# 1976 USAMO Problems/Problem 4

## Problem

If the sum of the lengths of the six edges of a trirectangular tetrahedron $PABC$ (i.e., $\angle APB=\angle BPC=\angle CPA=90^o$) is $S$, determine its maximum volume.

## Solution

Let the side lengths of $AP$, $BP$, and $CP$ be $a$, $b$, and $c$, respectively. Therefore $S=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}$. Let the volume of the tetrahedron be $V$. Therefore $V=\frac{abc}{6}$.

Note that $(a-b)^2\geq 0$ implies $\frac{a^2-2ab+b^2}{2}\geq 0$, which means $\frac{a^2+b^2}{2}\geq ab$, which implies $a^2+b^2\geq ab+\frac{a^2+b^2}{2}$, which means $a^2+b^2\geq \frac{(a+b)^2}{2}$, which implies $\sqrt{a^2+b^2}\geq \frac{1}{\sqrt{2}} \cdot (a+b)$. Equality holds only when $a=b$. Therefore $S\geq a+b+c+\frac{1}{\sqrt{2}} \cdot (a+b)+\frac{1}{\sqrt{2}} \cdot (c+b)+\frac{1}{\sqrt{2}} \cdot (a+c)$ $=(a+b+c)(1+\sqrt{2})$. $\frac{a+b+c}{3}\geq \sqrt{abc}$ is true from AM-GM, with equality only when $a=b=c$. So $S\geq (a+b+c)(1+\sqrt{2})\geq 3(1+\sqrt{2})\sqrt{abc}=3(1+\sqrt{2})\sqrt{6V}$. This means that $\frac{S}{3(1+\sqrt{2})}=\frac{S(\sqrt{2}-1)}{3}\geq \sqrt{6V}$, or $6V\leq \frac{S^3(\sqrt{2}-1)^3}{27}$, or $V\leq \frac{S^3(\sqrt{2}-1)^3}{162}$, with equality only when $a=b=c$. Therefore the maximum volume is $\frac{S^3(\sqrt{2}-1)^3}{162}$.

## Solution 2

Note that by AM-GM $$S = a + b + c + \sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{a^2 + c^2} \ge 3\sqrt{abc} + \sqrt{2ab} + \sqrt{2bc} + \sqrt{2ac} \ge 3\sqrt{abc} + 3\sqrt{2}\sqrt{abc},$$ so $$\sqrt{abc} \le \frac{S}{3 + 3\sqrt{2}} = \frac{S}{3} \cdot (\sqrt{2} - 1).$$ Proceed as before.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 