Yg'.1 = yg..1a, where
*5'.1 = 26.1Q, where

= X2:

Substituting, we get
AQ = + 9x1*(**,+26*2)4.0.

= + 44.1 3.1*14.0.2+ 44.1%3:582.1.0

= 43.. + 94.1%3.1% 1.2 14.1%3.5X2.0) by mere inspection of the table. We must now apply Rule 5 to each of these three terms thus :

a = P(43-) = p (x + 1) = 1,

P (25 — 2) = P(-2x + 3) = p' (2x — 3)

- p'(x – 3) Hence the first term = 43.1%5'.172'.1'.0",

which 43,1%5.1*2.0

by inspection. This term is now elementary, and we will denote it by E,.

Note.—A statement of the above form is said to be elementary when the application of Rule 5 to any of the variables will introduce either no fresh limits among the other variables, or only such as are implied in the limits already stated.

The second and third terms may be shown in like manner to be already elementary. We will denote them by E, and Ez respectively; so that we have

E, = 43.1753.132.0, E, = 44.1 3.1*1'.2 Ez = 44.1%3.52.01 and the required chance

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46 + do log, 2 = .196495 .... [This Solution, though, at the Proposer's request, subsequent in publication, was written before the second article on “Symbolical Language," which appeared on p. 100 of Vol. XXVIII. of the Reprint. A more regular order of the letters will be obtained by interchanging y. and z both in the Question and Solution. The present order was adopted because the Proposer was at first under the erroneous impression that the shortest solution would be obtained by making the coefficient of o vary first; and the first solution sent to the Editor was in accordance with this order of integration.]

5515. (By E. P. CULVERWELL, M.A.)—If y be one of the four normal distances of a point P from an ellipse, and p the parallel central perpendicular on a tangent line; prove that, if x{ipr)-1} vanishes, then P lies on the director circle of the ellipse; and state the corresponding theorem for an ellipsoid.

Solution by E. W. Symons; the PROPOSER; and others. Let the coordinates of P be (a, b), and of Q, the foot of the normal, be (ac', y); then we have

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*+ pr

goed = ('-)$+(1-B)* = 42

; = ipr; a*


and similarly for 봉 Substituting in the ellipse, we have

aRa(* + pr)* + 6*8* (a* + prjʻ = (*+pr)(8 + pr). If the coefficient of pr in this equation vanish, the sum of the reciprocals vanishes. That coefficient is

[pr][2 (a+b + + 3824) - 2 (do44 + b*af)]. Hence sum of reciprocals vanishes if a*8* [a* +82-(a + b)] = 0, that is to say, if a, ß lies on the director circle.

In the ellipsoid the sum of reciprocals evidently vanishes when (a, b) lies on an ellipsoid.

5467. (By CHRISTINE LADD.) If three conics touch each other and have a common focus, prove that the common tangent of any two will cut the directrix of the third in three points which lie on one straight line.

1. Solution by S. JOHNSTON; C. BICKERDIKE; and others. Reciprocating the theorem in Question 5356 with respect to any point, we have the theorem in question.

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II. Solution by H. W. HARRIS, D. EDWARDES, and others. First projecting and then reciprocating, we reduce the question to the following :-“Three conics touch each other at a common point C, and have two points A and B common to all; prove that the three polars of the line AB with respect to each of the conics lie on a right line." Take ABC for triangle of reference, and let the three conics be h+mn, 12 + m2 + ,

B 7
B 7

B 7 then, since they have a tangent at C (which we take as a, B) common,

4 la lz

mi ma тз But the three polars of AB, or y, have for their coordinates

(1, m, n), (12, ma, - na), (13, Mz, nz); and the condition that these should lie on a right line is

4 mm = 0,
1, т. п.

1 та 13
which is evidently satisfied if the relations above hold good.

we have

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5299. (By L. H. ROSENTHAL.)-Solve the simultaneous equations, 23 — ax2 + (1-2y) x + ayc = = 0.

.(1), cov-ary-vo-b + a) = 0..




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Solution by the PROPOSER; J. RICHARDS, M.A.; and others. To solve these equations, take the general biquadratic, viz.,

24 - ax3 + bx*— cx + d = 0, and assume it to be the product of two quadratic factors, viz.,

22 - 2x + and X? — vx+p; then, equating coefficients, we have the four equations,

λ+ ν= α, λν +μ + ρ = 0, μν + ρλ = 0, μρ = d...... (1, 2, 3, 4), between any three of which we can linearly eliminate either of the pairs 1, th, or v, p; the result being a relation between the other pair; and therefore, if we get two such relations from the four equations (1, 2, 3, 4), we shall have equations sufficient for the determination of these quantities. Now, eliminating v and p between (1), (2), (3), and (1), (2), (4), we have respectively 0

0, 1 0 i-a 1 u-b

1 μ- και

0 -d or, expanding, λ3 – αλ? + (0 - ) + αμ – = 0, and

μλ? αμλ- (μ2 όμ + d) = 0, respectively. Hence, from the foregoing, we see that the solutions of the two latter equations are 1 = a + B, u = ; where a, b, y, 8 are the roots of x4 – ax3 + bx2- cx+d 0. It is evident also from the above that there are six distinct in one of the two equations) pairs of equations of the form proposed, whose solutions are the same.

Otherwise, to solve these equations directly, assume x— ax + b = A, when the proposed equations become respectively, XA-Y (2x a) -4 = 0, ya-(y2 +d) 0

..(5, 6). Solve for y from (5) in terms of x and a, substitute for it in (6), and eliminate x between result, and x2 — ax + b-A = 0; we thus get for a the equation A3–64" + (ac-4d) A +4bd 02 – aRd = 0, the well-known equation for +70, where a, B, 7, 8 are the roots of

x4 - ax3 + bx2 — cx + d = 0........ therefore x? - ax + b + 38, and hence, substituting for a and 6 in terms of the roots of (7), we find corresponding to the root aß +yd, 1 = a+B and +8; similarly for the other two roots ay + B8 and a8 + By. Again, from (6), substituting for a, + go we get for y the values , yo, and similarly for the other values of A.

Thus we see that the equation whose roots are a + B for a biquadratic may be written in the convenient and I believe new form, (x2-ax+6)8 — 6 (x2 – ax + 5)2 + (ac— 4d) (x2- ax + b) + 4bd — -*—a?d = 0.

It is also evident that the equation in A is the same as the equation which would be obtained for u + p, from the four equations (1, 2, 3, 4), and thus answers an elimination question connected with those four equations.


; sub

Again, to obtain the equation in y whose roots are , &c., we have, eliminating a between (5) and (6) and solving for x, x =

y (ayc)

y2stituting in (6) and reducing, we get for the required equation,

(72d)2 (72by + d) + y2 (ay —c) (cy - ad) = 0. The same equations may be formed for an equation of the nth degree by supposing it to be the product of a quadratic and an equation of the (n-2)th degree; and we can get two determinants of the (n − 1)th and (n- 2jtli degrees respectively in 1 containing both , and u; between which eliminating either we get required equation in the other.

Similarly, by supposing an equation to be the product of two equations of the mth and (nm)th degrees, we can find the equations whose roots are Ema, Emaß, Emaßy, &c. &c. This method, however, becomes very troublesome in equations beyond the sixth degree.

5441. (By D. EDWARDES.)– Prove that sin 2a sin (B-y) cos (-a) + sin 2ß sin (w-a) cos (0-B) + sin 2ay sin (a - b) cos (0-7)

=tan (a + B + g-0). sin 2a sin (B -y) sin (e-a) + sin 2ß sin (y-a) sin (0 –B)

+ sin 29 sin (a–B) sin (8-9)

Solution by R. Tucker, M.A.; D. EDWARDES ; J. O. JELLY; and others. Let A =a+B+y = k; then we have

4 (numerator) - 4 {sin 2a sin (B—) cos (B+ y k) + ... +...} 2 sin 2a {sin (2B — K) – sin (29–K)} + ... + ...

cos (2a —2B+) - cos (2a + 2B – K) – cos (2a – 2y + x) + cos (2a + 2q -K) +cos (28–29 +K) - cos (2B + 29K)- cos (2B – 2a + k) + cos (2B + 2a – K) +cos (29—2a+k) -cos (2a + 2a – K) - cos (27-28 + x) + cos (2y + 2B – K) - 2 sin k {sin (2a – 2B) + sin (2B – 2y) + sin (2y—2a)}; 4 (denominator) = 2 sin 2a {cos (2y—k) —cos (2B—K)} +

– 2 cos x {sin (2a-2B) + ... + ...}; therefore &c.


+ ...

5475. (By C. TAYLOR, M. A.)—Prove the following construction for tangents to a conic :—Take a point T at a distance TN from the directrix, and divide ST in t so that St : ST = AX : TN, where A is the vertex, and X the foot of the directrix. About S draw a circle with radius SA, and from t draw tangents to the circle cutting the tangent at A in V, V'. Then TV, TV' will touch the conic.

Solution by R. TUCKER, M.A.; S. TEBAY, B.A.; and others.
We have LtSM = _ tSM',
LVSM = _ VSA = V'ST,
therefore TSt is a straight line;

ZVSZ, therefore VM is parallel to

hence we have

[For clearness' sake, the
figure has been drawn with
tangents on opposite sides of
the axis, T being the escribed
centre of the triangle VV't.]



5564. (By Professor BENJAMIN Peirce, F.R.S.)—Find the probabili. ties at a game of a given number of points, which is played in such a way that there is only one person who is the actual player, and when the player is successful he counts a point, but when he is unsuccessful he loses all the points he has made and adds one to his opponent's score.

Solution by SEPTIMUS TEBAY, B.A. The player may be successful atimes in succession, and then fail; after this he may be again successful az times, and then fail a second time; and so on for i periods. Now, in order that the player may be finally successful, there must be i-1 failures, enabling the player to score di points. Therefore, if n be the number of points in the game, we must have

aj + Ag + ... + ai = n-i +1. The equation aj + a2 + ... + di-1 = n-i+ 1 - ai, admits of (n-1-ai)!

solutions, each of which terminates with (4–2)! (0 – +1– ai)! di successes. Hence the number of points scored by the player on this hypothesis is ai


(1-2)! (n-i +1-a;)?; and putting n = 1, 2, ... n-i+1, the general term is


(n-1+2-) . x (x + 1) ... (x+i-3)

. x + x<— x ] (-2)!

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