of 60°. The curve r3 = a3 sin 30 becomes (by turning the prime radius through an angle of 30°),

23 =

a3 cos 30, one of the important class of curves "a" Higher Plane Curves, p. 76, &c.)

m

cos me. (See SALMON'S

5590. (By S. A. RENSHAW.)—If a quadrilateral be inscribed in a conic, show that a point may be found on each of its sides such that, the four points being joined, a quadrilateral inscribed in the former will be formed whose opposite sides produced will meet on the directrix.

Solution by G. MARRO; G. VINCENZO; and others.

Questo problema si risolve in un modo più generale, considerando cioé in véce di un quadrangolo inscritto in una conica e della direttrice di questa, un quadrangolo ed una retta qualunque. Siano infatti PQRS e d il quadrangolo e la retta data. Scelgo due punti U, U' rispettivamente su PQ, PR, indi pel punto M comune alla d e alla UU', fanno passare tante rette le guali segate con QS, RS determineranno rispettivamente due punteggiate prospettive QSA, B, C..., RSA', B', C'... Considero ponia i fasci UA, B, C..., U'A', B', C'..., essi sono prospettivi per avere l'elunto unito UU', quindi i raggi corrispondenti UA, U'A'... si segheranno su di una stessa retta u". Proietto il punto N comune alla u' e alla d, da U e da U'; e sego UN con QS nel punto T, e la U'N con RS in T; la TT' passerà per M, e cosi si sará formato un quadrangolo UUTT' inscritto nel dato, e in un due coppie di lati opposti UT, U'T'; UU', TT' si segano nei punti N ed M situati nella d che e' quanto si cercava. Se si fossero costanti i fasci UA', B', C'..., U' A, B, C..., e si fosse operato come prendentemente si avrebbe ottenuta un' altra soluzione del problema. Variando poi i punti U, U' si avramo altrettante soluzioni.

5605. (By J. C. MALET, M.A.)—If a quadric V intersects another quadric U in the planes L and M, and passes through the pole of L with respect to U prove that it will also pass through the pole of M with respect to U.

I. Solution by Professor TOWNSEND, F.R.S.

Denoting by P and Q the two poles in question, by R and S the intersections of the line PQ with L and M, and by X and Y its intersections with U; then, in order to establish the property, it is only necessary to

show that the three pairs of collinear points P and Q, R and S, X and Y are three pairs of conjugates of an involution; which, as XY is cut harmonically by P and R and by Q and S, is manifestly the case; and therefore, &c.

II. Solution by E. B. ELLIOTT, M.A.; W. S. F. LONG, B.A.; and others.
The equation V = 0 must be identical with U+kLM = 0.
Now let (x1,y1, Z1, W1), (X2, Y2, Z2, w2) be the poles of L and M respectively
with regard to U. Then this equation of V may be written

Now (x, y, z, w1) satisfies this equation; therefore, remembering EULER's theorem of homogeneous functions, we have

5497. (By Professor WOLSTENHOLME.)-A heavy uniform chain rests on a smooth arc of a curve in the form of the evolute of a catenary, a length of chain equal to the diameter of curvature at the vertex of the catenary hanging vertically below the cusp; prove (1) that the resolved vertical pressure on the curve per unit is equal to the weight of a unit-length of the chain; (2) that the resolved vertical tension is constant [the chain being, of course, fixed at its highest point]; and (3) that the former property is true for a uniform chain held tightly in contact with the curve whose intrinsic equation is s = a sin o (1 + cos2 p) -, where is measured upwards from the horizontal tangent, and the directrix (or straight line from which the tension is measured) is at a depth a、/2 below the vertex.

I. Solution by R. F. DAVIS, M.A.

Measurings from the cusp and from the vertical cuspidal tangent, the equation of the curve will be sa tan2 p. If then T be the tension at any point and R the normal resistance at that point per unit-length (weight

R sin w, which proves (1).

=

=

To prove (3), measure from the horizontal tangent at the vertex. Then dT wds sin p; Tdp = Rds + wds cos ; and (by the condition = w. Thus T wp sec p (1+cos" p), and

given) R cos

p-1.

Integrating, we have

consequently T

=

Αρ

аф

P =

=

2wa (1+cos2p)-, and T。 = √2.wa.

tegration gives sa sino (1 + cos2 ) -.

A second in

if the resolved vertical pressure per unit be кg; whence

= ∞ √o ( 1 + cos2 p) #

(sin2 o do

20 Jo (1+ cos p)}'

which is less than Tc and greater than c. Of course its value is at once found from any table of elliptic functions; but the general form of the curve is obvious, being something like a cycloid with the ordinates measured from the axis reduced in a certain ratio.

139. NOTE ON PROFESSOR MONCK'S SOLUTION OF QUESTION 5502. (See pp. 58, 59 of this Volume). By W. S. B. WOOLHOUSE, F.R.A.S.

It may not be out of place here to briefly indicate the source of defect in Professor MONCK's solution. After Professor MONCK has assumed one end, say 5, of the third chord to fall between 1 and 2, I would observe that, as the total positions of 5 must range over the circumference of the circle equally with the other points, it becomes then imperative to consider 1, 5, 2, 3, 4 as five random points in the circumference taken in order; so that, in assigning different positions to the other end, 6, of the third chord, the proper reasoning will be to regard as equal the separate chances that it will lie between 1 and 5, 5 and 2, 2 and 3, 3 and 4, and 4 and 1. This mode of procedure admits a free and equal range to the several points, and leads directly to the true results.

5572. (By ELIZABETH BLACKWOOD.)-The centre of a given circle is O, and P and Q are two random points within the circle; find the chance that the triangle POQ is less than a given area.

I. Solution by E. B. SEITZ.

Let OPx, OQ = y, r = radius, a = side of square equivalent to given area, 4 POQ = 0, sin-1

=

rx

sin-1

(

=

Then it is required to find the chance that xy sin <2a2. 2a2

so when x < = 1, for all admissible values of 0 and y.

2"

Φι

This will be

When >1,

=

2a2

x sin e

the favourable cases occur from y=0 to y=r for all values of @ from 0 to p, and for the supplementary values; and from y=0 to y for all values of @ from o to π, and for the supplementary values. Hence the required chance is

The chance will be the same if we restrict P and Q to the first quadrant. Let OP = x, OQ = y, POQ = 0, r = radius of circle, and a2 = given

area.

Employing the notation and principles of Mr. McCoL's Calculus of

Equivalent Statements (with which I have recently become acquainted), the required chance is

in which A

AQ [ d (x2) ƒ ão ƒ ã (y2) ÷ A

=

=

1.0 1.0 1.0, and Q = p′(xy sin 0 — a2) — Y2,•

The denominator of the above fraction is

8

so that the required chance = AQS 2; xdx [ do [y

Now

São Sydy.

AQ = 2.1.0 1.0 1.0, and 2.1 = Y2a+Y1,B,