# 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 $\Gamma$ with three fixed points A, B, and C on it. Also, let us fix a real number $\lambda \in (0,1)$. For a variable point $P\notin {A,B,C}$ on $\Gamma$, let M be the point on the segment CP such that $CM=\lambda .CP$. 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.

Throughout the solution, we denote by $\measuredangle (a,b)$ the directed angle between the lines a and b. Let D be the point on the segment AB such that $BD=\lambda .BA$. We will show that either Q = D, or $\measuredangle (DQ,QB)= \measuredangle (AB,BC)$; 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 ωA and ωB, respectively. The lines AP, BC, and MQ are pairwise radical axes of the circles $\Gamma$, ωA, and ωB, 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 $\Delta XBA$ 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 $\measuredangle (MX,XP)=\measuredangle (DY,YA)$. On the other hand, since the points A, Q, X, and Y lie on Ω, we have $\measuredangle (QY,YA)=\measuredangle (MX,XP)$. Therefore, $\measuredangle (QY,YA)= \measuredangle (DY,YA)$, so the points Y , D, and Q are collinear.

Finally, we have $\measuredangle (DQ,QB)=\measuredangle (YQ,QB)=\measuredangle (YA,AB)=\measuredangle (AB,BX)=\measuredangle (AB,BC)$, as desired.

Also Read : Geometry shortlist problem in IMO 2014

×