Prove that the midpoint of segment $P Q$ is the center of the inscribed circle of triangle $A B C$.

## Solution

Let $D, E, F$ be the feet of the perpendiculars dropped from points $P, Q, O$ to line $A B$ respectively. By the given condition, $P D=6, Q E=10$. $O F$ is the midline of trapezoid $D P Q E$, so the length of $O F$ is $1 / 2(P D+Q E)=1 / 2(6+10)=8$. Therefore, point $O$ is 8 units away from line $A B$. Similarly, we find that $O$ is $1 / 2(7+9)=8$ units away from line $B C$ and $1 / 2(12+4)=8$ units away from line $C A$. Thus, $O$ is the center of the inscribed circle of triangle $A B C$, and the radius of this circle is 8.

## Answer

## Problem
Inside triangle $A B C$, there are points $P$ and $Q$ such that point $P$ is at distances 6, 7, and 12 from the lines $A B, B C, C A$ respectively, and point $Q$ is at distances 10, 9, and 4 from the lines $A B, B C, C A$ respectively. Find the radius of the inscribed circle of triangle $A B C$.