## Problem 4

A tetrahedron ABCD has edges of total length 1. The angles at A (BAC etc) are all $90^{\circ}$. Find the maximum volume of the tetrahedron.


## Solution

Answer: $(2 \sqrt{2}-7) / 162$.

Let the edges at $A$ have lengths $x, y, z$. Then the volume is $x y z / 6$ and the perimeter is $x+y+z+\sqrt{ }\left(x^{2}+y^{2}\right)+$ $\sqrt{ }\left(y^{2}+z^{2}\right)+\sqrt{ }\left(z^{2}+x^{2}\right)=1$. By AM/GM we have $x+y+z \geq 3 k$, where $k=(x y z)^{1 / 3}$. Also $x^{2}+y^{2} \geq 2 x y$, so $\sqrt{ }\left(x^{2}+y^{2}\right)+\sqrt{ }\left(y^{2}+z^{2}\right)+\sqrt{ }\left(z^{2}+x^{2}\right) \geq \sqrt{ }(2 x y)+\sqrt{ }(2 y x)+\sqrt{ }(2 z x)$. By AM/GM that is $>=3 \sqrt{2}$ k. So we have $1 \geq$ $3(1+\sqrt{2}) \mathrm{k}$. Hence $\mathrm{k}^{3} \leq(5 \sqrt{2}-7) / 27$. Hence the volume is at most $(5 \sqrt{2}-7) / 162$. This is achieved if $\mathrm{x}=\mathrm{y}=\mathrm{z}$ $=(\sqrt{ } 2-1) / 3$. Then the other three sides are $(2-\sqrt{ } 2) / 3$ and the perimeter is 1 .

