The force on the mass $M$ due to each of the masses $m$ will be have a vertical component of
$\begin{align*}\frac{1}{2} M g\end{align*}$
This force must be normal to the surface of the triangle, so the horizontal component is
$\begin{align}\frac{1}{2} M g \label{eqn:1}\end{align}$
The force of friction is
$\begin{align*}\mu \left(m g + \frac{1}{2} \cdot M g\right)\end{align*}$
Setting this equal to (1) and solving for $M$ gives
$\begin{align*}M = \frac{2 \mu m}{1 - \mu}\end{align*}$
Therefore, answer (D) is correct.

Solution to 1992 Problem 6