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of 60°. The curve go3 : ai sin 38 becomes (by turning the prime radius through an angle of 30°), 73 = ao cos 30, one of the important class of curves cos mo. (See SALMON'S Higher Plane Curves, p. 76, &c.)

m

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 QS=A, B, C ..., RS = A', B', C ... Considero ponia i fasci UEA, B, C ..., U'=A', B', C ..., essi sono prospettivi per avere l’elunto unito UU', quindi i raggi corrispondenti UA, Ū'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 passera 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 U=A', B', C..., U'EA, 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 ca se ; and therefore, &c.

+yi

+yi

+ di

X

{

du } = 0.

X2

+22

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, 91, 91, w.), (x2, Y2, Z2, w,) be the poles of L and M respectively with regard to U. Then this equation of V may be written

dU
k

dU
dU

dU
dac dy dz dw

dU dU dU
+Y2

+ W2
dx

dz dy

dw Now (x1, Y , %1, wn) satisfies this equation; therefore, remembering Euler's theorem of homogeneous functions, we have

AU

dU

dU U,+2kU, { 12

} 0, dzi

dU that is

1+2k {x,

} = 0.

'dxi Therefore, multiplying through by U2, we have dU dU

dU dU Uk

}{

}

= 0; dya

dyi that is to say, (x2, Y2, 72, wa) satisfies V = 0.

dU

+ Yr dyr

+ Z2

dxi

+ W2

dwi

+

X2

dxa

+42

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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 0 (1 + cos2o) -1, where ø 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. Measuring s from the cusp and o from the vertical cuspidal tangent, the equation of the curve will be s= a tan2 p. If then T be the tension at any point and R the normal resistance at that point per unit-length (weight w) of chain, DT = wds cos Q, and Tdp + wds sin • Rds. Hence dᎢ

2wa tan o sec ; or T 2wa sec 0 (no constant being required for do To 2wa). Thus T cos o 2wa, which proves (2). Substituting for T, Rsin o = w, which proves (1).

To prove (3), measure ø from the horizontal tangent at the vertex. Then dT = wds sin p; Tdo = Rds + wds cos 0 ; and (by the condition given) R cos o Thus T = wp sec 0 (1 + cos* o), and

dp

tan o cos 20 (1 + cos2 p)-1. do

ds Integrating, we have pa = 2a cos 0 (1 + cos2 p)- $ ;

do consequently T = 2wa (1 + cos2 v)-1, and To = 72. wa. A second integration gives 8 = a sin q (1 + cos2o) -1.

pol

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II. Solution by the PROPOSER. The equations for the first are - kg cos o 0,

- R-kg sin o, ds

ds ♡ being the angle which the tangent at a point P makes with the vertical, or at the corresponding point p of the catenary with the horizon; also 3, the arc VCP, = a sec? o, whence dT

= kg. 2a sec o tan d, or T = kg. 2a seco,

Fig. 1. no constant being wanted, since T = kg.2a at the cusp C, where 0 = 0, or T cos 2kga, that is, the resolved vertical tension is constant.

do

2a seco Again, R = kg sin 0 + T

= kg ds

¢

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ку

= kg

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cosa o

1 φ

)

sin o

T = KgY,

sin

sin whence R sin p = kg, or the resolved vertical pressure per unit is the weight of a unit-length of the chain.

In general, if y be the height above some fixed horizontal line, we have dT dy

к = 0, ds ds no constant being needed if we take the

Fig. 2. line properly; and T R+ kg cos o

kg

+ kg cos , ds

cos o
if the resolved vertical pressure per unit be kg; whence

dsin o cos , or T (1+ cos2p)* = C = nge,
Tdo 1 + cosa o

do

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c

or

do

ds
c sin o cos •

2c sin o
sin
(1 + cos2 )!!

(1 + cos? )!' (1 + cosa o)" and when p=0, y = $c 12, or the directrix is at a depth $c v2 below the vertex. The length of the axis is c(1-$v2),

pfe sino o do and the base = 2c

Jo (1 + cos? p) which is less than £nc and greater than inc. 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.

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139. Nore 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 i. 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 0, and P and Q are two random points within the circle; find the chance that the triangle POQ is less than a given area.

y, p =

I. Solution by E. B. Seitz. Let OP = x, OQ

radius, a = side of square equivalent to

2a? given area, ZPOQ = 0, sin-1

= 0,

P1. Then it is required to find the chance that xy sin 0 < 2a”. This will be

2a? 80 when x < = x1, for all admissible values of 0 and y. When x > x1,

sin-1 (2)

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=

the favourable cases occur from y=0 to y=r for all values of 0 from 0 to Ø, and for the supplementary values ; and from y=0 to y

2a2

- 91, for all values of e from o to žr, and for the supplementary values. Hence the required chance is

X sin 0

2 p=

7374

..

27x dx de y dy

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II. Solution by the PROPOSER. The chance will be the same if we restrict Pand 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. McColi's Calculus of

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

AQ () ao a (92) = A ( (xa) ao a(v2), in which A = 41.001.0 X 7.0, and Q = p'(}xy sin 0 — ~~) = 42, The denominator of the above fraction is

d (x2) dd (y) so that the required chance = AQ .7 « dx [ s ydy.

Now AQ = 42.1.00.0 21.0, and Y24.11 Yz, a + y),B,

*81

"Y1

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