2
$\begingroup$

How do you prove this? And, more generally, can a $n$-simplex cast a hypercube shadow in one lower dimension?

$\endgroup$
3
  • 2
    $\begingroup$ The community here prefers/expects questions to include something of what the asker knows about a problem. (What have you tried? Where did you get stuck? etc) This information helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already understand or using techniques unfamiliar to you. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) Comments are easily overlooked, so please edit your question to add details. $\endgroup$ – Blue Feb 7 '20 at 4:01
  • $\begingroup$ Sorry about that. I was just curious, and genuinely had no idea how to approach the problem, so I couldn't really try anything. I literally built an origami tetrahedron and looked at it, thought "yeah, that's about right", and got stuck there. $\endgroup$ – Erez Israeli Miller Feb 10 '20 at 4:36
  • 1
    $\begingroup$ That's exactly the kind of context that should appear in the question. :) It's good (even gratifying) when a question is one of curiosity. Our issue with raw "problem statement questions" is that we can't distinguish sincere curiosity from lazy homework. Generally, the more you can say about a problem, the better. (Knowing that you tried origami tells us you're invested in the problem, and it can inspire us to get invested, too.) If you don't know how to proceed, that's fine; we're here to help. But we could use a sense of how to help (and, sometimes, how not to help). Cheers! $\endgroup$ – Blue Feb 10 '20 at 5:26
6
$\begingroup$

Answer is "yes". Take two non-adjacent edges of the tetrahedron and construct a plane, parallel to both of them. This will be your projection plane. The projection is supposed to be orthographic - all projection lines are perpendicular to the projection plane.

The tetrahedron will be viewed from above as (the invisible edge is drawn by dotted line): enter image description here Informal proof: these two edges are perpendicular, have the same length and intersect in the middle - so they are diagonals of a square.

As for your general question about $n$-simplex - I appreciate the @IvanNeretin answer, given in comments. The answer is "no" for all $n \gt 3$, because the $n$-simplex has $n+1$ vertices, so any its projection will have this number of vertices, or less. However the number of vertices of $(n-1)$-dimensional hypercube is $2^{n-1}$, which is more than $n+1$ for $n \gt 3$.

$\endgroup$
3
  • 2
    $\begingroup$ The answer for $n$-simplex with $n>3$ is obviously no, for it has fewer vertices than a cube in $n-1$ dimensions. $\endgroup$ – Ivan Neretin Feb 7 '20 at 9:21
  • $\begingroup$ @IvanNeretin - right, easy and simple! Thanks $\endgroup$ – HEKTO Feb 7 '20 at 17:59
  • $\begingroup$ Thank you, that's a very elegant answer. $\endgroup$ – Erez Israeli Miller Feb 20 '20 at 21:03
0
$\begingroup$

Yes.

Viewing direction is along center of two skewed sides onto a plane perpendicular to it in orthographic projection. It also passes through tetrahedron center. The viewing point is at infinity.

Spherical symmetry restraint of four connected lines of equal projected length makes for a square for such a viewing direction. Mathematica projection is slightly off due to finite viewing distance and due to angle made to such a direction.

Tetrahedron projection

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.