5575. (By the EDITOR.) If of four triangles A,B,C1, A,B,C2, ABC, ABC4, the first is in perspective with the second, the second with third, the third with fourth, and the fourth with first, in such a way that the vertices of the same letters are corresponding; and if the four centres of perspective lie in a straight line: prove that the four perspective lines meet in a point.

Solution by Prof. TOWNSEND, F.R.S.

The three triangles A,A,A, B,B,B, C,C,C, taken in pairs, having a common axis of perspec

tive PQS (see the Figure, which explains itself), so therefore, by a well known property of triangles in perspective, have the three A,B,C1, ABC2, and ABC; and the three triangles  ̧ÂА, ВВВ, С3C4C, taken in pairs, having a common axis of perspective QRS, so there

B

R

fore, by the same property, have the three A,B,C3, A4B4С4, and ABC; consequently, the axis of perspective of the pairs A,B,C, and A,B,C, coincides with that of either of them and ABC, and the axis of perspective of the pair ABC, and A4B4C4 coincides with that of either of them and ABC. Again, the three triangles A2B2C2, A,B,C3, and ABC, taken in pairs, having a common centre of perspective Q, their three axes of perspective, by the reciprocal property, are concurrent, and therefore, by the above, so are those of the three pairs of triangles A,B,C, and A,B,C, A,B,C and A,B,C3, A3B3C, and A,B,C,; and the three triangles A4B4C4, ABC, and ABC, taken in pairs, having a common centre of perspective S, their three axes of perspective, by the same property, are concurrent; and therefore, by the above, so are those of the three pairs of triangles A,B,C, and A,B,C3, A,B,C and A4B4C4, ABC and A,B,C,; therefore &c. N.B.-The converse of the above property is manifestly the reciprocal of the original, and may consequently be regarded as established with it by reciprocation, either of the property itself or of the process employed for its establishment as above.

Postscript.-The above demonstration, based essentially on the supposition, since seen to be unnecessary, that the planes of the three quadrilaterals AA,AA, B ̧‚ ̧Â, С1C2C3C4, containing as they always do the common line PQRS, were coincident, may be simplified for the more general case when they are not, as follows:-The three triangles AAA, BBB, CCC, taken in pairs, having a common axis of perspective PQS, the two planes A,B,C, and A,B,C consequently, by the property above referred to, intersect in the plane ABC; and the three triangles  ̧Â4A, BBB, CC4C, taken in pairs, having a common axis of perspective QRS, the two planes A,B,C, and A4B4C4 consequently, by the same well-known property, intersect in the plane ABC; therefore the four planes A,B,C1, A2B2C2, A,B,C3, A4B4C4 intersect at a common point O in the plane ABC;

and therefore &c.

N.B.-If A', B', C' be the triad of sixth vertices of the three complete quadrilaterals A‚ ̧Â4, BBB3B4, C1C2C3C4 in the above, it appears at once from the demonstration just given that the point O lies also in the plane A'B'C', and therefore that the six planes ABC, A,B,C2, A¿B¿Cз, A4B4C4, ABC, and A'B'C' are concurrent.

5585. (By R. E. RILEY, B.A.)-If a series of circles be described concentric with an ellipse of eccentricity e, prove that the chords of contact of tangents drawn to the ellipse from points in these circles envelop a series of concentric similar ellipses of eccentricity e (2-e2)*.

be respectively the projective equation of one of the concentric circles and the tangential equation of the given ellipse whose eccentricity is e.

Assume, on the circumference of the circle, a point, as pole, whose projective coordinates are (x, y); and let (s, v) be the tangential coordinates of its polar, that is to say, of the chord of the ellipse. Then, as in BOOTH's Geometrical Methods, Vol. I., Art. 31, we shall have, from (2), x = az, y = b2; hence, substituting these values of x and y in (1), the tangential equation of the envelop required is found to be a 2 +64 v2 = r2, which designates an ellipse whose eccentricity is e (2-e2)*.

x Ч
h kc

1, a like sub

+ 2=
k

1, which (BOOTH's Methods, Vol. I., p. 1) desig

a2

b2

and that is to say,

h

k

If the pole (x, y) move along the straight line + stitution would give us, for the envelop of the polar, the tangential a2 b2 equation h nates a point whose projective coordinates are third proportionals to (h, a) and to (k, b).

5478. (By C. B. S. CAVALLIN.)-Three straight lines are drawn at random across a triangle; show that the probability that each line cuts unequal pairs of sides is 16 (a + b + c)−4▲2, where A is the area of the triangle and a, b, c its sides.

Solution by E. B. ELLIOTT, M.A.

Following Professor CROFTON's theory, the number of lines crossing the triangle is measured by the perimeter a+b+c. Consequently the measure of the number of sets of three lines crossing it is (a+b+c)3.

Again, 2a measures the number of lines which meet the side a of the triangle. Therefore the measure of the number meeting b and c, that is to say, of all the lines which cross the triangle except those which meet a,

is a + +b+c= -2ab+c-a. So also c+a-b and a+b-c measure the

numbers of lines meeting c, a and a, b, respectively. Thus

measures the number of sets of three lines, one meeting b, c, a second c, a, and the third a, b. Hence the required probability is

(b + e − a) (c+a−b) (a+b−c) 1642

(a+b+c)3

=

(a+b+c) 4'

5496. (By Professor CROFTON, F.R.S.)-Prove that the mean value of the reciprocal of the distance of any two points within a circle of radius r is

Solution by E. B. SEITZ.

Let P and Q be any two points within the circle, centre O, and while P ranges over the circle, let Q be confined to the concentric circle whose circumference passes through P.

Let OA=r, OP=x, PQ=y,_4OPQ=0; then an element of the circle at P is 2πx dx, and at Q it is deydy; the limits of x are 0 and r, those of 0, - and T, and of y, 0 and 2x cos 0. Hence, doubling, since P may be confined to the concentric circle whose circumference passes through Q, we have

5458. (By Professor WOLSTENHOLME, M.A.)-Find the locus of the intersection of perpendicular tangents to a cardioid, and trace the resulting curve.

Solution by R. TUCKER, M.A.

In the Proceedings of the London Mathematical Society (Vol. IV., pp. 327-330), Professor WOLSTENHOLME has discussed this locus and drawn the figure. The locus is made up of a circle and a bicircular quartic.

[Prof. WOLSTENHOLME sent the accompanying figure, with the question, long before the above-cited article appeared in the Mathematical Society's Proceedings. In a note that accompanied the figure, it was stated that the locus of the intersection of tangents, inclined at a constant angle to each other, to any epi- or hypo-cycloid is a trochoid, a more general form whereof this is a particular case. In regard to the above form of the locus, it was added that it is rather too much flattened near the vertex; AB should be a little greater than OC; SB=3(2+ √√/3)a, OB= (5 + 3√3)a, which is a little greater than 10a, 10.2a nearly, therefore AB = 2.2α 1.1 OC. The Proposer stated, furthermore, that in all cases in which he had seen perpendicular tangents to a cardioid mentioned, he had only noticed the circular locus taken account of.]

=

5146. (By S. ROBERTS, M.A.)-Given a pencil of rays and a system of concentric circles; prove (1) that if one set of intersections range on a

straight line, the other intersections lie on a circular cubic, having a double point at the origin of the pencil and the double focus at the common centre of the circles; and (2) determine therefrom, with reference to a system of parabolas having the same focus and axis, the locus of the points the normals at which intersect in a fixed point.

I. Solution by E. B. ELLIOTT, M.A.; Prof. EVANS, M.A.; and others. 1. More generally, take any system of similar, similarly situate, and concentric conics Uz + U2 + c = 0 ...(1), c being arbitrary; and let the origin be the vertex of a pencil of lines. Let one set of intersections range on the straight line v1 = 1. Required the locus of the other set of intersections.

is that of the rays of the pencil corresponding to the conic (1). Eliminating, then, c between this and (1), the equation

is found as that of the locus of intersections of the pencil and system. This is, of course, divisible by v1-1, and there results

as the equation of the locus of the second intersections. It is a cubic having a double point at the origin, having v1+1 = 0 for one asymptote, and for the other two the asymptotes of u2+u1 = 0, that is to say, the common asymptotes of the system (1).

In particular, then, if the system (1) be one of concentric circles, the locus is a cubic having the asymptotes of that system for a pair of asymptotes; that is to say, a circular cubic with its double focus at the centre of the system, and having the origin, the vertex of the pencil, for a double point.

2. A circle whose centre is at the focus of a parabola, and which passes through any point on the curve, passes also through the point where the normal at that point meets the axis.

Thus, drawing normals to the system of parabolas from any point, and describing circles with the focus as centre to pass through all their feet, we have a pencil and a system of circles as above, the straight line which is the locus of one intersection being the axis.

Thus the locus of the feet of the normals, the second set of intersections, is a circular cubic having a double point at the point from which normals are drawn, and the focus of the system of parabolas for its double focus. Its third asymptote is a line parallel to the axis, and at the same distance on one side of the fixed point as the axis is on the other.

II. Solution by C. LEUDESDORF, M.A.; Prof. WOLSTENHOLME; and others. (1). Take the vertex of the pencil as origin, and let the equation of one of the circles be g2 – 2dr cos 0+ip = 0..... and let the fixed straight line be

..(1) ;