Difference between revisions of "1967 IMO Problems/Problem 2"
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==Solution== | ==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] | ||
Revision as of 12:16, 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 | ||
