**Problem 3 of Asian Pacific Mathematical Olympiad 1995**

Let PQRS be a cyclic quadrilateral such that the segments PQ and RS are not parallel. Consider the set of circles through P and Q, and the set of circles through R and S. Determine the set I of points of tangency of circles in these two sets.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Extend PQ and SR to intercept each other at J. Since PQRS is a cyclic quadrilateral ∠PSR + ∠PQR = 180° or ∠PSR = ∠RQJ

Similarly ∠SPQ = ∠QRJ

And the two triangles JPS and JRQ are similar, therefore

JQ / JR = JS / JP or JQ × JP = JS × JR

From J draw the two lines tangential to the bottom and top circles and assume that the two tangential points are different, respectively, are I’ and I” on the bottom and top circles as shown. We have

JI’² = JR × JS and JI”² = JQ × JP

With the assumption that the tangential JI’ and JI” do not coincide, the two circles are either overlap or not touching each other at all which is not true with the given condition of the problem. Therefore, for the two circles to be tangent I’ must coincide I” and also coincide with I.

or JI² = JR × JS which is a constant

So the set of points of tangency of the two circles is a circle with center at J and radius r = \sqrt{JR × JS}

or r = \sqrt{JQ × JP}

(There is a dynamic illustration along with a rephrased proof.)