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Consider two tetrahedrons one placed inside the other. Prove that the sum of all 6 sides of inner tetrahedron is at most the sum of the 6 sides of exterior tetrahedron.

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  • $\begingroup$ This looks like homework to me... $\endgroup$
    – zeb
    Dec 17, 2011 at 13:03
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    $\begingroup$ After thinking about it a bit, it probably isn't a homework question, because it is false... the outer tetrahedron can have side lengths close to 2, 2, 0, 1, 1, and 1, and the inner tetrahedron can have side lengths close to 2, 2, 2, 2, 0, and 0. $\endgroup$
    – zeb
    Dec 17, 2011 at 13:09
  • $\begingroup$ This is (if true) a nice problem, but it is more appropriate to an Olmpiad/Putnam competition than to research mathematics. You'd probably get both a more positive reception and better answers at artofproblemsolving.com/Forum/index.php . $\endgroup$ Dec 17, 2011 at 13:42
  • $\begingroup$ The origin of this question is the prior question mathoverflow.net/questions/79621/two-rectangular-parallelepiped $\endgroup$ Dec 17, 2011 at 19:08
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    $\begingroup$ Finding the maximal possible ratio of side sums (4/3) was a problem at the International mathematical olympiad long ago. $\endgroup$ Mar 7, 2012 at 17:55

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The question as stated is false, be we can salvage it by answering the following question: what's the biggest possible ratio of the sum of the inner tetrahedron's edges to the sum of the outer tetrahedron's edges?

To approach problems like this, there is the very useful trick of noticing that the distance between two points is a convex function of either point. If we assume the outer tetrahedron is fixed, and fix three of the vertices of the inner tetrahedron, then since the interior of the outer tetrahedron is convex, the sum of the side lengths of the inner tetrahedron, as a function of the last vertex, is maximized when the last vertex is at an extreme point of the outer tetrahedron. Repeating this argument for each vertex of the inner tetrahedron, we can reduce to the degenerate cases where the inner tetrahedron has all of its vertices on the vertices of the outer tetrahedron. We have several cases:

  • The inner tetrahedron and the outer tetrahedron are the same. Then the sums of the side lengths are equal.
  • The inner tetrahedron is a face of the outer tetrahedron, with one of the vertices repeated. Suppose the outer tetrahedron has vertices $ABCD$, and the inner tetrahedron has vertices $AABC$. Then the sum of the lengths of the inner tetrahedron is $0+2AB+2AC+BC \le \frac{4}{3}AB + \frac{2}{3}(AD+BD) + \frac{4}{3}AC + \frac{2}{3}(AD+CD) + BC$ $\le \frac{4}{3}(AB+AC+BC+AD+BD+CD)$, and we have equality if the outer tetrahedron is degenerate, with $B=C=D$.
  • The inner tetrahedron is a side of the outer tetrahedron. Suppose the outer tetrahedron has vertices $ABCD$, and the inner tetrahedron has vertices $AABB$. Then the sum of the side lengths of the inner tetrahedron is $0+0+4AB \le \frac{4}{3}AB + \frac{4}{3}(AC+BC) + \frac{4}{3}(AD+BD)$ $\le \frac{4}{3}(AB+AC+BC+AD+BD+CD)$, with equality when $C = D$ and $C$ is on the line segment $AB$.

Thus, the sum of the side lengths of the inner tetrahedron is at most $\frac{4}{3}$ times the sum of the side lengths of the outer tetrahedron.

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    $\begingroup$ Perhaps you'd like to mention, for the peace of mind, also the easy (not sharp) case $AAAB$. $\endgroup$ Feb 17, 2013 at 8:02
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The conjecture about the quotient of the sums of the lengths of edges for a tetrahedron inside another was formulated (a long time ago) by Jerzy Browkin, who saw that, somewhat paradoxically it can be greater than 1, and even arbitrarily close to 4/3, as his examples proved it. He asked if   $\frac 43$   is the least upper bound of such quotients.

The conjecture was solved in 1961 by Wlodzimierz Kuperberg and myself. We published it in 1962 or 1963, in Polish, in Wiadomosci Matematyczne (of Polish Mathematical Society). Years later an English translation was published in Alabama Journal of Mathematics (or similar). And a powerful complete multidimensional generalization was obtained and published by Carl Linderholm, An inequality for simplices .

Ancient Greeks could easily formulate and solve the given problem (in 3d) but (as long as we know) never did--it took Jerzy Browkin to raise it.

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    $\begingroup$ I have added the name (Carl Linderholm) of the mathematician who obtained the ultimate (finite-dimensional, geometric) generalization of the tetrahedra result + link. $\endgroup$ May 1, 2013 at 2:19
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Sum of the 6 edges of the inner tetrahedron can be larger than the sum of the 6 edges of the outer tetrahedron. Imagine the outer tetrahedron with 3 sides of 1000 and 3 sides of 1. The inner tetrahedron has 2 vertices close to the top and 2 vertices close to the bottom. Sum of outer = 1000+1000+1000+1+1+1=3003 Sum of inner = 990+990+990+990+1+1= 3962

enter image description here

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