Problem 4. In a right triangle $P Q R$, a circle is constructed on the leg $P R$ as its diameter, which intersects the hypotenuse $P Q$ at point $T$. A tangent to the circle is drawn through point $T$, which intersects the leg $R Q$ at point $S$. Find the length of $S Q$, if $P T=15$, and $Q T=5$.
Answer: 5.

Solution. The solution is based on two simple observations. First, $\angle P T R=90^{\circ}$, since it subtends the diameter. Second, $S T$ and $S R$ are tangents to the circle from the given conditions, so $S T=S R$. Therefore, in the right triangle $R T Q$, a point $S$ is marked on the hypotenuse $Q R$ such that $S T=S R$; it is well known that then $S$ is the midpoint of this hypotenuse, $Q S=Q R / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4a99e6e32d5aecd6d775g-09.jpg?height=697&width=1325&top_left_y=468&top_left_x=363)

The solution can be completed in several ways; we will present two of them.

First method, height of a right triangle and the Pythagorean theorem. In the right triangle $P Q R$, the height $R T$ divides the hypotenuse $P Q$ into segments $P T=15$ and $Q T=5$. We get that $R T^{2}=P T \cdot Q T=15 \cdot 5$; by the Pythagorean theorem $Q R^{2}=Q T^{2}+R T^{2}=5^{2}+15 \cdot 5=100, Q S=Q R / 2=5$.

Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $Q R$ and the secant $Q P$, we have: $Q R^{2}=Q T \cdot Q P=5 \cdot(5+15)=100, Q S=Q R / 2=5$.