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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Revision as of 14:10, 17 September 2012
Problem
If the sum of the lengths of the six edges of a trirectangular tetrahedron (i.e., ) is , determine its maximum volume.
Solution
Let the side lengths of , , and be , , and , respectively. Therefore . Let the volume of the tetrahedron be . Therefore .
Note that implies , which means , which implies , which means , which implies . Equality holds only when . Therefore
.
is true from AM-GM, with equality only when . So . This means that , or , or , with equality only when . Therefore the maximum volume is .
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 |