Problem 4, 1975 USA Math Olympiad: The Final Touch

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, with no apparent relation between the two.

According to the trigonometric solution, the optimal AB forms at P equal angles with PE and PF.

According to the synthetic solution, the optimal AB is defined by the circle C(S) with S on C(O) and OS||PQ.

Here we establish the connection between the two solutions. There are several ways to do that. Hubert Shutrick, in my view, came up with the shortest one. (The discussion is illustrated by the applet below. The function of controls at the bottom of the applet has been explained elsewhere.)

The notations are as follows: C(E), C(F) are the two given circles. C(O) is the locus of the centers S of the circumcircles ABQ; the one corresponding to the optimal AB is denoted C(S). This meets PQ in R. U is the intersection of C(O) with PQ. T is the other end of the diameter QT of C(O). W is the intersection of AE and BF. This is the common point of C(O) and C(S).


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, download and install Java VM and enjoy the applet.

What if applet does not run?

By the construction, SO is the external angle bisector for ΔQSR at S. Extend RS beyond S to meet C(O) in, say, Wo. The latter is the mirror image of Q in OS. It follows that SWo = SQ and, since Q is a common point of C(O) and C(S), Wo lies on C(S) and is, therefore common to C(O) and C(S), meaning that Wo = W. This makes RW a diameter of C(S). The inscribed angles RAW and RBW subtended by a diameter in C(S) are right,right,complementary,supplementary,vertical. Since, ∠RAW is the same as ∠RAE and ∠RBW the same as ∠RBF, RA and RB are tangent to C(E) and C(F) respectively. This condition implies the trigonometric solution: ∠EPA = ∠FPB.

Hubert Shutrick also suggested a different approach to the linking of the two solutions.

The line UPQ intersects EF orthogonally and EF bisects ∠QEP because of the isosceles triangles QEP and QFP. But, ∠QEP = ∠WEP + ∠QEW = 2∠EAP + ∠QUW and the angle between AB and EF is therefore ∠QUW/2 = ∠OSW (because EF bisects ∠PEQ = 2∠WAB + ∠QEW, the line parallel with AB through E makes an angle ∠QEW/2 with AB, but ∠QEW = ∠QUW); so SW is orthogonal to AB. This implies that AW = BW since SA = SB, which gives the trigonometry solution ∠EPA = ∠FPB.

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