Problem 4, 1975 USA Math Olympiad: Normals and Tangents

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 illustrated below with a Java applet. The connection between the two solutions is revealed elsewhere.

(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, 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.

QP meets C(O) in U. Any point S on the circle is the center of a circle through Q that is one of the AQB circles and it intersects QP at the point R such that QR is the base of an isosceles triangle with the point S as the other vertex and so QR is twice,twice,thrice,four timesQN where N is the foot of the perpendicular to PQ from S. This is clearly maximum when SN is tangent to the circle as illustrated and it occurs when the point bisects the arc from T to U, where QT is a diameter of C(O).

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