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1967 IMO Problems/Problem 2

Revision as of 23:40, 14 September 2024 by Pf02 (talk | contribs)

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.

Prob 1967 2 fig1.png

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 $CD$ are $= 1$, $P, R$ are midpoints of $AB$ and $Q = P$ (in which case the planes $ABC, ABD$ are perpendicular). In this case, $CD = \frac{\sqrt{6}}{2}$, as can be seen from an easy computation.


Solution 2

We begin with two simple propositions.

Proposition

Let $ABCD$ be a tetrahedron, and consider the transformations which rotate $\triangle ABC$ around $AB$ while keeping $\triangle ABD$ fixed. We get a set of tetrahedrons, two of which, $ABC_1D$ and $ABC_2D$ are shown in the picture below. The lengths of all sides except $CD$ are constant through this transformation.

Prob 1967 2 fig2.png

1. Assume that the angles between the planes $ABD$ and $ABC$, and $ABD$ and $ABC_1$ are both acute. If the perpendicular from $C_1$ to the plane $ABD$ is larger that the perpendicular from $C$ to the plane $ABD$ then the volume of $ABC_1D$ is larger than the volume of $ABCD$.

2. Furthermore, the tetrahedron $ABC_2D$ obtained when the position of $C_2$ is such that the planes $ABD$ and $ABC_2$ are perpendicular has the maximum volume of all tetrahedrons obtained from rotating $\triangle ABC$ around $AB$.

These statements are intuitively clear, since the volume $V$ of the tetrahedron $ABCD$ is given by

$V = \frac{1}{3} \cdot ($area of $\triangle ABD) \cdot ($height from $C)$.

A formal proof is very easy, and I will skip it.

Corollary

Given a tetrahedron $T$, and an edge $e_1$ of it, we can find another tetrahedron $U$ such that $\mathbf{volume}(U) > \mathbf{volume}(T)$, with an edge $f_1 > e_1$, and such that all the other edges of $U$ are equal to the corresponding edges of $T$, $\mathbf{unless}$ the edge $e_1$ stretches between sides of $T$ which are perpendicular. When we chose a bigger $f_1$, if $e_1 < 1$ we can choose $f_1 = 1$. Or, we can choose $f_1$ such that it stretches between sides which are perpendicular.

(By "stretches between two sides" I mean that the end points of the edge are the vertices on the two sides which are not common to the two sides. In the picture above, $DC_2$ stretches between the sides $ABD, AC_2B$ of $ABC_2D$.)

Lemma

Assume we have a tetrahedron $T$ with edges $e_1, \dots, e_6$, such that $e_2, \dots, e_6 \le 1$. If there is an edge $e_m < 1$ among $e_2, \dots, e_6$ then there is a tetrahedron $U$ with volume bigger than the volume of $T$, whose edges are equal to those of $T$, except for $e_m$, which is replaced by an edge of size $1$.

Proof

Case 1: If $T$ does not have any sides which are perpendicular, then the existence of $U$ follows from the corollary.

Case 2: Assume $T$ has exactly two sides which are perpendicular (like $ABC_2D$ in the picture above). If $C_2D$ were the only adge $< 1$, then all the other edges are $= 1$ (because they were assumed to be $\le 1$). In this case $\triangle ABD, \triangle AC_2B$ are equilateral with sides $= 1$, and the planes can not be perpendicular since $C_2D < 1$. So $e_m$ must be one of the other sides. Then again, the existence of $U$ follows from the corollary.

Case 3: Assume that three sides are perpendicular.

Prob 1967 2 fig3.png

Assume the perpendicular sides are the ones meeting at $A$, i.e. each pair of the planes meeting at $A$ are perpendicular. Since at least two of the edges $BD, BC, CD \le 1$ it follows that $AB, AC, AD < 1$ (the sides of the right angle in a right triangle are less than the hypotenuse). Just for the sake of notation, assume $e_m = AB < 1$. We can apply the corollary, and find a tetrahedron $U$ with volume bigger than the volume of $T$, with edges equal to those of $T$, except that $AB$ is replaced by an edge $= 1$.

Now the problem is very easy to prove. Pick a tetrahedron $T$ with edges $e_1, \dots, e_6$, such that $e_2, \dots, e_6 \le 1$. (To obtain this, take any acute tetrahedron, and scale it down as needed until five edges are $\le 1$.) Apply the lemma as many times as necessary (up to five times), successively replacing each edge $< 1$ by an edge $= 1$. We obtain a tetrahedron $U$ with five edges $= 1$, and one edge $f_1 = e_1$. If $f_1$ stretches between two perpendicular sides, we are done. If not, apply the corollary one more time to obtain a bigger tetrahedron in which $f_1$ is replaced by a larger edge which stretches between two perpendicular sides.

We obtain the same result as in the first solution: the largest tetrahedron is the one formed by two equilateral triangles with sides $= 1$, having one side in common, with the two planes containing the triangles perpendicular. An easy calculation shows that the edge which is $> 1$ is in fact of length $\frac{\sqrt{6}}{2}$, and the volume of this tetrahedron is $\frac{1}{8}$.

(Solution by pf02, September 2024)


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
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