Difference between revisions of "1976 USAMO Problems/Problem 4"

Revision as of 19:03, 3 July 2013

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

$S\geq a+b+c+\frac{1}{\sqrt{2}} \cdot (a+b)+\frac{1}{\sqrt{2}} \cdot (c+b)+\frac{1}{\sqrt{2}} \cdot (a+c)$

$=(a+b+c)(1+\sqrt{2})$ .

$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$ is true from AM-GM, with equality only when $a=b=c$ . So $S\geq (a+b+c)(1+\sqrt{2})\geq 3(1+\sqrt{2})\sqrt[3]{abc}=3(1+\sqrt{2})\sqrt[3]{6V}$ . This means that $\frac{S}{3(1+\sqrt{2})}=\frac{S(\sqrt{2}-1)}{3}\geq \sqrt[3]{6V}$ , 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 $\frac{S^3(\sqrt{2}-1)^3}{162}$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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 {{USAMO box|year=1976|num-b=3|num-a=5}}
+{{MAA Notice}}
 [[Category:Olympiad Inequality Problems]]
 [[Category:Olympiad Geometry Problems]]

1976 USAMO (Problems • Resources)
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Difference between revisions of "1976 USAMO Problems/Problem 4"

Revision as of 19:03, 3 July 2013

Problem

Solution

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