Problem 4, 1975 USA Math Olympiad and the Radical Axis

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 an elegant solution that employs some trigonometric identities.

The applet below illustrates the configuration and may suggest if not a purely synthetic solution then, perhaps, a different approach to solving the problem. The circles have centers E and F, respectively.

The applet also displays the circumcircle ABQ and marks point R where the latter meets line PQ.

(In the applet, the two circles are defined by three points each: P, Q, A, for one, and P, Q, B, for the other. Dragging either P or Q modifies the circles. Dragging either A or B may have different effect depending on which of the buttons at the bottom of the applet is checked. If it's "Adjust circles" then the circles will be modified. If the "Adjust chord" button is checked, A and B would be dragged over existing circles supplying a set of possible locations for the segment AB.

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?

Line PQ, being the radical axis of the two circles has the following feature. Let X be a generic point on the PQ. Find points of tangency At on C(E) and Bt on C(F) of the tangents from X. Since X is on the radical axis of C(E) and C(F), XAt = XBt, making triangle AtXBt isosceles. Let Pe and Pf be the points on C(E) and C(F) where AtBt meets the circles and consider triangles AtEPe and BtFPf. The two are isosceles, with sides EAt and FBt perpendicular to the tangents XAt and XBt, respectively. The base angles EAtPe and FBtPf are complementary to the base angles of the isosceles ΔAtXBt and are therefore equal.

Point R, the intersection of the circumcircle ABQ and PQ, has the distinction of being the unique point X on PQ for which AtBt passes through P.

In the circle ABQ, AB is a chord that crosses chord QR at P. AP×PB is one of the equal products from the Intersecting Chords theorem. PQ×PR is the other: PQ×PR = AP×PB. Since PQ is fixed for the given two circles, the product AP×PB is maximized together with PR. Combining the two observation, PR is maximum when AtBt passes through P.

Several properties of this configuration have been discovered by Hubert Shutrick who also supplied a synthetic solution.

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