95Author(s): William HooverSource: The American Mathematical Monthly, Vol. 5, No. 8/9 (Aug. - Sep., 1898), p. 206Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2969362 .

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206

-bc . .BCD=tanl c2as-ab

which will bea maximum when c2-a2-ab---O, or c2=a(a+b). Q. E. D.

95. Proposed by WILLIAM HOOVER, A. M., Ph. D., Professor of Mathematics and Astronomy in Ohio State University, Athens, Ohio.

At each point of a parabola is descrlbed the rectangular hyperbola of a four-pointic contact; prove that the locus of the center of the hyperbola is an equal parabola.

Solution by the PROPOSER.

The curve havinigfour-pointic contact with the parabola y2_z4ax ...... (1)

is y2 -4ax-A(yy'-2ax-2ax')2 0 ...... (2),

or, -4Xa2x2 +4aXy'xy+(1-Xy'2)y2-(4a+8a2AX')X+4aAx'y'y-4a2Xx'2=O. .(3).

If this be an equilateral hyperbola,

4Xa2- 1 Xy'2, or XA 1 . (y'2+ 42() ..... (4).

Substituting this in (2) and reducing,

ax2 --y'xy- (iy2 + (4a2 + y'2 + 2ax')x-x'y'y a xi'2 o . (5).

The center of this is given by

x=_-(y'2+8a2),/4a .... (6), yzzy". . (7). (x', y') being on (1), y'2=-4ax'... .(8).

Eliminating x',- y' from (6), (7), and (8), we have the required locus,

ys----4a(x + 2a) . (9).

96. Proposed by W. F. BRADBURY, A. M., Head Master, Cambridge Latin School, Cambridge, Mass. Isosceles tritangles are constructed externally on the three sides of a triangle as bas-

es', with the angles at the bases each 30?. The triangle formied by joining the remote ver- tices (the 1200 vertices) of these isosceles tritangles is equilateral. [Geometric-niot Trig- ononmetric-solution.]

Solution by J. K. ELLWOOD, A. M., Principal of Colfax School, Pittsburg, Pa.

The vertices 0, 01 02 of the isosceles trianigles are the centers of equi- lateral triangles described on the sides of ABC. Circumferences passed about these triangles intersect in a point, P.

Let P be the intersection of the, two circles, AFC and CEB. Join AP, BP, and CP. SinceA POF is inscribed, ZF+ ZAPC=18O0. But ZF==600.

Z APC=120?. Similarly, Z CPB=1200.

L APB-120'; and LAPB+ Z D=z1800.

APBD is inscribed, and P is in the circumference of ADB. Q. E. D.

Lines that join the centers of intersectilg circles bisect the common

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