# Problem 6. (4 points)

Quadrilateral $A B C D$ is circumscribed around a circle with center at point $O, K, L, M, N$ - the points of tangency of sides $A B, B C, C D$ and $A D$ respectively, $K P, L Q, M R$ and $N S$ - the altitudes in triangles $O K B, O L C, O M D, O N A . O P=15, O A=32, O B=64$.

Find the length of the segment $Q R$.
# Answer: 30

## Solution:

Triangles $O K A$ and $O N A$ are right triangles with a common hypotenuse and a leg equal to the radius of the circle, so they are congruent. Therefore, their altitudes fall on the same point of the common hypotenuse, meaning $K S$ is the altitude in triangle $O K A$. Thus, points $S$ and $P$ lie on a circle with diameter $O K$. Similarly, points $R$ and $S$ lie on a circle with diameter $O N$. Since the diameters of these circles are equal, the degree measures of arc $O S$ in these circles coincide. In the first circle, this arc is subtended by $\angle O P S$, and in the second by $\angle O R S$, so these angles are equal. (They are exactly equal, not supplementary to $180^{\circ}$, because points $P$ and $R$ lie on opposite sides of line $O S$, and the circles are symmetric with respect to it).

Similarly, $\angle O P Q = \angle O R Q$. Adding this to the previous equality, we get $\angle S P Q = \angle S R Q$. Similarly, $\angle P S R = \angle P Q R$, so quadrilateral $P R Q S$ is a parallelogram. Therefore, instead of the length of segment $Q R$, we can find the length of segment $P S$.

(For participants familiar with the concept of inversion: we can understand that the vertices of quadrilateral $P Q R S$ are the inverses of the vertices of quadrilateral $A B C D$ relative to our circle, meaning we have just repeated the proof of the theorem that a quadrilateral inverse to a circumscribed one is a parallelogram).

By the property of the altitude in a right triangle, $O K^{2} = O S \cdot O A$. Similarly, $O K^{2} = O P \cdot O B$, so $\frac{O S}{O B} = \frac{O P}{O A} = k$. Additionally, angle $\angle O$ is common in triangles $O B A$ and $O S P$, so they are similar with a ratio of $k$. Therefore, $P S = k \cdot A B = k(A K + K B) = k\left(\sqrt{O A^{2} - O K^{2}} + \sqrt{O B^{2} - O K^{2}}\right) = \frac{O P}{O A}\left(\sqrt{O A^{2} - O B \cdot O P} + \sqrt{O B^{2} - O B \cdot O P}\right) = 30$.