[ Geometry (miscellaneous).]

A sphere of radius $\sqrt{5}$ with center at point $O$ touches all sides of triangle $ABC$. The point of tangency $N$ bisects side $AB$. The point of tangency $M$ divides side $AC$ such that $AM=\frac{1}{2} MC$. Find the volume of the pyramid $OABC$, given that $AN=NB=1$.
The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semi-perimeter.

## Solution

Let the given sphere touch the side $BC$ of triangle $ABC$ at point $K$. Then

$$
BK = BN = 1, AM = AN = 1, CM = 2 \cdot AM = 2, CK = CM = 2
$$

The section of the sphere by the plane of triangle $ABC$ is the circle inscribed in triangle $ABC$, and the center $O_1$ of this circle is the orthogonal projection of the center $O$ of the sphere onto the plane of triangle $ABC$. Therefore, $OO_1$ is the height of the pyramid $OABC$.

Let $r$ be the radius of the circle inscribed in triangle $ABC$, $p$ be the semi-perimeter of the triangle, and $s$ be the area. Since triangle $ABC$ is isosceles, the segment $CN$ is its height. Then

$$
\begin{gathered}
CN = \sqrt{AC^2 - AN^2} = \sqrt{9 - 1} = 2\sqrt{2}, \\
s = \frac{1}{2} AB \cdot CN = 2\sqrt{2}, \quad r = s / p = 2\sqrt{2} / 4 = \sqrt{2} / 2
\end{gathered}
$$

From the right triangle $OO_1N$, we find that

$$
OO_1 = \sqrt{ON^2 - ON^2} = \sqrt{5 - 1/2} = 3 / \sqrt{2} \text{.}
$$

Therefore,

$$
V(OABC) = \frac{1}{3} s \cdot OO_1 = \frac{1}{3} 2\sqrt{2} \cdot 3 / \sqrt{2} = 2
$$

## Answer

2.