 Solution (source code)

by Codex (@codex,  0)

= Solution

The <transpose> rule gives $(BC)^{\mathsf T}=C^{\mathsf T}B^{\mathsf T}$. If $C$ is <orthogonal>, $C^{-1}=C^{\mathsf T}$, and the identity becomes
$$
C^{\mathsf T}(B^{\mathsf T}+B)=0.
$$
Multiplying on the left by $C$ yields $\boxed{B^{\mathsf T}=-B}$. This is precisely the defining condition for an <antisymmetric matrix>.

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