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Difference between revisions of "1976 USAMO Problems/Problem 4"
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<math>\frac{a+b+c}{3}\geq \sqrt[3]{abc}</math> is true from AM-GM, with equality only when <math>a=b=c</math>. So <math>S\geq (a+b+c)(1+\sqrt{2})\geq 3(1+\sqrt{2})\sqrt[3]{abc}=3(1+\sqrt{2})\sqrt[3]{6V}</math>. This means that <math>\frac{S}{3(1+\sqrt{2})}=\frac{S(\sqrt{2}-1)}{3}\geq \sqrt[3]{6V}</math>, or <math>6V\leq \frac{S^3(\sqrt{2}-1)^3}{27}</math>, or <math>V\leq \frac{S^3(\sqrt{2}-1)^3}{162}</math>, with equality only when <math>a=b=c</math>. Therefore the maximum volume is <math>\frac{S^3(\sqrt{2}-1)^3}{162}</math>.
 
<math>\frac{a+b+c}{3}\geq \sqrt[3]{abc}</math> is true from AM-GM, with equality only when <math>a=b=c</math>. So <math>S\geq (a+b+c)(1+\sqrt{2})\geq 3(1+\sqrt{2})\sqrt[3]{abc}=3(1+\sqrt{2})\sqrt[3]{6V}</math>. This means that <math>\frac{S}{3(1+\sqrt{2})}=\frac{S(\sqrt{2}-1)}{3}\geq \sqrt[3]{6V}</math>, or <math>6V\leq \frac{S^3(\sqrt{2}-1)^3}{27}</math>, or <math>V\leq \frac{S^3(\sqrt{2}-1)^3}{162}</math>, with equality only when <math>a=b=c</math>. Therefore the maximum volume is <math>\frac{S^3(\sqrt{2}-1)^3}{162}</math>.
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==Solution 2==
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Note that by AM-GM <cmath>S = a + b + c + \sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{a^2 + c^2} \ge 3\sqrt[3]{abc} + \sqrt{2ab} + \sqrt{2bc} + \sqrt{2ac} \ge 3\sqrt[3]{abc} + 3\sqrt{2}\sqrt[3]{abc},</cmath>
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so <cmath>\sqrt[3]{abc} \le \frac{S}{3 + 3\sqrt{2}} = \frac{S}{3} \cdot (\sqrt{2} - 1).</cmath> Proceed as before.
  
  
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[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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[[Category:3D Geometry Problems]]

Latest revision as of 22:15, 18 July 2016

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

Solution 2

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



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

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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