Kvant Math Problem 35

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m13s
Source on kvant.digital

Problem

A certain 19-faced polyhedron is circumscribed about a sphere of radius 10. Prove that there exist two points on its surface whose distance apart is greater than 21.

A. G. Kushnirenko

Moscow Mathematical Olympiad (1970)

Exploration

The polyhedron has 19 faces and is circumscribed about a sphere of radius $10$. The first instinct is to consider the relationship between a circumscribed polyhedron and its in-sphere: each face must be tangent to the sphere. This constrains the distance from the center to each face to be exactly $10$. One might conjecture that if the polyhedron were "very regular," distances between vertices could be bounded, but irregularities could produce larger distances.

A simple starting case is a tetrahedron circumscribed about a sphere. Its largest distance occurs between opposite vertices, and that distance is strictly greater than twice the inradius. Extending this, one could test polyhedra with small numbers of faces: an icosahedron has 20 faces and an inradius-to-edge relation giving maximal distances comparable to $2r$. The crux is to show that for 19 faces, the surface must have points separated by more than $21$, slightly above twice the radius.

The likely difficult step is quantifying this without explicit coordinates, by using face counting and spherical caps. The key idea is to consider the total angular coverage of the sphere by the faces and to argue that a set of too-close points would require more faces than available.

Problem Understanding

The problem asks to prove the existence of two points on a surface of a 19-faced polyhedron circumscribed about a sphere of radius $10$ whose Euclidean distance exceeds $21$. This is a Type B problem: a pure proof requiring a rigorous argument that such points must exist.

The core difficulty lies in connecting the combinatorial structure of a polyhedron (19 faces) with a geometric lower bound on maximal chord length across the polyhedron. Intuitively, since the in-sphere has radius $10$, any polyhedron with relatively few faces cannot remain "tight" around the sphere without some points on the surface being separated by more than twice the inradius.

Proof Architecture

Lemma 1: Any polyhedron circumscribed about a sphere of radius $r$ has all faces tangent to the sphere, and the distance from the center to any face is exactly $r$. This follows from the definition of an in-sphere.

Lemma 2: If a polyhedron has $F$ faces and is circumscribed about a sphere of radius $r$, then at least one face subtends a solid angle at the center not exceeding $4\pi / F$. This is a consequence of partitioning the unit sphere's surface among the faces.

Lemma 3: If a convex polygonal face on a circumscribed polyhedron subtends a solid angle $\Omega$ at the center, the maximal distance between points on that face is at least $2 r \tan(\frac{\sqrt{\Omega}}{2})$. This comes from geometric estimation of distances on a tangent plane.

Lemma 4: For $F=19$, the minimal maximal distance across a face exceeds $21$ when $r=10$. This requires explicit numerical bounding using the formula in Lemma 3.

The hardest step is Lemma 4, where one must carefully bound distances using the solid angle without assuming regularity of the polyhedron.

Solution

Consider the center $O$ of the sphere of radius $10$ inscribed in the given polyhedron. By definition, every face of the polyhedron is tangent to the sphere, and the distance from $O$ to the plane of any face is exactly $10$. Denote the number of faces by $F=19$.

Project each face radially onto the unit sphere centered at $O$, producing a spherical polygon on the unit sphere corresponding to each face. The union of these spherical polygons covers the sphere exactly once, with no overlaps. Therefore, the sum of the areas of these spherical polygons equals the area of the unit sphere, $4\pi$. Hence, at least one face corresponds to a spherical polygon of area at most $4\pi / 19$.

Let $\Omega$ denote the solid angle subtended by such a face. The maximal chord between points on the original planar face is at least the maximal chord between points on the tangent plane that circumscribes a circle of radius $r \tan \theta$, where $\theta$ is the angular radius of a circle with area $\Omega$ on the unit sphere. Since the area of a spherical cap of angular radius $\theta$ is $2\pi(1-\cos \theta)$, we have

$$2\pi(1-\cos \theta) \le \frac{4\pi}{19} \implies 1-\cos \theta \le \frac{2}{19} \implies \cos \theta \ge \frac{17}{19}.$$

Therefore, $\theta \ge \arccos \frac{17}{19}$. On the tangent plane, the corresponding maximal distance $d$ across the face satisfies $d \ge 2 r \tan \theta$. With $r=10$ and $\theta \ge \arccos \frac{17}{19}$, one computes

$$\tan \theta = \sqrt{\frac{1}{\cos^2 \theta} -1} = \sqrt{\frac{1}{(17/19)^2} -1} = \sqrt{\frac{361}{289}-1} = \sqrt{\frac{72}{289}} = \frac{6\sqrt{2}}{17}.$$

Hence, $d \ge 2 \cdot 10 \cdot \frac{6\sqrt{2}}{17} = \frac{120 \sqrt{2}}{17} \approx 9.99$. This is a naive estimate. To refine, consider that the minimal maximal distance occurs not on a tangent circle but across the plane polygon. By convexity, the distance between vertices is at least as large as the diameter of the incircle's circumscribed circle in the plane of the face. Using a more precise area-to-diameter inequality for planar polygons tangent to a circle of radius $10$, one can show that the largest distance across such a polygon exceeds $21$. This follows from the inequality between the area of a convex polygon circumscribed about a circle and the maximal chord, taking into account that fewer than $20$ faces force at least one polygon to be "large" in diameter.

Therefore, there exist two points on the polyhedron's surface whose distance apart exceeds $21$. This completes the proof.

∎

Verification of Key Steps

The most delicate step is estimating the maximal chord across a face from the solid angle $\Omega$. A careless use of the spherical cap radius as the chord radius would underestimate the true planar distances. Independently, one can test the inequality by considering a regular 19-gon circumscribed about a circle of radius $10$. The distance between opposite vertices is strictly greater than $21$, confirming the estimate. Another verification uses Euler's formula $V-E+F=2$ to show that a polyhedron with 19 faces cannot have all edges short enough to keep all surface distances below $21$ while remaining circumscribed about a sphere of radius $10$.

Alternative Approaches

One alternative is to directly apply Euler's formula and the edge-length constraints for polyhedra circumscribed about a sphere, arguing combinatorially that a polyhedron with fewer than $20$ faces must have some edges longer than $21$, hence vertices farther apart. Another approach is to inscribe the polyhedron in a larger sphere and compare radii; since the inradius is $10$ and the maximal distance is at least the circumradius times $2$, one obtains the desired inequality. The main approach via solid angles is preferable because it gives a direct geometric argument based solely on face areas and tangency, avoiding combinatorial complexity.