Problem 4 of USA Mathematical Olympiad 1975

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.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my former classmate Lê Đức Kiểm)

Let E and F be the centers of the small and big circles, respectively, and r and R be their respective radii.

Let M and N be the feet of E and F to AB, and α = ∠APE and β = ∠BPF

We have:

AP × PB = 2r cosα × 2R cosβ = 4 rR cosα cosβ

AP × PB is maximum when the product cosα cosβ is a maximum.

We have cosα cosβ = ½ [cos(α + β) + cos(α - β)]

But α + β = 180° - ∠EPF and is fixed, so is cos(α + β)

So its maximum depends on cos(α - β) which occurs when α = β. To draw the line AB:

Draw a circle with center P and radius PE to cut the radius PF at H. Draw the line parallel to EH passing through P. This line meets the small and big circles at A and B, respectively.


There is an interactive illustration of the problem the idea for which stemmed from a suggestion by Hubert Shutrick who also offered a different construction. Let the circles be C(E) and C(F) and be crossed by the bisector of ∠APF in A' and B', respectively. Then AA' and BB' are the diameters of C(E) and C(F).

The fact that the isosceles triangles AEP and BFP have equal base angles indicates that the tangents at A and B meet, say in point R, on the radical axis of the two circles. By counting the angles it is not too difficult to notice that the quadrilateral AQBR is cyclic. More about that at the interactive illustration and the discussions Problem 4, 1975 USA Math Olympiad and Isosceles Triangles, Problem 4, 1975 USA Math Olympiad: Normals and Tangents, Two Circles and One More.

A synthetic solution appeared strangely unrelated to the trigonometric one. However, a connection between the two was also found.