Subfield: Geometry


Question: Let $\Gamma$ and $I$ be the circumcircle and the incenter of an acute-angled triangle $A B C$. Two circles $\omega_{B}$ and $\omega_{C}$ passing through $B$ and $C$, respectively, are tangent at $I$. Let $\omega_{B}$ meet the shorter arc $A B$ of $\Gamma$ and segment $A B$ again at $P$ and $M$, respectively. Similarly, let $\omega_{C}$ meet the shorter arc $A C$ of $\Gamma$ and segment $A C$ again at $Q$ and $N$, respectively. The rays $P M$ and $Q N$ meet at $X$, and the tangents to $\omega_{B}$ and $\omega_{C}$ at $B$ and $C$, respectively, meet at $Y$.

Prove that the points $A, X$, and $Y$ are collinear.

Solution:
Let $A I, B I$, and $C I$ meet $\Gamma$ again at $D, E$, and $F$, respectively. Let $\ell$ be the common tangent to $\omega_{B}$ and $\omega_{C}$ at $I$. We always denote by $\sphericalangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\circ}$.

Step 1: We show that $Y$ lies on $\Gamma$.

Recall that any chord of a circle makes complementary directed angles with the tangents to the circle at its endpoints. Hence,

$$
\sphericalangle(B Y, B I)+\sphericalangle(C I, C Y)=\sphericalangle(I B, \ell)+\sphericalangle(\ell, I C)=\sphericalangle(I B, I C) .
$$

Therefore,

$$
\begin{array}{r}
\sphericalangle(B Y, B A)+\sphericalangle(C A, C Y)=\sphericalangle(B I, B A)+\sphericalangle(B Y, B I)+\sphericalangle(C I, C Y)+\sphericalangle(C A, C I) \\
=\sphericalangle(B C, B I)+\sphericalangle(I B, I C)+\sphericalangle(C I, C B)=0,
\end{array}
$$

which yields $Y \in \Gamma$.

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Step 2: We show that $X=\ell \cap E F$.

Let $X_{*}=\ell \cap E F$. To prove our claim, it suffices to show that $X_{*}$ lies on both $P M$ and $Q N$; this will yield $X_{*}=X$. Due to symmetry, it suffices to show $X_{*} \in Q N$.

Notice that

$$
\sphericalangle\left(I X_{*}, I Q\right)=\sphericalangle(C I, C Q)=\sphericalangle(C F, C Q)=\sphericalangle(E F, E Q)=\sphericalangle\left(E X_{*}, E Q\right) ;
$$



therefore, the points $X_{*}, I, Q$, and $E$ are concyclic (if $Q=E$, then the direction of $E Q$ is supposed to be the direction of a tangent to $\Gamma$ at $Q$; in this case, the equality means that the circle $\left(X_{*} I Q\right)$ is tangent to $\Gamma$ at $\left.Q\right)$. Then we have

$$
\sphericalangle\left(Q X_{*}, Q I\right)=\sphericalangle\left(E X_{*}, E I\right)=\sphericalangle(E F, E B)=\sphericalangle(C A, C F)=\sphericalangle(C N, C I)=\sphericalangle(Q N, Q I) \text {, }
$$

which shows that $X_{*} \in Q N$.

Step 3: We finally show that $A, X$, and $Y$ are collinear.

Recall that $I$ is the orthocenter of triangle $D E F$, and $A$ is symmetric to $I$ with respect to $E F$. Therefore,

$$
\sphericalangle(A X, A E)=\sphericalangle(I E, I X)=\sphericalangle(B I, \ell)=\sphericalangle(B Y, B I)=\sphericalangle(B Y, B E)=\sphericalangle(A Y, A E) \text {, }
$$

which yields the desired collinearity.
Perform an inversion centered at $I$; the images of the points are denoted by primes, e.g., $A^{\prime}$ is the image of $A$.

On the inverted figure, $I$ and $\Gamma^{\prime}$ are the orthocenter and the circumcircle of triangle $A^{\prime} B^{\prime} C^{\prime}$, respectively. The points $P^{\prime}$ and $Q^{\prime}$ lie on $\Gamma^{\prime}$ such that $B^{\prime} P^{\prime} \| C^{\prime} Q^{\prime}$ (since $B^{\prime} P^{\prime}=\omega_{B}^{\prime}$ and $C^{\prime} Q^{\prime}=\omega_{C}^{\prime}$ ). The points $M^{\prime}$ and $N^{\prime}$ are the second intersections of lines $B^{\prime} P^{\prime}$ and $C^{\prime} Q^{\prime}$ with the circumcircles $\gamma_{B}$ and $\gamma_{C}$ of triangles $A^{\prime} I B^{\prime}$ and $A^{\prime} I C^{\prime}$, respectively. Notice here that $\gamma_{C}$ is obtained from $\gamma_{B}$ by the translation at $\overrightarrow{B^{\prime} C^{\prime}}$; the same translation maps line $B^{\prime} P^{\prime}$ to $C^{\prime} Q^{\prime}$, and hence $M^{\prime}$ to $N^{\prime}$. In other words, $B^{\prime} M^{\prime} N^{\prime} C^{\prime}$ is a parallelogram, and $P^{\prime} Q^{\prime}$ partitions it into two isosceles trapezoids.

Point $X^{\prime}$ is the second intersection point of circles $\left(I P^{\prime} M^{\prime}\right)$ and $\left(I Q^{\prime} N^{\prime}\right)$ that is - the reflection of $I$ in their line of centers. But the centers lie on the common perpendicular bisector $p$ of $P^{\prime} M^{\prime}$ and $Q^{\prime} N^{\prime}$, so $p$ is that line of centers. Hence, $I X^{\prime} \| B^{\prime} P^{\prime}$, as both lines are perpendicular to $p$.

Finally, the point $Y$ satisfies $\sphericalangle(B Y, B I)=\sphericalangle(P B, P I)$ and $\sphericalangle(C Y, C I)=\sphericalangle(Q C, Q I)$, which yields $\sphericalangle\left(Y^{\prime} B^{\prime}, Y^{\prime} I\right)=\sphericalangle\left(B^{\prime} P^{\prime}, B^{\prime} I\right)$ and $\sphericalangle\left(Y^{\prime} C^{\prime}, Y^{\prime} I\right)=\sphericalangle\left(C^{\prime} Q^{\prime}, C^{\prime} I\right)$. Therefore,

$$
\sphericalangle\left(Y^{\prime} B^{\prime}, Y^{\prime} C^{\prime}\right)=\sphericalangle\left(B^{\prime} P^{\prime}, B^{\prime} I\right)+\sphericalangle\left(C^{\prime} I, C^{\prime} Q^{\prime}\right)=\sphericalangle\left(C^{\prime} I, B^{\prime} I\right)=\sphericalangle\left(A^{\prime} B^{\prime}, A^{\prime} C^{\prime}\right)
$$

which shows that $Y^{\prime} \in \Gamma^{\prime}$.

In congruent circles $\Gamma^{\prime}$ and $\gamma_{B}$, the chords $A^{\prime} P^{\prime}$ and $A^{\prime} M^{\prime}$ subtend the same angle $\angle A^{\prime} B^{\prime} P^{\prime}$; therefore, $A^{\prime} P^{\prime}=A^{\prime} M^{\prime}$, and hence $A^{\prime} \in p$. This yields $A^{\prime} X^{\prime}=A^{\prime} I$, and hence $\sphericalangle\left(I A^{\prime}, I X^{\prime}\right)=$ $\sphericalangle\left(X^{\prime} I, X^{\prime} A^{\prime}\right)$.

Finally, we have

$$
\begin{aligned}
\sphericalangle\left(Y^{\prime} I, Y^{\prime} A^{\prime}\right) & =\sphericalangle\left(Y^{\prime} I, Y^{\prime} B^{\prime}\right)+\sphericalangle\left(Y^{\prime} B^{\prime}, Y^{\prime} A^{\prime}\right) \\
& =\sphericalangle\left(B^{\prime} I, B^{\prime} P^{\prime}\right)+\sphericalangle\left(I A^{\prime}, I B^{\prime}\right)=\sphericalangle\left(I A^{\prime}, B^{\prime} P^{\prime}\right)=\sphericalangle\left(I A^{\prime}, I X^{\prime}\right)=\sphericalangle\left(X^{\prime} I, X^{\prime} A^{\prime}\right),
\end{aligned}
$$

which yields that the points $A^{\prime}, X^{\prime}, Y^{\prime}$, and $I$ are concyclic. This means exactly that $A, X$, and $Y$ are collinear.



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