Japanese Mathematical Olympiad, Final Round, 2019

Let us write $XY$ to indicate the length of the line segment $XY$. If $AB = AC$, the points $K$ and $M$ coincide and the 2 circles become tangent to each other. So, in the sequel, we assume that $AB \ne AC$.

Let $\Gamma$ be the ex-circle within $\angle A$ of the triangle $ABC$. Let $D$ be the point of tangency of the circle $\Omega$ on the side $BC$, and let $D'$ be the point for which $DD'$ is a diameter of $\Gamma$. Let also be the point on $BC$ which is the point of tangency of the circle $\Gamma$, and let $E'$ be the point for which $EE'$ is a diameter of $\Gamma$. Let $F$ be the point of intersection of $\Omega$ and the line $AD'$, different from $D'$, and let $G$ be the point of intersection of $\Gamma$ and the line $AE'$ different from $E'$.

Let $B', C'$ be the point of intersection of the line tangent to the circle $\Omega$ going through $D'$ and the line $AB, BC$, respectively. Then, we see that the triangles $ABC$ and $A'B'C'$ are similar since lines $BC$ and $B'C'$ are parallel. Since the point $E$ is the point of tangency of the ex-circle within $\angle A$ on the side $BC$, and since the point $D'$ is the point of tangency of the ex-circle within $\angle A$ on the side $B'C'$, under the similarity map between the triangles $ABC$ and $AB'C'$ the points $E$ and $D'$ correspond. Therefore, the 3 points $A, D', E$ lie on the same straight line. We have $\angle DFD' = 90^\circ$ since the line segment $DD'$ is a diameter of $\Gamma$, we also have $\angle DFE = 90^\circ$. In the same way, we get that the 3 points $A, D, E'$ lie on the same straight line and that $\angle DGE = 90^\circ$. Consequently, if we let $K'$ be the point of intersection of lines $DE$ and $EG$, then $K'$ is the orthocenter of the triangle $ADE$ and we see that the line $AK'$ is perpendicular to the side $BC$.

Since $BD = \frac{1}{2}(AB+BC=CA) = CE$, we see that $M$ is the mid-point of the line segment $DE$. Consequently, the powers of the point $M$ with respect to the two circles $\Omega$ and $\Gamma$ coincide. Also, from $\angle DFE = \angle DGE = 90^\circ$, we see that the four points $D, E, F, G$ lie on the circumference of the same circle. Therefore, from the theorem on power of points with respect to a circle, we conclude that $K'D \cdot K'F = K'E \cdot K'G$. Consequently, the powers of the point $K'$ with respect to the two circles $\Omega$ and $\Gamma$ coincide. Therefore, we see that $MK'$ is the radical axis of the two circles $\Omega$ and $\Gamma$, and if we let $I_A$ be the center of the circle $\Gamma$, then the line $MK'$ is perpendicular to the line $II_A$. Since the three points $A, I, I_A$ lie on a straight line, we conclude that the line $MK'$ is perpendicular to the line $AI$.

From the discussions made above, we can conclude that the points $K$ and $K'$ coincide, and therefore, the three points $K, D, F$ are collinear. Furthermore, both of the lines $AK$ and $DD'$ are perpendicular to the line $BC$, and they are parallel, from which it follows that the triangles $FKA$ and $FDD'$ are similar. Since $\angle AFK = 90^\circ$, the circle having the line segment $AK$ as its diameter is the circum-circle of the triangle $FKA$, while $\Omega$ is the circum-circle of the triangle $FDD'$, we see that these circles correspond to each other under the similarity map between the triangles $FKA$ and $FDD'$. Thus, we conclude that the two circle are tangent to each other at the point $F$.