Difference between revisions of "1976 USAMO Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
− | Note that <cmath>S = a + b + c | + | Note that <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. | so <cmath>\sqrt[3]{abc} \le \frac{S}{3 + 3\sqrt{2}} = \frac{S}{3} \cdot (\sqrt{2} - 1).</cmath> Proceed as before. | ||
Revision as of 18:05, 15 August 2014
Contents
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
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.