Problem 1914

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

Let $P X$ intersect $A B$ and $D' R'$ at $S$ and $T$, respectively. Let $U$ be the ideal common point of the parallel lines $B C$ and $D' R'$. 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 $V S \parallel B C$. It follows that $\Varangle(S V, V X)=\Varangle(B C, C X)= \Varangle(B A, A X)$, hence $A X S V$ is cyclic. Therefore, $\Varangle(P X, X A)=\Varangle(S V, V A)=\Varangle\left(R' T, T A\right)$. Since $\angle A P T=\angle A R' T=90^{\circ}$, the quadrilateral $A P R' T$ is cyclic. Hence,
$$\Varangle(X A, A P)=90^{\circ}-\Varangle(P X, X A)=90^{\circ}-\Varangle\left(R' T, T A\right)=\Varangle\left(T A, A R'\right)=\Varangle\left(T P, P R'\right) .$$
It follows that $P R'$ is tangent to the circle $(A P X)$.

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

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 $R'$ be the point such that $R' P$ is tangent to the circle $(A P X)$ and $A R' \perp B C$. Note that the line $A R'$ is symmetric to the line $A O$ with respect to $A J$.

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 $\Varangle(N J, J A)=\Varangle(K J, J T)$. Then
$$\Varangle(J N, N A)=\Varangle(K A, A M)=\Varangle(K X, X M)=\Varangle(M K, K X)=\Varangle(J K, K T) .$$
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. $\square$

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 $R^{*}=\phi\left(R'\right)$. Observe that $\phi(T K)$ is a circle with diameter $A P^{*}$. Let $A A'$ be a diameter of $\omega$. Then $P^{*} K \perp A K \perp A' K$, so $A'$ lies on $P^{*} K$. The triangles $A R' P$ and $A P^{*} R^{*}$ are similar and equioriented, hence
$$\Varangle\left(A A', A' P^{*}\right)=\Varangle\left(A A', A' K\right)=\Varangle(A X, X P)=\Varangle(A X, X P)=\Varangle\left(A P, P R'\right)=\Varangle\left(A R^{*}, R^{*} P^{*}\right),$$
so $A, A', R^{*}$, and $P^{*}$ are concyclic. Since $A'$ and $R^{*}$ lie on $A O$, we obtain $R^{*}=A'$. So $R'=\phi\left(A'\right)$, and $\phi\left(A'\right) P$ is tangent to the circle $(A P X)$.

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

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 $d r$.