# Can a tetrahedron lying completely inside another tetrahedron have a larger sum of edge lengths?

Find 2 tetrahedrons $ABCD$ and $EFGH$ such that

• $EFGH$ lies completely inside $ABCD$.

• The sum of edge lengths of $EFGH$ is strictly greater than the sum of edge lengths of $ABCD$.

I am completely stumped on this. Seems very counter intuitive to begin. I now have doubts if a solution exists or not.

Source : Here

A spire is a tetrahedron with one tiny face and three long edges. A splinter is a tetrahedron with two tiny opposite edges and four long edges.

From Cheng et al., "Sliver Exudation", Proc. J. ACM, 2000.

Take a spire of height $1$ and fit a splinter inside it. The sum of edge lengths of the spire is $\approx 3$ while that of the splinter is $\approx 4$.

• Nice to see some references. +1 Jun 5, 2013 at 3:56

$$A=(0,0,0), \;B = (1,0,0),\;C=(1,1,0) ,\;D=(-1,0,1)$$

$$E =(1,1-\epsilon,0),\; F = (1-\epsilon,0,\epsilon/2),\; G = (-1+\epsilon,0,1-\epsilon/2) ,\;H = (-1+\epsilon,\epsilon/2,1-\epsilon/2)$$

Let $\epsilon = 0.01$, sum of edge length of $ABCD$: $$|AB|+|AC|+|AD|+|BC|+|BD|+|CD| = 2+2\sqrt{2}+\sqrt{5}+\sqrt{6}\approx 9.51.$$

sum of edge length of $EFGH$: $$|EF|+|EG|+|EH|+|FG|+|FH|+|GH| \approx 10.3.$$

Roughly looks like the following:

You can perturb $EFGH$ by setting them completely inside $ABCD$, by adding another parameter $\delta\ll 1$ while having the sum of edge lengths changing by $O(\delta)$.

The backstory of this construction is that: If $ABCD$ has three long edges, we can make $EFGH$ having four long edges.

• +1 for the backstory. A good way to think of it. Mine is a similar, but different, backstory. Jun 5, 2013 at 4:02
• @RossMillikan For I have been working on numerical pde on simplicial mesh for a while, the first thing popped in my mind was: hey, the interior tetrahedron is a typical tetrahedron we should avoid in Delaunay triangulation, while the exterior tetrahedron was on the edge of being ok and bad. Jun 5, 2013 at 4:12
• Fantastic answer. Could have never come up with this solution on my own. Jun 5, 2013 at 5:39

Think of a tetrahedron composed of $(-k,0,0), (-k,\epsilon,0), (0,\epsilon,0), (0,0,k)$ where $k$ is large and $\epsilon \ll 1$ You can move $(0,\epsilon,0)$ to the right (and down a bit) and increase the sum of the sides. You can then move all the points inward by $\epsilon^2$ to get strictly inside.

• Lovely way to think. Jun 5, 2013 at 5:39