Subfield: Geometry


Question: Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other.

Solution:
Let $A^{\prime}$ be the point diametrically opposite to $A$ on $\Gamma$. Since $\angle A Q A^{\prime}=90^{\circ}$ and $\angle A Q H=90^{\circ}$, the points $Q, H$, and $A^{\prime}$ are collinear. Similarly, if $Q^{\prime}$ denotes the point on $\Gamma$ diametrically opposite to $Q$, then $K, H$, and $Q^{\prime}$ are collinear. Let the line $A H F$ intersect $\Gamma$ again at $E$; it is known that $M$ is the midpoint of the segment $H A^{\prime}$ and that $F$ is the midpoint of $H E$. Let $J$ be the midpoint of $H Q^{\prime}$.

Consider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\angle H K T=\angle H Q K$ and we are to prove that $\angle M K T=\angle C F K$. Thus it remains to show that $\angle H Q K=\angle C F K+\angle H K M$. Due to $\angle H Q K=90^{\circ}-\angle Q^{\prime} H A^{\prime}$ and $\angle C F K=90^{\circ}-\angle K F A$, this means the same as $\angle Q^{\prime} H A^{\prime}=$ $\angle K F A-\angle H K M$. Now, since the triangles $K H E$ and $A H Q^{\prime}$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\angle K F A=\angle H J A$, and analogously one may obtain $\angle H K M=\angle J Q H$. Thereby our task is reduced to verifying

$$
\angle Q^{\prime} H A^{\prime}=\angle H J A-\angle J Q H .
$$

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Figure 1

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Figure 2

To avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\angle Q^{\prime} H A^{\prime}=\angle J Q H+\angle H J Q$ and $\angle H J A=\angle Q J A+\angle H J Q$, we just have to show that $2 \angle J Q H=\angle Q J A$. To this end, it suffices to remark that $A Q A^{\prime} Q^{\prime}$ 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 $Q^{\prime} A$.
We define the points $A^{\prime}$ and $E$ and prove that the ray $M H$ 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 $M H$ with $\Gamma$, and $E$ is the point on $\Gamma$ with the property $\angle H E A^{\prime}=90^{\circ}$ (see Figure 3).

In the circles $K Q H$ and $E A^{\prime} H$, the line segments $H Q$ and $H A^{\prime}$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A^{\prime} H$. Their pairwise radical axes are the lines $Q K$, $A^{\prime} E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\angle Q K H=$



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Figure 3

$\angle H E A^{\prime}=90^{\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K=S E=S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$.

The circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have

$$
S M \cdot S F=S H^{2}=S K^{2}
$$

So, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.