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Difference between revisions of "1976 USAMO Problems/Problem 4"
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Let the side lengths of <math>AP</math>, <math>BP</math>, and <math>CP</math> be <math>a</math>, <math>b</math>, and <math>c</math>, respectively. Therefore <math>S=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}</math>. Let the volume of the tetrahedron be <math>V</math>. Therefore <math>V=\frac{abc}{6}</math>.
 
Let the side lengths of <math>AP</math>, <math>BP</math>, and <math>CP</math> be <math>a</math>, <math>b</math>, and <math>c</math>, respectively. Therefore <math>S=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}</math>. Let the volume of the tetrahedron be <math>V</math>. Therefore <math>V=\frac{abc}{6}</math>.
  
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Note that <math>(a-b)^2\geq 0</math> implies <math>\frac{a^2-2ab+b^2}{2}\geq 0</math>, which means <math>\frac{a^2+b^2}{2}\geq ab</math>, which implies <math>a^2+b^2\geq ab+\frac{a^2+b^2}{2}</math>, which means <math>a^2+b^2\geq \frac{(a+b)^2}{2}</math>, which implies <math>\sqrt{a^2+b^2}\geq \frac{1}{\sqrt{2}}*(a+b)</math>. Equality holds only when <math>a=b</math>. Therefore
+
Note that <math>(a-b)^2\geq 0</math> implies <math>\frac{a^2-2ab+b^2}{2}\geq 0</math>, which means <math>\frac{a^2+b^2}{2}\geq ab</math>, which implies <math>a^2+b^2\geq ab+\frac{a^2+b^2}{2}</math>, which means <math>a^2+b^2\geq \frac{(a+b)^2}{2}</math>, which implies <math>\sqrt{a^2+b^2}\geq \frac{1}{\sqrt{2}} \cdot (a+b)</math>. Equality holds only when <math>a=b</math>. Therefore
  
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<math>S\geq a+b+c+\frac{1}{\sqrt{2}}*(a+b)+\frac{1}{\sqrt{2}}*(c+b)+\frac{1}{\sqrt{2}}*(a+c)</math>
+
<math>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)</math>
  
 
<math>=(a+b+c)(1+\sqrt{2})</math>.
 
<math>=(a+b+c)(1+\sqrt{2})</math>.

Revision as of 17:01, 22 February 2012

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 $\frac{S^3(\sqrt{2}-1)^3}{162}$.

See also

1976 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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