Blocks on two inclines of a free wedge

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This article is about the following scenario. A freely movable triangular wedge placed on a fixed frictionless horizontal floor has two inclines $I_1$ and $I_2$ making angles $\alpha_1$ and $\alpha_2$ with the horizontal. There are blocks of masses $m_1$ and $m_2$, resting on the two inclines $I_1$ and $I_2$ respectively. All surfaces of contact are frictionless.

Summary of cases

Case	What happens
$\! m_1 \sin(2\alpha_1) = m_2 \sin(2\alpha_2)$	The wedge remains stationary and the blocks both slide down their respective inclines.
$\! m_1 \sin(2\alpha_1) > m_2 \sin(2\alpha_2)$	The horizontal push exerted by $m_1$ exceeds that by $m_2$, causing the wedge to move in the push direction of $m_1$. Both $m_1$ and $m_2$ slide down.
$\! m_1 \sin(2\alpha_1) < m_2 \sin(2\alpha_2)$	The horizontal push exerted by $m_2$ exceeds that by $m_1$, causing the wedge to move in the push direction of $m_2$. Both $m_1$ and $m_2$ slide down.