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>. | ||
− | 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}} | + | 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 |
− | <math>S\geq a+b+c+\frac{1}{\sqrt{2}} | + | <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 (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 .
See also
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |