Difference between revisions of "1967 IMO Problems/Problem 2"

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Prove that iff. one edge of a tetrahedron is less than <math>1</math>; then
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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>.
its volume is less than or equal to <math>\frac{1}{8}</math>.
 
  
{{solution}}
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==Solution==
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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:Olympiad Geometry Problems]]
 
[[Category:3D 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 $1$; then its volume is less than or equal to $\frac{1}{8}$.

Solution

Assume $CD>1$ and let $AB=x$. Let $P,Q,R$ be the feet of perpendicular from $C$ to $AB$ and $\triangle ABD$ and from $D$ to $AB$, respectively.

Suppose $BP>PA$. We have that $CP=\sqrt{CB^2-BT^2}\le\sqrt{1-\frac{x^2}4}$, $CQ\le CP\le\sqrt{1-\frac{x^2}4}$. We also have $DQ^2\le\sqrt{1-\frac{x^2}4}$. So the volume of the tetrahedron is $\frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1-\frac{x^2}4\right)$.

We want to prove that this value is at most $\frac18$, which is equivalent to $(1-x)(3-x-x^2)\ge0$. This is true because $0<x\le 1$.

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