[Pythagorean Theorem (direct and inverse) $]$ Law of Cosines $\quad]$

A circle inscribed in a right triangle with legs of 6 and 8 touches the hypotenuse at point $M$. Find the distance from point $M$ to the vertex of the right angle.
Let the circle inscribed in triangle $ABC$ touch side $AB$ at point $M$. Then $AM = p - BC$, where $p$ is the semiperimeter of triangle $ABC$.

## Solution

Let the circle inscribed in the right triangle $ABC$ touch the hypotenuse $AB$ at point $M$. If $AC = 6$ and $BC = 8$, then

$$
AB = \sqrt{AC^2 + BC^2} = 10, \cos \angle A = \frac{AC}{AB} = \frac{3}{5}
$$

Let $p$ be the semiperimeter of triangle $ABC$. Then

$$
AM = p - BC = 12 - 8 = 4
$$

By the Law of Cosines in triangle $AMC$, we find that

$$
CM^2 = AC^2 + AM^2 - 2 \cdot AC \cdot AM \cdot \cos \angle A = 36 + 16 - 2 \cdot 6 \cdot 4 \cdot \frac{3}{5} = \frac{116}{5}
$$

Therefore,

$$
CM = \sqrt{\frac{116}{5}} = 2 \sqrt{\frac{29}{5}}
$$

## Answer

$2 \sqrt{\frac{29}{5}}$