1967 IMO Problems/Problem 2

Prove that if one and only one edge of a tetrahedron is greater than $1$ , then its volume is $\le \frac{1}{8}$ .

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]

1967 IMO (Problems) • Resources
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1967 IMO Problems/Problem 2

Solution

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