# One tetrahedron inside another tetrahedron

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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This looks like homework to me... –  zeb Dec 17 '11 at 13:03
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. –  zeb Dec 17 '11 at 13:09
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 . –  David Speyer Dec 17 '11 at 13:42
The origin of this question is the prior question mathoverflow.net/questions/79621/two-rectangular-parallelepiped –  Zsbán Ambrus Dec 17 '11 at 19:08
Finding the maximal possible ratio of side sums (4/3) was a problem at the International mathematical olympiad long ago. –  mikhail skopenkov Mar 7 '12 at 17:55

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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Perhaps you'd like to mention, for the peace of mind, also the easy (not sharp) case $AAAB$. –  Wlodzimierz Holsztynski Feb 17 '13 at 8:02
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.