7. On the edges $A C, B C, B S, A S$ of a regular triangular pyramid $S A B C$ with vertex $S$, points $K, L, M, N$ are chosen respectively. It is known that points $K, L, M, N$ lie in the same plane, and $K L = M N = 14, K N = L M = 25$. In the quadrilateral $K L M N$, there are two circles $\Omega_{1}$ and $\Omega_{2}$, where circle $\Omega_{1}$ is tangent to sides $K N, M N$, and $L M$, and circle $\Omega_{2}$ is tangent to sides $K L, K N$, and $L M$. The right circular cones $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ with bases $\Omega_{1}$ and $\Omega_{2}$ respectively are located inside the given pyramid, with the vertex $P$ of cone $\mathcal{F}_{1}$ lying on edge $A B$, and the vertex $Q$ of cone $\mathcal{F}_{2}$ lying on edge $C S$.

a) Find $\angle S A B$.

b) Find the length of segment $C Q$.
Answer: a) $\angle S A B=\arccos \frac{3}{5} ;$ b) $C Q=\frac{77}{5}$.

Solution. The opposite sides of quadrilateral $K L M N$ are pairwise equal, so it is a parallelogram. Since the plane $(K L M N)$ intersects the planes $(A B C)$ and $(A B S)$ along parallel lines $K L$ and $M N$, these lines are parallel to the line of intersection of these planes, i.e., $A B$. Similarly, $N K\|L M\| S C$. In a regular triangular pyramid, skew edges are perpendicular to each other, so $S C \perp A B$, and $K L M N$ is a rectangle. Therefore, the radii of circles $\Omega_{1}$ and $\Omega_{2}$ are both 1.

From this, it also follows that the rectangle $K L M N$ is symmetric with respect to the plane $\alpha$, which contains the edge $S C$ and the midpoint of $A B$. Then the cones $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ are also symmetric with respect to this plane. Therefore, $P$ is the midpoint of $A B$.[^3]

Let $X$ and $Y$ be the midpoints of sides $K L$ and $M N$ respectively, and $O_{1}$ and $O_{2}$ be the centers of circles $\Omega_{1}$ and $\Omega_{2}$ respectively; these four points lie on the axis of symmetry of the rectangle $K L M N$, parallel to $K N$, and thus in the plane $\alpha$. Moreover, $X Y \| S C$, so triangles $P C S$ and $P X Y$ are similar.

Let $A B=B C=C A=2 a, S A=S B=S C=\ell, \nu=a / \ell$. Then $C P=a \sqrt{3}, S P=\sqrt{\ell^{2}-a^{2}}$. Since $X Y=K N=25$, from the similarity we get $\frac{X P}{C P}=\frac{X Y}{C S}$, i.e., $\frac{X P}{a \sqrt{3}}=\frac{25}{\ell}, X P=\frac{25 \sqrt{3} a}{\ell}=25 \nu \sqrt{3}$. Similarly, $\frac{Y P}{S P}=\frac{X Y}{C S}, \frac{Y P}{S P}=\frac{25}{\ell}, Y P=\frac{25 \sqrt{\ell^{2}-a^{2}}}{\ell}=25 \sqrt{1-\nu^{2}}$. On the other hand, since the cone $\mathcal{F}_{1}$ is a right cone, we have $P O_{1} \perp X Y$, and $Y O_{1}=\frac{1}{2} M N=7, X O_{1}=X Y-Y O_{1}=18$. Therefore, $7^{2}-18^{2}=O_{1} Y^{2}-O_{1} X^{2}=\left(O_{1} Y^{2}+O_{1} P^{2}\right)-\left(O_{1} X^{2}+O_{1} P^{2}\right)=P Y^{2}-P X^{2}=25^{2}\left(1-\nu^{2}-3 \nu^{2}\right)$, or $-11 \cdot 25=25^{2}\left(1-4 \nu^{2}\right)$, from which $\nu=\frac{3}{5}$. Thus, $\angle S A B=\arccos \frac{A P}{A S}=\arccos \nu=\arccos \frac{3}{5}$.

Therefore, $\ell=\frac{5 a}{3}$, and from the similarity we have

$$
\frac{14}{2 a}=\frac{K L}{A B}=\frac{C X}{C P}=1-\frac{X P}{C P}=1-\frac{X Y}{C S}=1-\frac{25}{\ell}=1-\frac{15}{a}
$$

from which $a=22$ and $\ell=\frac{110}{3}$. Let $P O_{1}$ intersect $S C$ at point $H$. Then $P H$ is the height of triangle $S C P$, and (since $X Y \| C S$) $\frac{C H}{C S}=\frac{X O_{1}}{X Y}=\frac{18}{25}$. Thus, $C H=\frac{18}{25} S C=\frac{132}{5}$. Since $O_{2} Q \perp X Y, H O_{1} O_{2} Q$ is a rectangle, so $H Q=O_{1} O_{2}=11$. Therefore, $C Q=C H-H Q=\frac{77}{5}$.

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