Difference between revisions of "1967 IMO Problems/Problem 2"
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− | Prove that | + | Prove that iff. one edge of a tetrahedron is less than <math>1</math>; then its volume is less than or equal to <math>\frac{1}{8}</math>. |
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+ | ==Solution== | ||
+ | Assume <math> CD>1</math> and let <math> AB=x</math>. Let <math> P,Q,R</math> be the feet of perpendicular from <math> C</math> to <math> AB</math> and <math> \triangle ABD</math> and from <math> D</math> to <math> AB</math>, respectively. | ||
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+ | Suppose <math> BP>PA</math>. We have that <math> CP=\sqrt{CB^2-BT^2}\le\sqrt{1-\frac{x^2}4}</math>, <math> CQ\le CP\le\sqrt{1-\frac{x^2}4}</math>. We also have <math> DQ^2\le\sqrt{1-\frac{x^2}4}</math>. So the volume of the tetrahedron is <math> \frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1-\frac{x^2}4\right)</math>. | ||
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+ | We want to prove that this value is at most <math> \frac18</math>, which is equivalent to <math> (1-x)(3-x-x^2)\ge0</math>. This is true because <math> 0<x\le 1</math>. | ||
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+ | The above solution was posted and copyrighted by jgnr. The original thread can be found here: [https://aops.com/community/p1480514] | ||
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+ | == See Also == {{IMO box|year=1967|num-b=1|num-a=3}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 13:16, 29 January 2021
Prove that iff. one edge of a tetrahedron is less than ; then its volume is less than or equal to .
Solution
Assume and let . Let be the feet of perpendicular from to and and from to , respectively.
Suppose . We have that , . We also have . So the volume of the tetrahedron is .
We want to prove that this value is at most , which is equivalent to . This is true because .
The above solution was posted and copyrighted by jgnr. The original thread can be found here: [1]
See Also
1967 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |