## [

[ Quadrilateral Pyramid ]

Edge $P A$ of the pyramid $P A B C$ is perpendicular to the base plane $A B C$ and equals 1. In triangle $A B C$, the angle at vertex $A$ is a right angle, and each of the legs $A B$ and $A C$ equals 2. Points $M$ and $N$ are the midpoints of $A C$ and $B C$ respectively. Find the radius of the sphere inscribed in the pyramid PMNC.

#
If $r$ is the radius of the sphere inscribed in the pyramid $P M N C$, $S$ is the total surface area of this pyramid, and $V$ is its volume, then $r=\frac{3 V}{S}$. Since $M N$ is the midline of triangle $A B C$, then $M N \| A B$, and since $A B$ is perpendicular to the plane $A P C$, then $M N \perp P M$, so $P M$ is the height of triangle $M P N$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_2c7701213788eaf04894g-28.jpg?height=109&width=1140&top_left_y=2158&top_left_x=465)

Note that $A N$ is the height of triangle $A B C$. Then

$$
A N=\frac{A B \cdot A C}{B C}=2 \cdot \frac{2}{2 \sqrt{2}}=\sqrt{2}
$$

By the theorem of three perpendiculars, $P N \perp B C$, so $P N$ is the height of triangle $B P C$ and

$$
\begin{gathered}
P N=\sqrt{P A^{2}+A N^{2}}=\sqrt{1+2}=\sqrt{3} \\
S_{\triangle C P N}=\frac{1}{2} S_{\triangle B P C}=\frac{1}{2} \cdot \frac{1}{2} B C \cdot P N=\frac{1}{2} \cdot \frac{1}{2} \cdot 2 \sqrt{2} \cdot \sqrt{3}=\frac{\sqrt{6}}{2}
\end{gathered}
$$

Thus,

$$
S=S_{\triangle P M N}+S_{\triangle C P N}+S_{\triangle C P M}+S_{\triangle C M N}=
$$

$$
\begin{gathered}
=S_{\triangle P M N}+S_{\triangle C P N}+\frac{1}{2} S_{\triangle A P C}+\frac{1}{4} S_{\triangle A B C}= \\
=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}+\frac{1}{2} \cdot \frac{1}{2} \cdot 2 \cdot 1+\frac{1}{4} \cdot \frac{1}{2} \cdot 2 \cdot 2=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}+\frac{1}{2}+\frac{1}{2}=\frac{\sqrt{2}+\sqrt{6}+2}{2}
\end{gathered}
$$

and since

$$
V=\frac{S_{\triangle C M A}}{S_{\triangle A B C}} \cdot V_{P A B C}=\frac{1}{4} V_{P A B C}=\frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot 2 \cdot 2 \cdot 1=\frac{1}{6}
$$

then

$$
r=\frac{3 V}{S}=\frac{1}{\sqrt{2}+\sqrt{6}+2}
$$

## Answer

## $\frac{1}{\sqrt{2}+2+\sqrt{6}}$.