... {r (a cos 0 + b sîn 0) + ah + bk }2 + 2r (d cos 0 + e sin 0) + 2dh + 2ek +ƒ = 0; ... r2 (a cos 0 + b sin 0)2+2r {(ah +bk) (a cos e + b sin ◊) + (d cos ◊ + e sin 0)} +2dh+2ek+f+ (ah + bk)2

If TPQ be a tangent, we have

(a cos 0 + b sin 0) { (ah + bk)2 +2dh + 2ek +ƒ}

=

= 0.

{(ah+bk) (a cos 0 + b sin e) + d cos e + e sin 0}2;

.. (a cos 0+b sin 0)2 (2dh + 2ek +f)

=

2 (ah+bk) (a cos 0 + b sin ◊) (d cos ✪ + e sin ✪) + (d cos 0+e sin 0)2;

.. (a+b tan 0)2 (2dh+2ek +ƒ)

=

2 (ah+bk) (a + b tan 0) (d+e tan 0) + (d + e tan 0)2.

But if 01, 02 are the two angles thus found,

.. b2 (2dh+2ek +ƒ) −2be (ah +bk) — e2 + a2 (2dh + 2ek +f)

therefore the directrix is

— 2ad (ah+bk) — ď2 = 0 ;

2bx (bd—ae) + 2ay (ae — bd) + b2ƒ— e2 — d2 + a3ƒ = 0,

or

bx-ay

=

e2+d2 - (a2+b) ƒ
2 (bd-ae)

5320. (By J. J. WALKER, M.A.) If normals to the ellipse b2x2 + a2y2 — a2b2 = 0 be drawn from any point on the curve

(a2x2+b2y2-c4)3 + 54a2b22x2y2 = 0,

prove that they form an harmonic pencil.

Solution by C. LEUDESDORF, M.A.

The four points where the normals drawn from (X, Y) to the ellipse cut the ellipse are given as its intersections with the conic

c2xy-a2yX+b2xY = 0.

Eliminating y between this and the equation to the ellipse, we find c2x2-2c2a2x3X+ (b2Y2 + a2X2—c1) a2x2+2c2a1Xx— a¤X2

Since the equation to the normal at (x', y') is a2xy' — b2yx' = c2x'y', the roots of (1) are times the intercepts made on the axis of x by the four

a2

c2

normals that can be drawn through (X, Y). If, then, these normals form a harmonic pencil, the invariant T of (1) must vanish, so that

—2a€ (b2Y2 + a2X2-4)3 + 108a10c4X4-108a8c8X2

— 36a8c4X2 (b2Y2 + a2X2 — c1) — 72a8c4X2 (b2Y2+a2X2—c1) = 0,

which reduces to (a2X2+b2Y2-c4)3+54a2b2c2X2Y2 = 0.

5173. (By H. T. GERRÁNS, B.A.)-Find the sums of the infinite series

Solution by Rev. J. L. KITCHIN, M.A.; R. TUCKER, M.A.; and others.

= [ xdx = [ + log (x2 + x + 1) −log (1−x) — √/3 tan-1 2x+1
+ [+ tan-12x+1].

1-x3

and when x = 0, S= 0; therefore C√3.

Now the given series

=

— [+ { + log (2* + x2 + 1) −log (1—29) —√/3 tan-1 242+1

24

2. 2 =

√3

x3 (1 − })—x13 (} −3) + ... = (x2 + 1) tan-1x-x.

ΠΑ

+C;

5101. (By A. MARTIN, M.A.)-An auger-hole is made through the centre of a sphere; show that the average of the volume removed is, in parts of the volume of the sphere, 1-π.

Solution by the PROPOSER.

Let = radius of auger-hole, r = radius of the sphere, and V = volume removed. The volume removed consists of a cylinder, radius x and altitude 2 (2-2), and two equal spherical segments, base diameters 2x and heights r-(2-2). The volume of the cylinder is 2x2 (r2—x2)3, and the volume of the two segments is

wx2 {r− (r2—x2)$} +zπ{r−(r2—x2)*}3; therefore V= {r3— (x2—x2)3}. The average volume removed

=

5367. (By ELIZABETH BLACKWOOD.)-If X, Y, Z be random points taken respectively in a sphere, in a great circle of the sphere, in a radius of the sphere; show that the respective chances of X, Y, Z being farthest from the centre are as 3, 2, 1.

Solution by the PROPOSER.

The problem may be enunciated analytically as follows:-Given that

all values of x, y, z between 0 and 1 are equally probable, and no other values possible, find the respective chances of x, y, or z being the greatest of the three variables. The respective chances are easily seen to be

d

dz =

」ae)」「a(6)」adak,」「(8) 」t (e) 」 66 ¥,

0

dz =

0

5445. (By C. H. PILLAI)-A point D is taken in the side BC produced of an equilateral triangle ABC, and CE is drawn cutting AD in E, and making the angle ACE = ADC: prove that

(1) AE+EC = BE; (2) AE.EC BE2 ~ BC2; (3) AE: EC=BC: CD.

=

Solution by J. O'REGAN; A. W. CAVE; and many others.

2. BE2-BC2

=

=

(BE+EC)2-AC2

(AE2 + EC2 + 2AE. EC)

— (AE3 + EC2+AE. EC) = AE. EC.

3. Since ACE ADC, and the angle CAE is common to the two triangles ACE and ACD, the two triangles are similar, and therefore AE: EC = AC: CD BC: CD.

=

133. ON THE RANDOM CHORD QUESTION BY HELEN THOMSON. [See Reprint, Vol. XXVIII., pp. 106-110, Vol. XXIX., pp. 17-20.]

Unto Fitz-James made answer Roderick Dhu,
"Your hasty threat you will as quickly rue.
To fence in logic is 't that you aspire?
Abracadabra! why, you're in the mire.

A joust of this sort you can ill afford.

Here's 'random' attribute of subject 'chord';

What was 't you took at random? Points, you said.
Then where's your logic ? 'Tis knocked on the head.
My line's at random, and my line's a chord."
Thus Roderick broke Fitz-James's vaunted sword.

5480. (By NILKANTHA SARKAR, B.A.)-C is the centre of an ellipse, CB its semi-minor axis; also a circle is drawn concentric with the ellipse, and touching its two directrices, and meeting CB produced in A; determine (1) the eccentricity of the ellipse in order that CA may be bisected in B, and (2) whether the ellipse 2x + 3y2=c2 answers the condition.

Solution by SIMONELLI RUGGERO; CECCHINO VINCENZO; and others.

Sia R il raggio del cerchio. Rappresentandoci questo raggio la distanza del centro C del ellisse, dalle direttrici, avremo R a (a2-b2)-.

Supponendo preso ad arbitrio l'asse 2a, avremo per le condizioni del problema, b = 4R, quindi b2 = a2, che sostituito nell' espressione dell' eccentricità, avremo e = 2.

Perciò l'eccentricità di tutte le infinite elissi che soddisfano alle condizioni richieste, è costante ed egua le ad //2.

L'equazione proposta non soddisfa al problema, essendo e =

3.

5569. (By Professor TANNER, M.A.)—Solve the functional equation

I. Solution by Prof. MOREL; Prof. EVANS, M.A.; and others.

Cette égalité devant avoir lieu quelsque soient x et y, faisons successivement y = x, x2, an-1; il vient

Multiplions cette dernière égalité par x, et nous aurons

xp(xn) = nx" p (x). On peut l'écrire

(2)

= n.

$ (x).

xn

Ꮖ

Mais le premier membre, dans lequel n et z sont quelconques, ne change pas, quand on change n en z, et z en n.

On en déduit

Comme le multiplicateur de z est constant, et que d'autre part on a on en déduit (x) = cx loga x.

%=

loga x,

Réciproquement, je dis que toute fonction de la forme cr logar répond à la question. Car on a cxy log xy = cy. x log x + cx. y log y ; donne bien la relation

(xy)

=

· xp (y) +yo (x).

ce qui

dx

[Professor TANNER'S solution is (x) = Ax log x + Σi A¡ . * dti where A, Ai, ti are quantities that are the same for all subjects of p, and the summation includes as many values of i as you please.]

II. Solution by Prof. TOWNSEND, F.R.S.; R. E. RILEY, B.A.; and others. Differentiating with respect to x and to y successively, and eliminating p(xy), we get, immediately,

which, being true for all values of x and y, shows that each side arbitrary constant a; hence, at once, by integration,

• (x) = ax log x, o (y) = ay log y,

= an

therefore

axy log xy

the constant of integration vanishing by virtue of the given relation.

III. Solution by J. L. MACKENZIE, J. O'REGAN, and others.

¥ (e°+)

=

↓ (e°) +¥ (co), or ƒ(0+¢) = ƒ (0) +ƒ ($)• Differentiating with respect to e, ƒ' (0 + ¢) = ƒ′ (0);

==

Similarly f'(+) = ƒ'(p; hence ƒ' (0) = ƒ' (p), or ƒ'(0) = k, where k is a constant. Hence ƒ (0) k9+k'; and from the equation ƒ(0 + q) = ƒ (0) +f(p) we see that k (x) = k log x, and finally

Hence ƒ (0)

=3

ke,

5456. (By Professor MINCHIN, M.A.)-If a body P moves in a plane orbit, so that the direction of its resultant acceleration is always a tangent to a given curve, prove that, if Q is the point of contact of this tangent, p the perpendicular from Q on the tangent at P, w the angle subtended at P by the radius of curvature at Q, 0 the angle which PQ makes with a fixed

[* On this Mr. MACKENZIE remarks that he believes that "no such addition as Prof. TANNER has here given to the value of p (x) can be correct, except on the supposition that is not an independent variable, but is limited to certain periodical values."]