8,9The legs of a right triangle are 15 and 20. Find the distance from the center of the inscribed circle to the height dropped to the hypotenuse.

#
The desired distance is equal to the distance between the base of the given height and the point of tangency of the inscribed circle with the hypotenuse.

## Solution

Let $O$ be the center of the inscribed circle, $Q$ and $P$ be the points of tangency with the smaller leg $BC$ and the hypotenuse $AB$, $CM$ be the height of the triangle, and $OK$ be the desired distance. If $r$ is the radius of the inscribed circle, then

$$
r=\frac{BC + AC - AB}{2}=\frac{20 + 15 - 25}{2}=5
$$

Therefore,

$$
\begin{gathered}
CQ = r = 5, \quad BP = BQ = BC - CQ = 15 - 5 = 10, \\
BM = \frac{BC^2}{AB} = \frac{225}{25} = 9, \\
OK = PM = BP - BM = 10 - 9 = 1.
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-22.jpg?height=283&width=669&top_left_y=2351&top_left_x=703)

## Answer

1. 

In a right triangle, a height is drawn from the vertex of the right angle. A circle is constructed on this height as a diameter. It is known that this circle intercepts segments of 12 and 18 on the legs. Find the legs.

## Hint

Connect the endpoints of the specified segments with the bases of the perpendicular.

## Solution

Let the given circle intersect the legs $AC$ and $BC$ of triangle $ABC$ at points $D$ and $E$, respectively, different from point $C$, with $CD = 12$ and $CE = 18$. If $M$ is the foot of the perpendicular dropped from vertex $C$ to $AB$, then

$$
\angle MDC = \angle MEC = 90^\circ, \quad DM = CE = 18, \quad ME = CD = 12.
$$

From the right triangles $AMC$ and $BMC$, we find that

$$
DM^2 = AD \cdot DC, \quad ME^2 = BE \cdot EC
$$

Thus,

$$
AD = \frac{DM^2}{DC} = \frac{18^2}{12} = 27, \quad BE = \frac{ME^2}{EC} = \frac{12^2}{18} = 8
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-23.jpg?height=393&width=1054&top_left_y=1210&top_left_x=502)

Answer

39 and 26.