56th International Mathematical Olympiad Shortlisted Problems

Consider any point $T$ such that $TK$ is tangent to the circle $KQH$ at $K$ with $Q$ and $T$ lying on different sides of $KH$ (see Figure 1). Then $\angle HKT=\angle HQK$ and we are to prove that $\angle MKT=\angle CFK$. Thus it remains to show that $\angle HQK=\angle CFK+\angle HKM$. Due to $\angle HQK=90^{\circ }-\angle Q^{\prime }H{A}^{\prime }$, and $\angle CFK=90^{\circ }-\angle KFA$, this means the same as $\angle Q^{\prime }H{A}^{\prime }=\angle KFA-\angle HKM$. Now, since the triangles $KHE$ and $AH{Q}^{\prime }$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\angle KFA=\angle HJA$, and analogously one may obtain $\angle HKM=\angle JQH$. Thereby our task is reduced to verifying $$\angle Q^{\prime }H{A}^{\prime }=\angle HJA-\angle JQH.$$ Figure 1 Figure 2 To avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\angle Q^{\prime }H{A}^{\prime }=\angle JQH+\angle HJQ$ and $\angle HJA=\angle QJA+\angle HJQ$, we just have to show that $2\angle JQH=\angle QJA$. To this end, it suffices to remark that $AQ{A}^{\prime }{Q}^{\prime }$ is a rectangle and that $J$, being defined to be the midpoint of $H{Q}^{\prime }$, has to lie on the mid parallel of $Q{A}^{\prime }$ and $Q^{\prime }A$.

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Alternative solution.

We define the points $A^{\prime }$ and $E$ and prove that the ray $MH$ passes through $Q$ in the same way as in the first solution. Notice that the points $A^{\prime }$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A^{\prime }$ is the second intersection of the line $MH$ with $\Gamma$, and $E$ is the point on $\Gamma$ with the property $\angle HE{A}^{\prime }=90^{\circ }$ (see Figure 3).

In the circles $KQH$ and $E{A}^{\prime }H$, the line segments $HQ$ and $H{A}^{\prime }$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $MH$. Let $R$ be the radical center of the circles $ABC,KQH$ and $E{A}^{\prime }H$. Their pairwise radical axes are the lines $QK$, $A^{\prime }E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $HR$; by $\angle QKH=\angle HE{A}^{\prime }=90^{\circ }$, the quadrilateral $HERK$ is cyclic and its circumcenter is $S$; hence we have $SK=SE=SH$. The line $BC$, being the perpendicular bisector of $HE$, passes through $S$. The circle $HMF$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $HMF$ we have $$SM\cdot SF=S{H}^2=S{K}^2.$$ So, the power of $S$ with respect to the circles $KQH$ and $KFM$ is $S{K}^2$. Therefore, the line segment $SK$ is tangent to both circles at $K$.

Figure 3