A sphere with center at point $O$ passes through the vertices $A, B$, and $C$ of the tetrahedron $A B C D$ and intersects the lines $A D, B D$, and $C D$ at points $K, L$, and $M$ respectively. It is known that $A D=10, B C: B D=3: 2$ and $A B: C D=4 \sqrt{3}$ $: 11$. The projections of point $O$ onto the planes $A B D, B C D$, and $C A D$ are the midpoints of the edges $A B, B C$, and $A C$ respectively. The distance between the midpoints of the edges $A B$ and $C D$ is 13. Find the perimeter of triangle $K L M$.
The section of the given sphere by the plane of face $ABD$ is a circle with its center at the midpoint of segment $AB$, passing through points $A, B, K$, and $L$ (Fig.1). Therefore, $AB$ is the diameter of this circle. The segment $AB$ is seen from points $K$ and $L$ at a right angle, so $AL$ and $BK$ are altitudes of triangle $ABD$. Similarly, $BM$ and $CL$ are altitudes of triangle $BCD$, and $CK$ and $AM$ are altitudes of triangle $ACD$. The line $CD$ is perpendicular to two intersecting lines $AM$ and $BM$ in the plane $ABM$, so the line $CD$ is perpendicular to this plane. Therefore, $CD \perp AB$. Similarly, $AD \perp BC$ and $BD \perp AC$. We extend the given tetrahedron $ABCD$ to a parallelepiped $APBQSDRC$ (where $AS\|PD\|BR\|QC$) by drawing through its opposite edges pairs of parallel planes (Fig.2). Since $PQ \| CD$, then $PQ \perp AB$, so the parallelogram $APBQ$ is a rhombus. Similarly, all six faces of the parallelepiped $APBQSDRC$ are rhombuses. Therefore, all edges of the parallelepiped are equal. Let their lengths be denoted by $a$. By the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
AB^2 + CD^2 = 4a^2, \quad AC^2 + BD^2 = 4a^2, \quad AD^2 + BC^2 = 4a^2
$$

Therefore,

$$
AB^2 + CD^2 = AC^2 + BD^2 = AD^2 + BC^2
$$

Let $BC = 3x, BD = 2x, AB = 4y\sqrt{3}, CD = 11y$. Since the edge of the parallelepiped $APBQSDRC$ is equal to the segment connecting the centers of opposite faces, $a = 13$. Therefore,

$$
100 + 9 \times 2 = 4 \cdot 169, \quad 48y^2 + 121y^2 = 4 \cdot 169, \quad AC^2 + 4x^2 = 4 \cdot 169
$$

From which we find that

$$
\begin{gathered}
x = 8, \quad y = 2, \quad AC = 2\sqrt{105}, \\
BC = 24, \quad BD = 16, \quad AB = 8\sqrt{3}, \quad CD = 22
\end{gathered}
$$

Consider triangle $ABD$. Let $\angle ADB = \alpha$. Since $BK$ and $AL$ are altitudes of triangle $ABD$, triangle $KDL$ is similar to triangle $BDA$ with a similarity ratio of $|\cos \alpha|$. Therefore, $KL = AB|\cos \alpha|$. By the cosine rule, we find that

$$
\cos \alpha = \frac{AD^2 + BD^2 - AB^2}{2 \cdot AD \cdot BD} = \frac{100 + 256 - 192}{2 \cdot 10 \cdot 16} = \frac{164}{4 \cdot 80} = \frac{41}{80}
$$

Thus, $KL = 8\sqrt{3} \cdot \frac{41}{80} = \frac{41\sqrt{3}}{10}$. Similarly, we find that

$$
LM = BC|\cos \angle BDC| = \frac{24(256 + 454 - 576)}{2 \cdot 16 \cdot 22} = \frac{24 \cdot 164}{2 \cdot 16 \cdot 22} = \frac{3 \cdot 41}{22}
$$

$$
KM = AC \cdot |\cos \angle ADC| = \frac{2\sqrt{105}(100 + 484 - 420)}{2 \cdot 10 \cdot 22} = \frac{2\sqrt{105} \cdot 164}{2 \cdot 10 \cdot 22} = \frac{41\sqrt{105}}{55}
$$

Therefore, the perimeter of triangle $KLM$ is

$$
\frac{41\sqrt{3}}{10} + \frac{41}{22} + \frac{41\sqrt{105}}{55} = 41\left(\frac{\sqrt{105}}{55} + \frac{\sqrt{3}}{10} + \frac{3}{22}\right)
$$

## Answer

$41\left(\frac{\sqrt{105}}{55} + \frac{\sqrt{3}}{10} + \frac{3}{22}\right)$