the foregoing probabilities, it is evident that those probabilities will remain unaltered when all the combinations are included in one total.

14,23

45

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III. Solution by ELIZABETH BLACKWOOD. Assuming that a random chord means the line joining random points on the circumference, let the random points taken in order be A, B, A', B', so that the intersecting chords are AÃ' and BB'. The circumference is thus divided into 4 random parts, namely the arcs AB, BA', A'B', B’A; the point C is equally likely to fall on any one of these, and the respective chances will remain the same if we restrict C to one of them, say to the arc AB.

Let P1, P2, P3 be the respective chances required. Of the four random arcs into which the circumference is divided the only favourable position of C for the chance P, is the arc AB. Hence P. = 1. There are two favourable positions of C for the chance pa, namely, the arcs BA' and B'A. Hence pz = . The only favourable position of Cofor the chance P3 is the arc A'B'. Hence P3 = 4. Thus P1, P2, P3 = 1, 1, 1.

a

135. NOTE ON MR. WOOLHOUSE'S SOLUTION OF HIS QUESTION 5502 (See

pp. 59–61 of this volume). By PROFESSOR Monck, M.A. The chance that a random point P will be in the arc a is undoubtedly

and the chance that a second random point P' will also lie in a +B+y+8 the same arc being identical, the chance that the chord PP will lie in it

a? is

But if we are to count P'P as a distinct chord from (a +B++8)2

a? PP, the chance that P'P will be in the arc a is also

and

(a + B+9+8) therefore the chance that our geometrical chord (which includes both PP'

aa

2a2 and P'P) will lie in the arc a is not

but

(a +B++8) (a + B + 7 + 8)2 On the other hand, if we are to reckon PP' and P'P as a single chord, the chance that one extremity will lie in the arc a, and the other in the arc B, 2aß

αβ is not

but
(a + B +7+8)2 (a + B+y+8)**

136. MISS BLACKWOOD'S REPLY TO HELEN THOMSON'S VERSES ON

“ RANDOM CHORDS.” (See p. 40 of this volume.)
With arrows fashioned by the Muse's hand
Your Roderick's foe you venture to withstand ?
So be it, Helen; be it even so ;
Another string is fastened to my bow;
Another arrow fits unto this string,
Expressly sharpened for your Muse's wing.
But ere my bow-string's awful twang you hear
I'd like to ask one question, Helen dear.
Whence comes your“ attribute of subject chord?
To trace this attribute let's trace the word.
The archer and the minstrel both require
(For trusty bow, or spirit-stirring lyre)
Their “ chords” well fastened at the ends and tight,-
Else, sure, nor shaft nor music would go right.
My chord is tight, my arrow right will go-
No sluggish shaft, as you will quickly know.
All proper chords are finitefinite, mark;
Each chord is finite, fastened to its arc.
The arc comes first, and then the chord co nes next ;

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