Given a triangular pyramid $ABCD$. A point $F$ is taken on edge $AC$, such that $CF: FA=2: 9$, and a point $M$ is taken on edge $CD$, such that $AM$ is the bisector of angle $DAC$. A plane is drawn through points $F, M$, and the centroid of triangle $DAB$, intersecting edge $DB$ at point $N$. It is known that $CA: AD = DN: NB + 1$. It is also known that the ratio of the area of triangle $ADB$ to the sum of the areas of all faces of the pyramid $ABCD$ is $p$, and the perpendicular dropped from vertex $C$ to the plane $ABD$ is $h$. A plane is drawn through point $N$, parallel to the plane $ACB$, intersecting edges $CD$ and $DA$ at points $K$ and $L$ respectively. Find the radius of the sphere inscribed in the pyramid $DKLN$.

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Let $E$ be the midpoint of edge $AB$, $P$ be the point of intersection of the medians of triangle $ABD$. Let lines $FM$ and $AD$ intersect at point $T$, line $TP$ intersect edge $AB$ at point $Q$, and lines $MN$ and $BC$ intersect at point $G$. Then line $QF$ passes through point $G$, and quadrilateral $MNQF$ is the specified section. Denote $\frac{DN}{NB}=k$. By the property of the angle bisector of a triangle,

$$
\text { times } \frac{CM}{MD}=\frac{AC}{AD}=\frac{DN}{MB}+1=k+1
$$

By Menelaus' theorem in triangles $BDE$, $BDC$, and $ABC$, we find that

$$
\frac{DN}{NB} \cdot \frac{BQ}{QS} \cdot \frac{EP}{PD}=1, \frac{AG}{QB} \cdot \frac{GG}{GC} \cdot \frac{GF}{FA}=1, \frac{DM}{NB} \cdot \frac{BG}{GC} \cdot \frac{GM}{MD}=1
$$

or

$$
k \cdot \frac{BQ}{QB} \cdot \frac{1}{2}=1, \frac{AQ}{QQ} \cdot \frac{BG}{GC} \cdot \frac{2}{9}=1, k \cdot \frac{BG}{GC} \cdot(k+1)=1
$$

From the second and third equalities, we find that

$$
\frac{AQ}{QB}=\frac{9(k(k+1)}{2}
$$

Let $AQ=x$, $EQ=y$. Since $E$ is the midpoint of $AB$, then

Therefore,

$$
\frac{AQ}{QB}+2 \cdot \frac{QE}{QB}=\frac{\pi}{x+2 \psi}+\frac{2 y}{x+2 \psi}=1
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-21.jpg?height=95&width=1123&top_left_y=244&top_left_x=-1)

$$
\frac{9 k(k+1)}{2}+k=1, \text { or } 9 k^2+11 k-2=0
$$

from which we find that $k=\frac{\sqrt{193-11}}{18}$. Let $s$ be the area of face $ABD$, $S$ be the sum of the areas of all faces of the pyramid, $V$ be its volume, $r$ be the radius of the sphere inscribed in the pyramid $DKLN$, $S_1$ be the total surface area of the pyramid $DKLN$, and $V_1$ be its volume. Then

$$
\begin{gathered}
V=\frac{1}{3} s h, V_1=V\left(\frac{k}{(k+1}\right)^3=\frac{1}{3} s h\left(\frac{k}{k+1}\right)^3, S_1=S\left(\frac{k}{k+1}\right)^2, \\
r=3 \cdot \frac{V_1}{S_1}=\frac{k}{(k+1)}^3 \cdot \frac{k h}{S\left(\frac{k}{k-1}\right)^2}= \\
=\frac{k}{k+1} \cdot h \cdot \frac{k}{S}=\frac{k}{k+1} \cdot h p=\frac{h p(\sqrt{193}-11)}{\sqrt{193}+7}=\frac{h p(15-\sqrt{103)}}{8}
\end{gathered}
$$

## Answer

$\frac{\operatorname{mon}(15-\sqrt{189})}{8}$.