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Every triangle $A$ can be oriented in $\mathbb{R}^3$ so that its orthogonal projection (shadow) onto the $xy$-plane is an equilateral triangle $Q$:


                   


Q. Can every $(d{-}1)$-simplex $A$ in $\mathbb{R}^d$ be oriented so that its orthogonal projection to $\mathbb{R}^{d-1}$ is a regular $(d{-}1)$-simplex $Q$?

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  • $\begingroup$ Try projecting a rectangular simplex onto a regular simplex. Then consider how to transform any simplex into a rectangular simplex. $\endgroup$ – The Masked Avenger Apr 11 '15 at 4:04
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I think the answer is no for $d\ge 4$.

Looking things backward, you start from a regular simplex $\Delta$ in $\mathbb{R}^{d-1}\times\{0\}$ and you choose a height for each of its vertex, defining a $(d-1)$ simplex in $\mathbb{R}^d$ as the convex hull of the points at the given height above each vertex of $\Delta$; you want to realize every possible $(d-1)$ simplex geometry that way.

On the one hand, you only have $d$ heights to choose, and with the vertical translation invariance this leaves you with $d-1$ degrees of freedom. On the other hand, the moduli space of $(d-1)$ simplices (up to similarity), if I am not mistaken (edit: I was, now corrected), has dimension $d(d-1)/2-1$. For $d=3$ you get an equality, but as soon as $d>3$ there are too many simplices for them to be projected on only one shadow.

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    $\begingroup$ The simplex is determined by only $\left( \begin{array}{c} d \\ 2 \end{array}\right)$ distances, satisfying some inequalities, so the number of degrees of freedom is $d(d-1)/2$. Still, this argument applies. $\endgroup$ – Will Sawin Apr 11 '15 at 13:45
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    $\begingroup$ Oh right! I performed a product of degrees of freedom instead of a sum. That's kind of embarrassing, but at least it will give some indulgence next time a student uses the first operation he can think of to answer a question. $\endgroup$ – Benoît Kloeckner Apr 11 '15 at 16:20
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No, already for $d=4$. Take, for example, a $3$-simplex with vertices $A=(0,0,0)$, $B=(100,0,0)$, $C=(0,100,0)$, $D=(0,0,1)$, embedded in $\mathbb{R}^4$. Every projection of the triangle $ABC$ has diameter at least $50\sqrt{2}$, but every projection of the segment $AD$ has length at most $1$.

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