One tetrahedron inside another tetrahedron - MathOverflow

One tetrahedron inside another tetrahedron

2011-12-17T12:40:04Z

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.

Answer by zeb for One tetrahedron inside another tetrahedron

2011-12-17T13:45:36Z

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.

Answer by Wlodzimierz Holsztynski for One tetrahedron inside another tetrahedron

2013-02-17T06:39:35Z

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.