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]

Remarks (added by pf02, September 2024)

The solution above is essentially correct, and it is nice, but it is so sloppily written that it borders the incomprehensible. Below I will give an edited version of it for the sake of completeness.

Then, I will give a second solution to the problem.

A few notes which may be of interest.

The condition that one side is greater than $1$ is not really necessary. The statement is true even if all sides are $\le 1$ . What we need is that no more than one side is $> 1$ .

The upper limit of $1/8$ for the volume of the tetrahedron is actually reached. This will become clear from both solutions.

Solution

Assume $CD > 1$ and assume that all other sides are $\le 1$ . Let $AB = x$ . Let $P, Q, R$ be the feet of perpendiculars from $C$ to $AB$ , from $C$ to the plane $ABD$ , and from $D$ to $AB$ , respectively.

At least one of the segments $AP, PB$ has to be $\ge \frac{x}{2}$ . Suppose $PB \ge \frac{x}{2}$ . (If $AP$ were bigger that $\frac{x}{2}$ the argument would be the same.) We have that $CP = \sqrt{BC^2 - PB^2} \le \sqrt{1 - \frac{x^2}{4}}$ . By the same argument in $\triangle ABD$ we have $DR \le \sqrt{1 - \frac{x^2}{4}}$ . Since $CQ \perp$ plane $ABD$ , we have $CQ \le CP$ , so $CQ \le \sqrt{1 - \frac{x^2}{4}}$ .

The volume of the tetrahedron is

$V = \frac{1}{3} \cdot ($ area of $\triangle ABD) \cdot$ height from $C = \frac{1}{3} \cdot \left( \frac{1}{2} \cdot AB \cdot DR \right) \cdot CQ \le \left( \frac{1}{6} \cdot x \cdot \sqrt{1 - \frac{x^2}{4}} \cdot \sqrt{1 - \frac{x^2}{4}} \right) = \frac{x}{6} \left( 1 - \frac{x^2}{4} \right)$ .

We need to prove that $\frac{x}{6} \left( 1 - \frac{x^2}{4} \right) \le \frac{1}{8}$ . Some simple computations show that this is the same as $(1 - x)(3 - x - x^2) \ge 0$ . This is true because $0 < x \le 1$ , and $-x^2 - x + 3 \ge 0$ on this interval.

Note: $V = \frac{1}{8}$ is achieved when $x = 1$ and all inequalities are equalities. This is the case when all sides except $AD$ are $= 1$ , $P, R$ are midpoints of $AB$ and $Q = P$ (in which case the planes $ABC, ABD$ are perpendicular). In this case, $AD = \frac{\sqrt{6}}{2}$ , as can be seen from an easy computation.

Solution 2

TO BE CONTINUED. DOING A SAVE MIDWAY SO I DON'T LOOSE WORK DONE SO FAR.

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

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Solution

Remarks (added by pf02, September 2024)

Solution

Solution 2

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