Problem 3. The perimeter of triangle $ABC$ is 1. Circle $\omega$ touches side $BC$, the extension of side $AB$ at point $P$, and the extension of side $AC$ at point $Q$. The line passing through the midpoints of $AB$ and $AC$ intersects the circumcircle of triangle $APQ$ at points $X$ and $Y$. Find the length of segment $XY$.

(D. Brodsky)
Answer: $\frac{1}{2}$.

Solution 1. Let the point of tangency of $\omega$ with segment $B C$ be $R$. Consider segments $A P$ and $A Q$. On one hand, they are equal. On the other hand, $B R = B P$ and $C R = C Q$, so

$$
A P + A Q = A B + B P + C Q + A C = A B + B C + A C
$$

Therefore, the lengths of segments $A P$ and $A Q$ are equal to half the perimeter of triangle $A B C$.

Let the midpoints of segments $A B$ and $A C$ be $M$ and $N$ respectively. Mark points $X^{\prime}$ and $Y^{\prime}$ on line $M N$ such that $X^{\prime} M = M A$ and $Y^{\prime} N = N A$ (see Fig. 1). Note that the length of segment $X^{\prime} Y^{\prime}$ is also equal to half the perimeter of triangle $A B C$, i.e., $A P = X^{\prime} Y^{\prime}$. Since $X^{\prime} M = M A$, it follows that $M P = M Y^{\prime}$. Therefore, triangles $A X^{\prime} M$ and $P M Y^{\prime}$ are isosceles, so $A X^{\prime} P Y_{1}$ is an isosceles trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b4f6750e3c0642144d1g-4.jpg?height=658&width=648&top_left_y=91&top_left_x=401)

Fig. 1: to the solution of problem 3

As is known, an isosceles trapezoid can be inscribed in a circle, so point $P$ lies on the circumcircle $\gamma$ of triangle $A X^{\prime} Y^{\prime}$. Similarly, point $Q$ lies on $\gamma$, so $\gamma$ is circumscribed around triangle $P A Q$. Therefore, $X^{\prime} = X$, $Y^{\prime} = Y$, and $X Y = 1 / 2$.

Solution 2. Mark point $I$ - the center of circle $\omega$. Since $\angle A P I = \angle A Q I = 90^{\circ}$, points $A, P, I, Q$ lie on the same circle $\gamma$, constructed on segment $A I$ as the diameter. Let $M$ and $N$ be the midpoints of segments $A B$ and $A C$ respectively, and $X^{\prime}$ and $Y^{\prime}$ be the points of intersection of lines $B I$ and $C I$ with line $M N$ respectively. We will prove that $X^{\prime}$ and $Y^{\prime}$ lie on $\gamma$.

Note that angles $P B I$ and $M B X^{\prime}$ are equal as vertical angles, and angles $C B I$ and $M X^{\prime} B$ are equal because lines $B C$ and $M N$ are parallel. Since $B I$ is the angle bisector of $\angle P B C$, $\angle M B X^{\prime} = \angle M X^{\prime} B$, so triangle $B M X^{\prime}$ is isosceles and $B M = M X^{\prime}$. In triangle $A B X^{\prime}$, median $X^{\prime} M$ is equal to half the side, so this triangle is right-angled with a right angle at $X^{\prime}$. Thus, $\angle A X^{\prime} I = 90^{\circ}$, so $X^{\prime}$ lies on $\gamma$, hence $X^{\prime} = X$. Similarly, triangle $A C Y^{\prime}$ is right-angled and $Y^{\prime} = Y$. Then $X Y = X M + M N + N Y = 1 / 2$.

Remark. Essentially, the solution proves the known fact that the projection of vertex $A$ onto the bisector of an internal or external angle lies on the midline of the triangle.

Solution 3. In the notation of the previous solution, let $a, b$, and $c$ be half the lengths of sides $B C, A C$, and $A B$ respectively, and $x$ and $y$ be the lengths of segments $M X$ and $N Y$ respectively. Write the powers of points $M$ and $N$ relative to $\gamma$ in two ways:

$$
\left\{\begin{array}{l}
M X \cdot M Y = M A \cdot M P, \\
N Y \cdot N X = N A \cdot N Q
\end{array} \Leftrightarrow \quad \left\{\begin{array}{l}
x(a + y) = c(a + b) \\
y(a + x) = b(a + c)
\end{array}\right.\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b4f6750e3c0642144d1g-5.jpg?height=605&width=601&top_left_y=91&top_left_x=420)

Fig. 2: to the solution of problem 3

Subtract the second equation from the first:

$$
a x - a y = a c - a b \Leftrightarrow y = x + b - c
$$

Substitute into the first equation of the system:

$$
x(a + x + b - c) = c(a + b) \Leftrightarrow x^2 + x(a + b - c) - c(a + b) = 0
$$

By Vieta's theorem, the quadratic equation in $x$ has two roots: $c$ and $-(a + b)$. Since $x$ is positive, $x = c$. Similarly, $y = b$. Therefore, $X Y = a + b + c = 1 / 2$.

Solution 4. As in the second solution, note that circle $\gamma$ is constructed on $A I$ as the diameter. Let $\omega^{\prime}$ be the image of circle $\omega$ under a homothety with center $A$ and coefficient $1 / 2$. Since $A I$ is the diameter of $\gamma$, circles $\gamma$ and $\omega^{\prime}$ have a common center.

Under the same homothety, triangle $A B C$ transforms into triangle $A M N$, so $\omega^{\prime}$ will be tangent to lines $A P, A Q$, and $X Y$. Thus, $X Y, A P$, and $A Q$ are three chords of circle $\gamma$ that are tangent to the concentric circle $\omega^{\prime}$. Therefore, their lengths are equal. As proven in the first solution, the length of segment $A P$ is equal to half the perimeter of triangle $A B C$, so the length of $X Y$ is also equal to half the perimeter, i.e., $1 / 2$.