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.)


 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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 ∠LQM + ∠LSM = 180° so that the quadrilateral LSMQ is cyclic.

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 ∠ALQ = ∠APQ and ∠BMQ = ∠BPQ.

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, LQ ⊥ AS and, therefore, QS is a diameter of C(O) which, in turn, implies that MQ ⊥ BS so, similarly, BQ is a diameter of C(F). E, O, F being the centers of C(E), C(O), C(F), divide the corresponding diameters,angles,circles,diameters,triangles in half making, say, triangles ASQ and EOQ similar,equal,identical,congruent,similar. Since the latter is isosceles, so is the former, implying AS = QS. By the same token, BS = QS, so that, finally AS = BS.

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 ∠PQM = ∠PQL, meaning that QP is the bisector of angle LQM.

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