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Say you are given triangle ABC in $\mathbb{R}^n$, and its translation A'B'C' such that A'B'C' is not coplanar with ABC. Must it be the case that the convex hull of ABC and A'B'C' is a triangular prism?

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  • $\begingroup$ @Logan, thanks. For a minute I thought I had a counterexample to this. Oops. $\endgroup$ – Dan Moore May 24 '12 at 13:05
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Yes.

Note that for such a translation to exist, $n \ge 3$. Set A as the origin. Let $e_1, \ldots, e_n$ be the standard basis for $\mathbb{R}^n$. Perform a linear transformation sending $\vec{AB}$ to $e_1$, $\vec{AC}$ to $e_2$, and $\vec{AA'}$ to $e_3$. It should be clear that the convex hull of $\{ 0, e_1, e_2, e_3, e_1+e_3, e_2+e_3\}$ is a triangular prism, and since invertible linear transformations preserve the combinatorial type of polytopes and commute with taking convex hulls, the original polytope was also a triangular prism.

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