## Problem 3

A tetrahedron has at most one edge longer than 1 . What is the maximum total length of its edges?


## Solution

Answer: $5+\sqrt{3}$.

Suppose $\mathrm{AB}$ is the edge which may be longer than 1. Then if we rotate the triangle $\mathrm{ACD}$ about $\mathrm{CD}$ until $\mathrm{A}$ lies in the same plane as $B C D$ and is on the opposite side of $\mathrm{CD}$ to $\mathrm{B}$, then we maximise $\mathrm{AB}$ without changing any of the other side lengths. So the maximal configuration must be planar.

Now suppose we regard $\mathrm{C}$ and $\mathrm{D}$ as fixed and the other points as variable. Suppose $\mathrm{CD}=2 \mathrm{x}(<=1)$. Then $\mathrm{A}$ and $\mathrm{B}$ must both lie inside the circle center $\mathrm{C}$ radius 1 and inside the circle center $\mathrm{D}$ radius 1 and hence inside their intersection which is bounded by the two arcs XY (assuming they meet at $\mathrm{X}$ and $\mathrm{Y}$ ). Obviously we maximise $\mathrm{AC}+\mathrm{AD}+\mathrm{BC}+\mathrm{BD}$ by taking $\mathrm{A}$ at $\mathrm{X}$ and $\mathrm{B}$ at $\mathrm{Y}$ (or vice versa). We claim that that choice also maximises $\mathrm{AB}$. Suppose that is true. Then it also maximises $\mathrm{AC}+\mathrm{AD}+\mathrm{BC}+\mathrm{BD}+\mathrm{AB}$ at $4+2\left(1-\mathrm{x}^{2}\right)^{1 / 2}$. So we now have to vary $\mathrm{CD}$ to maximise $2 \mathrm{x}+4+2\left(1-\mathrm{x}^{2}\right)^{1 / 2}$. We show that $\mathrm{x}+\left(1-\mathrm{x}^{2}\right)^{1 / 2}$ is increasing for $\mathrm{x}$ $<=1 / 2$ and hence that the maximum is at $x=1 / 2$. Put $x=\sin$. Then we have $x+\left(1-x^{2}\right)^{1 / 2}=\sqrt{2} \sin (t+\pi / 4)$ which is indeed increasing for $\mathrm{x} \leq \pi / 6$.

It remains to prove the claim. Take the circle diameter XY. Then the two ares both lie inside this circle. [Two circles intersect in at most two points, so each arc must lie entirely inside or entirely outside the circle center $\mathrm{O}$ and it obviously does not lie outside. ] But AB lies inside a chord of this circle The length of the chord cannot exceed the diameter of the circle (which is $\mathrm{XY}$ ) and hence $\mathrm{AB} \leq \mathrm{XY}$.

