document7
TRANSCRIPT
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 .
Accessed: 16/05/2014 23:32
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp
.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].
.
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.
http://www.jstor.org
This content downloaded from 195.78.108.79 on Fri, 16 May 2014 23:32:11 PMAll use subject to JSTOR Terms and Conditions
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).
This content downloaded from 195.78.108.79 on Fri, 16 May 2014 23:32:11 PMAll use subject to JSTOR Terms and Conditions