Problem:
Given circles ω1​ and ω2​ intersecting at points X and Y, let ℓ1​ be a line through the center of ω1​ intersecting ω2​ at points P and Q and let ℓ2​ be a line through the center of ω2​ intersecting ω1​ at points R and S. Prove that if P,Q,R and S lie on a circle then the center of this circle lies on line XY.
Solution:
Let ω denote the circumcircle of P,Q,R,S and let O denote the center of ω. Line XY is the radical axis of circles ω1​ and ω2​. It suffices to show that O has equal power to the two circles; that is, to show that
OO12​−O1​S2=OO22​−O2​Q2 or OO12​+O2​Q2=OO22​+O1​S2
Let M and N be the intersections of lines O2​O,ℓ1​ and O1​O,ℓ2​. Because circles ω and ω2​ intersect at points P and Q, we have PQ⊥OO2​ (or ℓ1​⊥OO2​ ). Hence
O O_{1}^{2}-O Q^{2}=\left(O M^{2}+M O_{1}^{2}\right)-\left(O M^{2}+M Q^{2}\right)=\left(O_{2} M^{2}+M O_{1}^{2}\right)-\left(O_{2} M^{2}+M Q^{2}\right)=O_{2} O_{1}^{2}-O_{2} Q^
or
O2​O12​+OQ2=OO12​+O2​Q2
Likewise, we have O2​O12​+OS2=OO22​+O1​S2. Because OS=OQ, we obtain that OO12​+O2​Q2=OO22​+O1​S2, which is what to be proved.
Solution 2. We maintain the notations of the first solution. Three pairs of circles (ω,ω1​), (ω1​,ω2​),(ω2​,ω) meet at three pairs of points (R,S),(X,Y),(P,Q), respectively; that is, lines RS,XY,PQ are the respective radical axes of these pairs of circles. We consider two cases.
In the first case, we assume that these three radical axes are not parallel. They must be concurrent at the radical center, denoted by H, of these three circles. In particular, it follows that H,X,Y lie a line, denoted by ℓ, and ℓ⊥O1​O2​. On the other hand, O1​M⊥O2​O and O2​N⊥O1​O. Hence H is the orthocenter of triangle OO1​O2​, from which it follows that OH⊥O1​O2​. Therefore, O lies on ℓ; that is, X,P,Q are collinear.
In the second case, we assume that these three radical axes are parallel. We will then deduce the above configurations. Let O3​ be the midpoint of segment XY. From right triangles O1​O3​Q,O1​O3​X,O1​O2​Q, we have
O3​Q2=O1​Q2+O1​O32​=O2​Q2−O1​O22​+O1​X2−XO32​
which is a expression symmetric about circles ω1​ and ω2​. Hence we can easily obtain that O3​Q2=O3​S2 and that O3​ is the circumcenter of isosceles trapezoid PQSR; that is, O3​=O, completing the proof.
This problem was suggested by Ian Le. The solutions were contributed by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions
