Problem 4. In a right triangle $X Y Z$, a circle is constructed on the leg $X Z$ as its diameter, which intersects the hypotenuse $X Y$ at point $W$. A tangent to the circle is drawn through point $W$, which intersects the leg $Z Y$ at point $V$. Find the length of $V Y$, if $X W=12$, and $Y W=4$.
Answer: 4.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_4a99e6e32d5aecd6d775g-15.jpg?height=702&width=1330&top_left_y=620&top_left_x=360)

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 $X Y Z$, the height $Z W$ divides the hypotenuse $X Y$ into segments $X W=12$ and $Y W=4$. We get that $Z W^{2}=X W \cdot Y W=12 \cdot 4$; by the Pythagorean theorem, $Y Z^{2}=Y W^{2}+Z W^{2}=4^{2}+12 \cdot 4=64, Y V=Y Z / 2=4$.

Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $Y Z$ and the secant $Y X$, we have: $Y Z^{2}=Y W \cdot Y X=4 \cdot(4+12)=64, Y V=Y Z / 2=4$.