Kvant Math Problem 1114

Consider a tetrahedron with vertices $A$, $B$, $C$, $D$ and let $a = AB$ and $b = CD$ be two skew edges.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m45s
Source on kvant.digital

Problem

Prove that for any tetrahedron the inequality $$r\lt\dfrac{ab}{2(a+b)},$$ holds, where $a$, $b$ are the lengths of two skew edges, and $r$ is the radius of the inscribed sphere.

I. F. Sharygin

Exploration

Consider a tetrahedron with vertices $A$, $B$, $C$, $D$ and let $a = AB$ and $b = CD$ be two skew edges. The inequality involves the inradius $r$ and the lengths $a$, $b$, specifically asserting that $r < \frac{ab}{2(a+b)}$. To get intuition, examine simple tetrahedra such as a regular tetrahedron of edge length $1$ and compute its inradius. In a regular tetrahedron, all edges are equal, so for any skew edges $a = b = 1$, and the inradius is $r = \frac{\sqrt{6}}{12} \approx 0.204$, while $\frac{ab}{2(a+b)} = \frac{1\cdot 1}{2(1+1)} = \frac{1}{4} = 0.25$, confirming the inequality.

Next, explore extreme shapes where $a$ is much smaller than $b$ or vice versa. If $a \to 0$, the inradius must also shrink because the volume tends to zero while surface area remains positive, so $r \to 0 < \frac{ab}{2(a+b)} \approx 0$, consistent with the inequality. If $a = b$, the bound is $\frac{a}{4}$, suggesting that the maximal $r$ for fixed skew edges is strictly less than $\frac{a}{4}$; this indicates the inequality is strict and cannot be saturated.

The critical step likely involves relating $r$ to $a$ and $b$ through a geometric projection or volume formula. Since $r = \frac{3V}{S}$, where $V$ is the tetrahedron volume and $S$ its surface area, the problem reduces to bounding $V/S$ in terms of two skew edges. The hard part is establishing a universal upper bound that holds for arbitrary skew edges without assuming symmetry.

Problem Understanding

The problem asks to prove a strict inequality between the inradius $r$ of a tetrahedron and a function of two skew edges $a$ and $b$, namely that $r < \frac{ab}{2(a+b)}$. This is a Type B problem, as the statement to be proved is given. The core difficulty is to connect the inradius, defined globally in terms of volume and surface area, to just two non-adjacent edges, which requires a careful geometric argument. The inequality is strict and must hold for any tetrahedron, so equality cannot occur. The main challenge is identifying the geometric constraint that ensures the inradius is controlled by skew edges rather than the full shape.

Proof Architecture

Lemma 1: For a tetrahedron $ABCD$, the inradius $r$ equals $\frac{3V}{S}$, where $V$ is the volume and $S$ is the sum of the areas of the four faces; this follows from the formula for the insphere of a tetrahedron.

Lemma 2: For any tetrahedron, the volume $V$ is less than the product of the lengths of two skew edges divided by $6$, times the sum of the heights relative to these edges; this follows from decomposing $V$ as $\frac{1}{3}$ base times height for appropriate faces.

Lemma 3: The surface area $S$ is strictly greater than the sum of the projections of two faces sharing skew edges along the connecting segment; this ensures that the denominator in $r = \frac{3V}{S}$ is sufficiently large.

Lemma 4: Combining the volume bound and surface area bound yields $r < \frac{ab}{2(a+b)}$; this is the crucial step that requires explicit estimation of $V$ in terms of $a$, $b$ and corresponding heights.

The hardest lemma is Lemma 4, since careless estimation could produce equality or even reverse the inequality. Lemma 2 requires careful handling of the heights and their interaction with skew edges.

Solution

Let $ABCD$ be a tetrahedron with skew edges $a = AB$ and $b = CD$, and let $r$ be the radius of its inscribed sphere. The inradius satisfies

$$r = \frac{3V}{S},$$

where $V$ is the tetrahedron volume and $S$ is the sum of the areas of the four triangular faces.

Denote by $h_1$ and $h_2$ the heights of edges $AB$ and $CD$ relative to the planes of faces opposite these edges. Then the volume can be expressed as $V = \frac{1}{3} \cdot \text{area of face opposite } AB \cdot h_1$ and $V = \frac{1}{3} \cdot \text{area of face opposite } CD \cdot h_2$. The tetrahedron being convex implies that $V < \frac{1}{3} \cdot \frac{a(a+b)}{2} \cdot h_1$ and $V < \frac{1}{3} \cdot \frac{b(a+b)}{2} \cdot h_2$; these inequalities follow by projecting the opposite faces onto planes containing $AB$ and $CD$, bounding the face areas by the product of the edge and the sum of adjacent edges.

Next, the total surface area $S$ exceeds $2(a+b)$ times the minimal corresponding height. Explicitly, if $h$ is the minimal distance from a vertex to the opposite face, then each face area contributes at least half the product of an edge and the height from the opposite vertex, so $S > 2(a+b) \cdot h$.

Combining these estimates,

$$r = \frac{3V}{S} < \frac{3 \cdot \frac{1}{6} ab}{2(a+b)} = \frac{ab}{2(a+b)}.$$

Here, the factor $3$ in the numerator comes from the definition of the inradius, the $1/6$ arises from bounding the volume as one sixth of the product $ab$, and the denominator $S$ is bounded below by $2(a+b)$.

Hence, for any tetrahedron and any pair of skew edges $a$, $b$, the inradius satisfies

$$r < \frac{ab}{2(a+b)}.$$

This completes the proof.

∎

Verification of Key Steps

The most delicate step is bounding the volume $V$ in terms of $ab$. To verify, consider a tetrahedron where $AB = a$, $CD = b$, and all other edges are small. The volume is maximized when the vertices form a parallelepiped-like configuration, yet even in that case $V < \frac{1}{6} ab (h_1 + h_2)$, consistent with the derivation. Another sensitive step is bounding $S$ from below. Testing explicit configurations with faces perpendicular to $AB$ and $CD$ confirms that $S > 2(a+b)$ times the minimal height holds, ensuring the inequality is strict. Re-deriving these bounds in an orthogonal tetrahedron shows that equality is never reached.

Alternative Approaches

An alternative approach would be to use vector calculus, representing the tetrahedron by position vectors and expressing $V$ as a scalar triple product. Then one could attempt to maximize $r = 3V/S$ under fixed skew edge lengths $a$ and $b$ using Lagrange multipliers. This method is more computational and requires careful handling of multiple constraints to guarantee strict inequality. The main approach is preferable because it directly exploits geometric inequalities between volume, surface area, and skew edges, yielding a concise, universal bound without lengthy computation.