[ Geometry (other) ]

A sphere with radius $3 / 2$ has its center at point $N$. From point $K$, located at a distance of $3 \sqrt{5} / 2$ from the center of the sphere, two lines $K L$ and $K M$ are drawn, touching the sphere at points $L$ and $M$ respectively. Find the volume of the pyramid $K L M N$, given that $M L=2$.
From the right triangle $K L N$, we find that

$$
K L=\sqrt{K N^{2}-L N^{2}}=\sqrt{45 / 4-9 / 4}=3
$$

Therefore, $K M=K L=3$. In the isosceles triangles $L N M$ and $L K M$, the medians $N P$ and $K P$ are altitudes, so

$$
\begin{gathered}
N P^{2}=\sqrt{N L^{2}-L P^{2}}=\sqrt{9 / 4-1}=\sqrt{5} / 2 \\
K P^{2}=\sqrt{K L^{2}-L P^{2}}=\sqrt{9-1}=\sqrt{8} / 2
\end{gathered}
$$

Thus, $S(L N M)=\frac{1}{2} L M \cdot N P=\frac{1}{2} 2 \cdot \sqrt{5} / 2=\sqrt{5} / 2$.

Let $K H$ be the height of the pyramid $K L M N$. Since $K L=K M$, point $H$ is equidistant from points $L$ and $M$. Therefore, point $H$ lies on the line $N P$, and $K H$ is the height of triangle $K N P$. Let $P H=x$. By the Pythagorean theorem,

$$
K N^{2}-N H^{2}=K P^{2}-P H^{2}, \text { or } 45 / 4-(\sqrt{5} / 2+x)^{2}=8-x^{2},
$$

from which $x=2 / \sqrt{5}$. Therefore,

$$
K H=\sqrt{K P^{2}-P H^{2}}=\sqrt{8-4 / 5}=6 / \sqrt{5}
$$

Thus,

$$
V(K L M N)=\frac{1}{3} S(L N M) \cdot K H=\frac{1}{3}(\sqrt{5} / 2) \cdot 6 / \sqrt{5}=1 .
$$

## Answer

1.