135Author(s): William HooverSource: The American Mathematical Monthly, Vol. 7, No. 11 (Nov., 1900), p. 264Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2968402 .

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264

GEOMETRY.

135. Proposed by WILLIAM HOOVER, A.M., Ph.D., Professor of Mathematics and Astronomy, Ohio Univer- sity, Athens, 0.

If a hyperbola be described touching the four sides of a quadrilateral which is in- scribed in a circle, and one focus lie on the circle, the other focus will also lie on the circle.

Solution by the PROPOSER.

Using quadrilinear notation, the equation to the circle circumscribing the quadrilateral whose sides are given by a-0O, /3O, r-0z, 6dO, is ;rAJd.... (1).

Now, it is well known that if the coordinates of one focus of a conic tan- gent to a given line be a', fl', y', d', those of the other focus are proportional to 1/a, 1//e', l/r'7 1/d'.

But by the problem, a'., ji', ", d6 is on (1); then a'r'zj9'6. . . . (2), or

1 1 a r' E .*(3).

Substituting the reciprocals in] (1) gives (3) also, and proves the theorem.

136. Proposed by J. OWEN MAHONEY, B. E., M. -Sc., Professor of Mathematics, Central High School, Dal- las, Tex.

Construct a triangle having given the base, the median line to the base, and the difference of the base angles.

I. Solution by B. L. REMICK, Instructor of Mathematics, Bradley Institute, Peoria, Ill.

Let LMh=a-the base, CF'=m-=median to the base, a-i =difference of base angles.

Then vertex P of required triangle lies on circle about C (mid point of LM) as center with. radius m ; it also lies on the locus of point of inter- section of straight lines through L, M forming an-

gles with base having the required constant differ- w ence. We propose to show that this latter locus is an equilateral hyperbola and that our problem has therefore four solutions corresponding to the four common points of the circle and hyperbola.

Z RPS between the perpendicular and anigle bisector- (a-i') by a well known result in geomietry ; and hence we have to consider the locus of intersec- tion of two straight lines passing through two given points L, M so that the alngle bisector remnains patallel to itself.

Let coordinates of P be (x1, yi) Equation PL is y1x-xly =. Equiation PM is y1x+'a-x1)1-ay1jO.

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