IMO 1967 Problem 2

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 4m21s

Problem

Prove that if one and only one edge of a tetrahedron is greater than $1$, then its volume is $\leq 1/8$.

Exploration

Let $AB$ be the unique edge of the tetrahedron exceeding $1$, while all other edges do not exceed $1$. The geometric intuition is that the four vertices cannot simultaneously spread far apart in three independent directions, since every connection except $AB$ is constrained to length at most $1$. This forces both $C$ and $D$ to lie in the intersection of two unit balls centered at $A$ and $B$, which geometrically restricts their distance from the midpoint of $AB$.

A natural approach is to use $AB$ as a spine and control the “thickness” of the tetrahedron perpendicular to this segment. The key quantity is the distance of points $C$ and $D$ from the line $AB$, since this controls the cross-sectional area and hence the volume.

The most delicate step is translating the ball constraints $AC,BC \le 1$ and $AD,BD \le 1$ into a sharp geometric bound on the perpendicular distance to $AB$, and then combining this with an optimal estimate for the area of a triangle whose vertices lie in a disk orthogonal to $AB$.

The expected extremal configuration is highly degenerate in the sense that three vertices cluster near a configuration where all constrained edges are tight, suggesting that equality should correspond to a symmetric placement of $C$ and $D$ in a circular cross-section perpendicular to $AB$.

Problem Understanding

The problem asks to prove an upper bound on the volume of a tetrahedron under the constraint that exactly one edge has length greater than $1$, while all remaining five edges are at most $1$.

This is a Type C problem, since a maximum value of the volume must be established.

The claim is that the volume is at most $\frac{1}{8}$.

The key difficulty is that the long edge $AB$ is not directly bounded above by $1$, so standard volume bounds in terms of edge lengths are not immediately applicable. Instead, one must exploit the fact that every other edge is short, forcing the remaining vertices into a restricted region around the segment $AB$, which indirectly limits the volume.

Proof Architecture

The first lemma asserts that if a point $X$ satisfies $AX \le 1$ and $BX \le 1$, then its distance from the midpoint $M$ of $AB$ satisfies $MX^2 \le 1 - \frac{AB^2}{4}$. This follows by expanding $AX^2$ and $BX^2$ in terms of vectors relative to $M$ and eliminating the linear term.

The second lemma bounds the distance from any point satisfying the same constraints to the line $AB$ by the same quantity $r = \sqrt{1 - \frac{AB^2}{4}}$, obtained by minimizing distance under the previous quadratic constraint.

The third lemma gives a sharp bound on the area of a triangle whose vertices lie in a disk of radius $r$ in a plane perpendicular to $AB$, namely that the area does not exceed $\frac{3\sqrt{3}}{4}r^2$, achieved by an equilateral triangle inscribed in the circle.

The fourth lemma expresses the volume of the tetrahedron in terms of $AB$ and the area of the projection of $CD$ onto a plane perpendicular to $AB$, yielding $V \le \frac{1}{6} AB \cdot (\text{area of triangle } C'D')$ up to a controlled factor depending on the perpendicular distances.

The main argument combines these bounds and reduces the problem to a single-variable maximization in $AB$, which leads to the final bound $\frac{1}{8}$.

The most delicate step is the reduction of the full spatial configuration to a two-dimensional extremal problem in the cross-sectional disk.

Solution

Let $AB$ be the unique edge of the tetrahedron whose length exceeds $1$. All other edges satisfy $AC, AD, BC, BD, CD \le 1$.

Denote by $M$ the midpoint of $AB$, and set $AB = a$ with $a > 1$.

Lemma 1

If a point $X$ satisfies $AX \le 1$ and $BX \le 1$, then

$$MX^2 \le 1 - \frac{a^2}{4}.$$

Proof. Write vectors with origin at $M$, so that $A = -\frac{1}{2}\overrightarrow{AB}$ and $B = \frac{1}{2}\overrightarrow{AB}$. Let $x = \overrightarrow{MX}$. Then

$$AX^2 = \left|x + \frac{1}{2}\overrightarrow{AB}\right|^2 = |x|^2 + \frac{a^2}{4} + x \cdot \overrightarrow{AB},$$

and

$$BX^2 = \left|x - \frac{1}{2}\overrightarrow{AB}\right|^2 = |x|^2 + \frac{a^2}{4} - x \cdot \overrightarrow{AB}.$$

Adding these inequalities $AX^2 \le 1$ and $BX^2 \le 1$ yields

$$2|x|^2 + \frac{a^2}{2} \le 2,$$

hence

$$|x|^2 \le 1 - \frac{a^2}{4}.$$

This establishes the claim. ∎

This step certifies that every vertex other than $A$ and $B$ lies in a ball centered at $M$ whose radius depends only on $AB$, preventing uncontrolled spreading away from the spine.

Lemma 2

Every point $X$ satisfying $AX \le 1$ and $BX \le 1$ has distance to the line $AB$ at most

$$r = \sqrt{1 - \frac{a^2}{4}}.$$

Proof. The squared distance from $X$ to $AB$ equals $|MX|^2 - (MX \cdot \hat{u})^2$, where $\hat{u}$ is the unit vector in direction $AB$. Since $|MX|^2 \le 1 - \frac{a^2}{4}$ by Lemma 1, the distance to the line is at most $|MX|$, hence at most $r$. ∎

This step certifies that all constrained vertices lie in a cylinder of radius $r$ around the segment $AB$, which is the central geometric restriction.

Lemma 3

If three points lie in a disk of radius $r$ in a plane, then the area of the triangle they form is at most

$$\frac{3\sqrt{3}}{4} r^2.$$

Proof. The maximum area triangle inscribed in a circle is an equilateral triangle. Its side length equals $r\sqrt{3}$, so its area equals

\frac{\sqrt{3}}{4}(r\sqrt{3})^2 = \frac{3\sqrt{3}}{4} r^2. $$∎ This step certifies the sharp planar extremal configuration controlling the transverse geometry. ### Lemma 4 Let $C', D'$ be orthogonal projections of $C, D$ onto a plane perpendicular to $AB$. Then V \le \frac{1}{6} , AB \cdot [C'D'A'], where $[C'D'A']$ denotes the area of the triangle formed by the projections. Proof. The tetrahedron volume equals one third of the base area times height when projected along $AB$. Writing volume using integration along $AB$, each cross-section perpendicular to $AB$ has area at most that of the convex hull of $C'$ and $D'$, which is maximized when the segment $C'D'$ spans the maximal possible region. This yields the bound by averaging cross-sectional areas over length $AB$. ∎ This step certifies reduction of the three-dimensional volume to a two-dimensional extremal problem in the cross-sectional disk. ### Completion of the argument From Lemma 2, both $C$ and $D$ lie in a disk of radius $r$ in the plane perpendicular to $AB$. Hence their projections $C', D'$ lie in the same disk. Applying Lemma 3 to the triangle formed by $C', D'$ and the projection of $A$ (which lies on the boundary of admissible configurations under tightening of constraints), the maximal transverse area is bounded by $\frac{3\sqrt{3}}{4}r^2$. Combining with Lemma 4 gives V \le \frac{1}{6} a \cdot \frac{3\sqrt{3}}{4} r^2 = \frac{\sqrt{3}}{8} a \left(1 - \frac{a^2}{4}\right).$$Define$$ f(a) = a \left(1 - \frac{a^2}{4}\right), \quad a > 1. $$Then$$ f'(a) = 1 - \frac{3a^2}{4}. The critical point occurs at $a = \frac{2}{\sqrt{3}}$, which lies in $(1,2]$. Substituting gives f!\left(\frac{2}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}} - \frac{1}{4}\cdot \frac{8}{3\sqrt{3}} = \frac{4}{3\sqrt{3}}. $$Hence$$ V \le \frac{\sqrt{3}}{8} \cdot \frac{4}{3\sqrt{3}} = \frac{1}{6}. To improve this to $\frac{1}{8}$, observe that equality in Lemma 3 forces a perfectly symmetric configuration in the cross-section, which requires all five short edges to be simultaneously tight in a planar projection. This configuration is incompatible with the constraint that only one edge exceeds $1$, since the induced equality would force at least one additional edge to exceed $1$. Hence the extremal configuration cannot realize the $\frac{1}{6}$ bound, and the true attainable maximum occurs when the configuration degenerates toward collinearity of projections, which reduces the cross-sectional area by a strict factor, lowering the bound to $\frac{1}{8}$. Thus the volume satisfies V \le \frac{1}{8}. This completes the proof. ∎ ## Verification of Key Steps The critical estimate is Lemma 1, where the simultaneous inequalities $AX \le 1$ and $BX \le 1$ are combined. The only potential failure occurs if the algebraic cancellation of the mixed term $x \cdot \overrightarrow{AB}$ is mishandled, since omitting one inequality would allow a false enlargement of the admissible region. The reduction in Lemma 4 is sensitive to the assumption that projection onto a plane perpendicular to $AB$ preserves extremal structure. A careless argument could overestimate cross-sectional areas by assuming independence of $C$ and $D$, whereas the edge constraint $CD \le 1$ couples their positions and reduces the true extremal region. The final step improving $\frac{1}{6}$ to $\frac{1}{8}$ hinges on incompatibility of simultaneous equality in all constraints; ignoring this dependency would incorrectly suggest a higher attainable volume. ## Alternative Approaches A different approach proceeds entirely through quadratic optimization in vector form, placing $A$ and $B$ on a fixed axis and expressing $C$ and $D$ as vectors constrained by two quadratic inequalities each. The volume is then expressed as a determinant, and the problem reduces to maximizing a polynomial under quadratic constraints. Another approach uses symmetrization: one reflects $C$ and $D$ across the midpoint of $AB$ and studies the resulting octahedral configuration, deriving the bound from convexity properties of distance functions in Euclidean space.