# How to proving That as P varies and point Q lies on a fixed circle (Solution Problem 4 : geometry shortlist IMO 2014)

Problem 4:

Consider a fixed circle with three fixed points A, B, and C on it. Also, let us fix a real number . For a variable point on , let M be the point on the segment CP such that . Let Q be the second point of intersection of the circumcircles of the triangles AMP and BMC. Prove that as P varies, the point Q lies on a fixed circle.

Solution.

Also Read:

Throughout the solution, we denote by the directed angle between the lines a and b. Let D be the point on the segment AB such that . We will show that either Q = D, or ; this would mean that the point Q varies over the constant circle through D tangent to BC at B, as required.

Denote the circumcircles of the triangles AMP and BMC by ω_{A} and ω_{B}, respectively. The lines AP, BC, and MQ are pairwise radical axes of the circles , ω_{A}, and ω_{B}, thus either they are parallel, or they share a common point X.

Assume that these lines are parallel (see Figure 1). Then the segments AP, QM, and BC have a common perpendicular bisector; the reﬂection in this bisector maps the segment CP to BA, and maps M to Q. Therefore, in this case Q lies on AB, and BQ / AB = CM / CP = BD / AB; so we have Q = D.

Now assume that the lines AP, QM, and BC are concurrent at some point X (see Figure 2). Notice that the points A, B, Q, and X lie on a common circle Ω by Miquel’s theorem applied to the triangle XPC. Let us denote by Y the symmetric image of X about the perpendicular bisector of AB. Clearly, Y lies on Ω, and the triangles Y AB and are congruent. Moreover, the triangle XPC is similar to the triangle XBA, so it is also similar to the triangle Y AB.

Next, the points D and M correspond to each other in similar triangles Y AB and XPC, since BD / BA= CM / CP = λ. Moreover, the triangles Y AB and XPC are equi-oriented, so . On the other hand, since the points A, Q, X, and Y lie on Ω, we have . Therefore, , so the points Y , D, and Q are collinear.

Finally, we have , as desired.

**Also Read : Geometry shortlist problem in IMO 2014**