Subfield: Geometry


Question: Let $\omega$ be the circumcircle of a triangle $A B C$, and let $\Omega_{A}$ be its excircle which is tangent to the segment $B C$. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_{A}$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_{A}$ at $X$ and $Y$, respectively. The tangent line at $P$ to the circumcircle of the triangle $A P X$ intersects the tangent line at $Q$ to the circumcircle of the triangle $A Q Y$ at a point $R$. Prove that $A R \perp B C$.

Solution:
Let $D$ be the point of tangency of $B C$ and $\Omega_{A}$. Let $D^{\prime}$ be the point such that $D D^{\prime}$ is a diameter of $\Omega_{A}$. Let $R^{\prime}$ be (the unique) point such that $A R^{\prime} \perp B C$ and $R^{\prime} D^{\prime} \| B C$. We shall prove that $R^{\prime}$ coincides with $R$.

Let $P X$ intersect $A B$ and $D^{\prime} R^{\prime}$ at $S$ and $T$, respectively. Let $U$ be the ideal common point of the parallel lines $B C$ and $D^{\prime} R^{\prime}$. Note that the (degenerate) hexagon $A S X T U C$ is circumscribed around $\Omega_{A}$, hence by the Brianchon theorem $A T, S U$, and $X C$ concur at a point which we denote by $V$. Then $V S \| B C$. It follows that $\sphericalangle(S V, V X)=\sphericalangle(B C, C X)=$ $\sphericalangle(B A, A X)$, hence $A X S V$ is cyclic. Therefore, $\sphericalangle(P X, X A)=\sphericalangle(S V, V A)=\sphericalangle\left(R^{\prime} T, T A\right)$. Since $\angle A P T=\angle A R^{\prime} T=90^{\circ}$, the quadrilateral $A P R^{\prime} T$ is cyclic. Hence,

$$
\sphericalangle(X A, A P)=90^{\circ}-\sphericalangle(P X, X A)=90^{\circ}-\sphericalangle\left(R^{\prime} T, T A\right)=\sphericalangle\left(T A, A R^{\prime}\right)=\sphericalangle\left(T P, P R^{\prime}\right)
$$

It follows that $P R^{\prime}$ is tangent to the circle $(A P X)$.

Analogous argument shows that $Q R^{\prime}$ is tangent to the circle $(A Q Y)$. Therefore, $R=R^{\prime}$ and $A R \perp B C$.

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Let $J$ and $r$ be the center and the radius of $\Omega_{A}$. Denote the diameter of $\omega$ by $d$ and its center by $O$. By Euler's formula, $O J^{2}=(d / 2)^{2}+d r$, so the power of $J$ with respect to $\omega$ equals $d r$.



Let $J X$ intersect $\omega$ again at $L$. Then $J L=d$. Let $L K$ be a diameter of $\omega$ and let $M$ be the midpoint of $J K$. Since $J L=L K$, we have $\angle L M K=90^{\circ}$, so $M$ lies on $\omega$. Let $R^{\prime}$ be the point such that $R^{\prime} P$ is tangent to the circle $(A P X)$ and $A R^{\prime} \perp B C$. Note that the line $A R^{\prime}$ is symmetric to the line $A O$ with respect to $A J$.
<img_3803>

Lemma. Let $M$ be the midpoint of the side $J K$ in a triangle $A J K$. Let $X$ be a point on the circle $(A M K)$ such that $\angle J X K=90^{\circ}$. Then there exists a point $T$ on the line $K X$ such that the triangles $A K J$ and $A J T$ are similar and equioriented.

Proof. Note that $M X=M K$. We construct a parallelogram $A J N K$. Let $T$ be a point on $K X$ such that $\sphericalangle(N J, J A)=\sphericalangle(K J, J T)$. Then

$$
\sphericalangle(J N, N A)=\sphericalangle(K A, A M)=\sphericalangle(K X, X M)=\sphericalangle(M K, K X)=\sphericalangle(J K, K T) .
$$

So there exists a spiral similarity with center $J$ mapping the triangle $A J N$ to the triangle $T J K$. Therefore, the triangles $N J K$ and $A J T$ are similar and equioriented. It follows that the triangles $A K J$ and $A J T$ are similar and equioriented.

<img_3163>



Back to the problem, we construct a point $T$ as in the lemma. We perform the composition $\phi$ of inversion with centre $A$ and radius $A J$ and reflection in $A J$. It is known that every triangle $A E F$ is similar and equioriented to $A \phi(F) \phi(E)$.

So $\phi(K)=T$ and $\phi(T)=K$. Let $P^{*}=\phi(P)$ and $R^{*}=\phi\left(R^{\prime}\right)$. Observe that $\phi(T K)$ is a circle with diameter $A P^{*}$. Let $A A^{\prime}$ be a diameter of $\omega$. Then $P^{*} K \perp A K \perp A^{\prime} K$, so $A^{\prime}$ lies on $P^{*} K$. The triangles $A R^{\prime} P$ and $A P^{*} R^{*}$ are similar and equioriented, hence

$\sphericalangle\left(A A^{\prime}, A^{\prime} P^{*}\right)=\sphericalangle\left(A A^{\prime}, A^{\prime} K\right)=\sphericalangle(A X, X P)=\sphericalangle(A X, X P)=\sphericalangle\left(A P, P R^{\prime}\right)=\sphericalangle\left(A R^{*}, R^{*} P^{*}\right)$,

so $A, A^{\prime}, R^{*}$, and $P^{*}$ are concyclic. Since $A^{\prime}$ and $R^{*}$ lie on $A O$, we obtain $R^{*}=A^{\prime}$. So $R^{\prime}=\phi\left(A^{\prime}\right)$, and $\phi\left(A^{\prime}\right) P$ is tangent to the circle $(A P X)$.

An identical argument shows that $\phi\left(A^{\prime}\right) Q$ is tangent to the circle $(A Q Y)$. Therefore, $R=$ $\phi\left(A^{\prime}\right)$ and $A R \perp B C$.