7Author(s): William HooverSource: The American Mathematical Monthly, Vol. 1, No. 4 (Apr., 1894), p. 131Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2968781 .

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131

by area;. hence D = - x 6 x 4 x 2-133 48' = 1 /43.225= 65.74Z7223-. 13 13a

7. Proposed by WLLIAM HOOVER, A. M., Ph. D., Professor of Mathematios and Astronomy in the Ohio University, Athen. Ohio.

Through each point of the straight line w,=jy? h is drawn a chord of the parabola y2 =4a, which is bisected in the point. Prove that this chord touches the parabola (y+ 2am)2 = 8a(x-h I).

Solution by the Proposer. Let the chord be gx +fy= .... (1). This cuts the curve V2 24at.... (2), in the points whose co-ordinates

are given by the equatiorns

x2 2q+24) x +1 =0.... (3), and 2 2

p4 af 4a, y2 +-y- =0.. .(4).

9 9

The middle of the chord is then (?f2 2a9)

If this point be on the line x=m?iy+*. .. (5), g+ 2qaf e_ 2arn. +A,, (6)

q q or, 9+22f 2 =-2amfj+g .... ().

Making this homogeneous by aid of (1),

(h-z)%-(Y+2 em)9.2a 0 ....(8), a quad-

ratic in the tindeterimiined constants and giving the envelope (y+2am)9

=8(i Or- h). A190n .dlv.d b.I E. P )?wyf Al1R"d Thimp. (0 B. 11. ZXprr. nt .1. PF.Ii. Scehffer.

8. Proposed by ADOLPH BAILOFF, Durand, Wisconsin, If the two exterior angles at the base of a triangle are equal, the triangle is

isosceles.

Solution by MISS GRACI H. GRIDLEY, Student in Kidder Institute, Kidder, Missouri; and P. S. BERG, Apple Cr"k, Ohio. Let the exterior angles, ABD and A CE of the trianale ABC be

equal. To prove the triangle isosceles.

PROOF: <A CE+ <A CF=2rt. <8. Also <ABD+ <ABE=2rt. <8.

<A CE+ <A CFE <ABD+ <JABF (Things equal to the same thing are equal to each

other). But <ACE=<AB_D. (HYP.)

.<A CF- ABF. (If equals be subtracted from equals, the remain-

ders are equal).

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