Given a regular triangular pyramid $S A B C$. Point $S$ is the apex of the pyramid, $S A=2 \sqrt{3}, B C=3, B M$ is the median of the base of the pyramid, $A R$ is the height of triangle $A S B$. Find the length of segment $M R$.

Answer

## $\frac{9}{4}$

Given a unit cube $A B C D A 1 B 1 C 1 D 1$. Find the radius of the sphere passing through points $A, B, C 1$ and the midpoint of edge $B 1 C 1$.
Let $M$ be the midpoint of edge $B_1C_1$ (Fig.1). The center $O$ of the sphere passing through points $A, B, C_1$, and $M$ is equidistant from the endpoints of segment $AB$, so point $O$ lies in the plane perpendicular to segment $AB$ and passing through its midpoint. Similarly, point $O$ lies in the plane perpendicular to segment $MC_1$ and passing through its midpoint, and in the plane perpendicular to diagonal $BC_1$ of face $BB_1C_1C$ and passing through its midpoint, i.e., through the center $Q$ of square $BB_1C_1C$. Consider the orthogonal projection of the cube onto the plane $BB_1C_1C$. Let $O'$ be the projection of the center of the sphere, $N$ be the midpoint of $BC$, $K$ be the midpoint of $CN$, $L$ be the midpoint of $MC_1$, and $H$ be the midpoint of $AB$. Then $O'K$ is the midline of triangle $NQC$, so

$$
O'K = \frac{1}{2} QN = \frac{1}{2} \cdot \frac{1}{2} BB_1 = \frac{1}{4}
$$

If $P$ is the orthogonal projection of point $O$ onto the plane $ABCD$ (Fig.2), then $OP = O'K = \frac{\mathbf{1}}{\mathbf{4}}$. Let $R$ be the desired radius. Then

$$
\begin{gathered}
R = OA = \sqrt{OP^2 + PH^2 + AH^2} = \sqrt{\left(\frac{1}{4}\right)^2 + \left(\frac{3}{4}\right)^2 + \left(\frac{1}{2}\right)^2} = \\
= \sqrt{\frac{1}{16} + \frac{9}{16} + \frac{4}{16}} = \sqrt{\frac{14}{16}} = \frac{\sqrt{14}}{4}
\end{gathered}
$$

## Answer

$\frac{\sqrt{14}}{4}$