Problem 4, 1975 USA Math Olympiad and Isosceles Triangles
The following problem has been offered at the 1975 USA Mathematics Olympiad:
Two given circles intersect in two points P and Q. Show how to construct a segment AB passing through P and terminating on the two circles such that AP×PB is a maximum. |
The problem admits a simple trigonometric solution. However, Hubert Shutrick discovered several engaging properties of this configuration which led to a synthetic solution of the problem. The applet below illustrates some of these properties. Hubert's solution is presented on a separate page.
(The function of controls at the bottom of the applet has been explained elsewhere.)
What if applet does not run? |
We shall denote the given circles C(E) and C(F). Let C(O) be the circumcircle of EFQ, with center O.
Besides point Q, C(O) meets C(E) in L and C(F) and M. Let S be the intersection of AL and BM. We first show that S lies on C(O).
∠ABS = ∠PBM = ∠PQM, ∠BAS = ∠PAL = ∠PQL. |
However, in ΔABS, ∠ABS + ∠BAS + ∠ASB = 180°. It follows that
Conversely, if S is on C(O) and SL and SM meet C(E) in A and C(F) in B then AB passes through P. Indeed, ∠QLS is supplementary,complementary,supplementary,obtuse,270° of ∠QMS and
Now observe, that, since the arcs LQ and MQ are fixed for a given pair of circles C(E) and C(F), inscribed angles QAS and ASQ do not depend on the position of A. Similarly, angles QBS and BSQ do not depend on the position of point B. For convenience then choose A so that to have AQ a diameter of C(E). (In passing, this makes AB the longest segment through P with the endpoints on C(E) and C(F).) In this case,
ΔASB is also isosceles so that C(O) is the locus of the centers of the circumcircles ABQ.
Also, we see that ∠ABS = ∠BAS, from which
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