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>. | ||
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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>. | ||
− | ==See | + | ==Solution 2== |
+ | 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> | ||
+ | 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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+ | {{alternate solutions}} | ||
+ | |||
+ | ==See Also== | ||
{{USAMO box|year=1976|num-b=3|num-a=5}} | {{USAMO box|year=1976|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 22:15, 18 July 2016
Contents
[hide]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 .
Solution 2
Note that by AM-GM so 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.