Problem 1321

The first solution. Let's assume that $A B > A C$ (see Fig. 3). Let $B B_{1}$ and $C C_{1}$ be the altitudes of triangle $\triangle A B C$, $H$ be the point of their intersection, $O$ and $I$ be the centers of the inscribed and circumscribed circles of triangle $A B C$, and $r$ be the radius of its inscribed circle; let $\angle B A C = \alpha$. Note that $A B_{1} = A B \cos \alpha, A C_{1} = A C \cos \alpha$. Therefore, triangle $A B_{1} C_{1}$ is similar to triangle $A B C$ with a similarity coefficient $k = \cos \alpha$.

Let point $L$ be symmetric to point $I$ with respect to $B^{\prime} C^{\prime}$.[^0]

Lemma. Points $L$ and $I$ are corresponding points in triangles $A B_{1} C_{1}$ and $A B C$.

Proof. Since $A I \perp B^{\prime} C^{\prime}$, point $L$ lies on the bisector $A I$. Therefore, it is sufficient to prove that $\frac{A L}{A T} = k$. Let $M$ be the midpoint of segment $B^{\prime} C^{\prime}$. Note that right triangles $A C^{\prime} I$ and $C^{\prime} M I$ are similar, so $\angle M C^{\prime} I = \angle C^{\prime} A I = \alpha / 2$.

We have $A I = \frac{r}{\sin (\alpha / 2)}, A L = A I - L I = A I - 2 M I = \frac{r}{\sin (\alpha / 2)} - 2 r \sin (\alpha / 2)$, thus $\frac{A L}{A I} = 1 - 2 \sin^2 (\alpha / 2) = \cos \alpha = k$, which is what we needed to prove.

!

Fig. 3

!

Fig. 4

Points $A, B_{1}, C_{1}$, and $H$ lie on a circle with diameter $A H$, so $E$ is the center of this circle. Therefore, points $E$ and $O$ in triangles $A B_{1} C_{1}$ and $A B C$ are also corresponding; hence $\angle O I A = \angle E L A$. Since point $F$ is symmetric to $E$ with respect to $B^{\prime} C^{\prime}$, segments $F I$ and $E L$ are also symmetric, and $\angle F I A = \angle E L I$. Thus, $\angle O I A + \angle F I A = \angle E L A + \angle E L I = 180^{\circ}$, which means that points $O, I, F$ lie on the same line.

The second solution. We use the same notation as in the first solution. Let $\ell_{1}, \ell_{2}$, and $\ell_{3}$ be the external bisector of angle $B A C$, the perpendicular bisector of segment $A I$, and the line $B^{\prime} C^{\prime}$, respectively. Clearly, lines $\ell_{1}, \ell_{2}$, and $\ell_{3}$ are parallel. Let $O^{\prime}$ be the point symmetric to point $O$ with respect to $\ell_{1}$. We will prove the following two statements: (1) points $O^{\prime}$, $A$, and $E$ lie on the same line; (2) the ratio of distances between points $O^{\prime}, A, E$ is equal to the ratio of distances between lines $\ell_{1}, \ell_{2}, \ell_{3}$.

!

Fig. 5

!

Fig. 6

(1). Note that $\angle O A B = \left(180^{\circ} - \angle A O B\right) / 2 = 90^{\circ} - \angle A C B = \angle C A H$. Therefore, rays $A O$ and $A H$ form angles with line $\ell_{1}$ equal to $\angle O A B + \left(180^{\circ} - \angle B A C\right) / 2$; hence rays $A O^{\prime}$ and $A H$ are oppositely directed (see Fig. 4).

(2). Let $A_{2}$ and $A_{3}$ be the points of intersection of line $A I$ with $\ell_{2}$ and $\ell_{3}$, respectively, and let $T, K, L$ be the midpoints of side $B C$ and arcs $B C, B A C$ of the circumscribed circle, respectively (see Fig. 5). We have $\angle L B K = \angle A B^{\prime} I = 90^{\circ}, \angle B L K = \angle B A K = \angle B^{\prime} A I$, so right triangles $L B K$ and $A B^{\prime} I$ are similar. In these triangles, points $T$ and $A_{3}$ are the feet of corresponding altitudes, and points $O, A_{2}$ are the midpoints of the hypotenuses, so $\frac{O L}{O T} = \frac{A_{2} A}{A_{2} A_{3}}$. On the other hand, points $T$ and $O$ map to $A$ and $H$ respectively under a homothety with center at the centroid and coefficient -2; therefore, $O T = \frac{A H}{2} = A E$, and from symmetry $A O^{\prime} = A O = O L$. Thus, $\frac{A O^{\prime}}{A E} = \frac{O L}{O T} = \frac{A_{2} A}{A_{2} A_{3}}$.

Now it is not difficult to complete the statement of the problem. We will show that the points symmetric to points $O^{\prime}, A$, and $E$ with respect to lines $\ell_{1}, \ell_{2}, \ell_{3}$ respectively (which are points $O, I, F$) lie on the same line. Let the lines $O I$ and $A O^{\prime}$ intersect at point $X$. Let $F^{\prime}$ be the point of intersection of lines $E F$ and $O I$. Finally, let $S_{1}$ and $S_{3}$ be the midpoints of segments $O O^{\prime}$ and $F^{\prime} E$, respectively (see Fig. 6). Triangles $X O O^{\prime}, X I A$, and $X F^{\prime} E$ are similar, so their medians $X S_{1}, X A_{2}, X S_{3}$ lie on the same line, and from similarity we get $\frac{S_{1} A_{2}}{A_{2} S_{3}} = \frac{O^{\prime} A}{A E} = \frac{A_{2} A}{A_{2} A_{3}}$. This means that $A_{3} S_{3} \| A S_{1}$. Therefore, $S_{3}$ lies on line $\ell_{3}$, from which $F^{\prime} = F$, which is what we needed to prove.

Remark. In the last part of the solution, essentially, the following fact is proved. Let point $X$ move along some line $m$ at a constant speed, and let line $\ell$ move in the plane, remaining parallel to itself. Then the point symmetric to $X$ with respect to $\ell$ also moves along some line.

[^0]: Words satisfying such conditions are called Davenport-Schinzel sequences.