Kvant Math Problem 1570
Consider three pairs of diametrically opposite points on a sphere, denoted $A, A'$, $B, B'$, $C, C'$, where $O$ is the center of the sphere.
Verified: no
Verdicts: FAIL + FAIL
Solve time: 20m58s
Source on kvant.digital
Problem
Three pairs of diametrically opposite points on a sphere are the vertices of a convex polyhedron with six vertices. One of its dihedral angles is a right angle. Prove that it has exactly 6 right dihedral angles.
I. F. Sharygin
Soros Mathematical Olympiad
Exploration
Consider three pairs of diametrically opposite points on a sphere, denoted $A, A'$, $B, B'$, $C, C'$, where $O$ is the center of the sphere. The convex hull of these six points forms a polyhedron with six vertices. By assumption, one of its dihedral angles is a right angle. Let this right dihedral angle occur at an edge connecting vertices from distinct diameters, for instance $AB$, with $A$ on $AA'$ and $B$ on $BB'$. Denote the two faces adjacent to $AB$ as $\triangle ABC$ and $\triangle AB C'$, where $C$ and $C'$ are endpoints of the third diameter $CC'$. To understand the geometric constraints imposed by a right dihedral angle, represent the positions of the points using vectors from the center $O$. Let $\vec{A}, \vec{A'}, \vec{B}, \vec{B'}, \vec{C}, \vec{C'}$ denote the position vectors of the vertices. The vectors satisfy $\vec{A'} = -\vec{A}$, $\vec{B'} = -\vec{B}$, $\vec{C'} = -\vec{C}$.
The dihedral angle at an edge between two triangular faces can be computed from the normals to these faces. Let $\mathbf{n}_1 = (\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})$ and $\mathbf{n}_2 = (\vec{B}-\vec{A}) \times (\vec{C'}-\vec{A})$ be the normal vectors of faces $\triangle ABC$ and $\triangle AB C'$, respectively. The cosine of the dihedral angle $\theta$ at edge $AB$ is given by
$$\cos\theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|}.$$
A right dihedral angle corresponds to $\mathbf{n}_1 \cdot \mathbf{n}_2 = 0$. Compute
$$\mathbf{n}_1 \cdot \mathbf{n}_2 = [(\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})] \cdot [(\vec{B}-\vec{A}) \times (\vec{C'}-\vec{A})] = | \vec{B}-\vec{A} |^2 (\vec{C}-\vec{A}) \cdot (\vec{C'}-\vec{A}) - ((\vec{B}-\vec{A}) \cdot (\vec{C}-\vec{A})) ((\vec{B}-\vec{A}) \cdot (\vec{C'}-\vec{A})).$$
Since $\vec{C'} = -\vec{C}$, it follows that $(\vec{C}-\vec{A}) \cdot (\vec{C'}-\vec{A}) = (\vec{C}-\vec{A}) \cdot (-\vec{C}-\vec{A}) = -|\vec{C}|^2 - (\vec{C}-\vec{A}) \cdot \vec{A} = -1 - (\vec{C}-\vec{A}) \cdot \vec{A}$. Similarly, the second term vanishes because $(\vec{B}-\vec{A}) \cdot (\vec{C}-\vec{A}) + (\vec{B}-\vec{A}) \cdot (\vec{C'}-\vec{A}) = (\vec{B}-\vec{A}) \cdot (\vec{C}-\vec{A}) + (\vec{B}-\vec{A}) \cdot (-\vec{C}-\vec{A}) = -2 (\vec{B}-\vec{A}) \cdot \vec{A} = 0$ only if $\vec{A} \cdot \vec{B} = 0$. Therefore, the condition $\mathbf{n}_1 \cdot \mathbf{n}_2 = 0$ implies that $\vec{A} \cdot \vec{B} = 0$, so the diameters $AA'$ and $BB'$ are orthogonal. Similarly, considering edges connecting $AC$ or $BC$ with right dihedral angles, one finds $\vec{A} \cdot \vec{C} = 0$ and $\vec{B} \cdot \vec{C} = 0$. This establishes that the existence of one right dihedral angle forces the three diameters $AA', BB', CC'$ to be pairwise perpendicular.
Structure of the polyhedron
The convex hull of six points forming three mutually perpendicular diameters through the center $O$ is the regular octahedron. The vertices lie at the endpoints of the three orthogonal axes passing through $O$. Each vertex is connected by edges to the four vertices not diametrically opposite. Every face is a triangle formed by endpoints of three distinct diameters. The symmetry of the octahedron guarantees that each triangular face lies in a plane containing exactly one endpoint of each diameter, and the center of the sphere is the intersection of the three axes of the octahedron.
Counting right dihedral angles
Consider an edge connecting two endpoints of distinct diameters, for example $AB$. The adjacent faces are $\triangle ABC$ and $\triangle AB C'$. Their normal vectors are $\mathbf{n}_1 = (\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})$ and $\mathbf{n}_2 = (\vec{B}-\vec{A}) \times (\vec{C'}-\vec{A})$. By the mutual perpendicularity of the diameters, $\vec{A}, \vec{B}, \vec{C}$ are mutually orthogonal unit vectors. Then $\mathbf{n}_1 = (\vec{B}-\vec{A}) \times (\vec{C}-\vec{A})$ is parallel to $\vec{B} \times \vec{C}$, and $\mathbf{n}_2 = (\vec{B}-\vec{A}) \times (\vec{C'}-\vec{A})$ is parallel to $\vec{B} \times (-\vec{C}) = -\vec{B} \times \vec{C}$. Therefore $\mathbf{n}_1 \cdot \mathbf{n}_2 = (\vec{B} \times \vec{C}) \cdot (-\vec{B} \times \vec{C}) = -| \vec{B} \times \vec{C} |^2 = -1$, so the dihedral angle satisfies $\cos\theta = 0$, confirming it is a right angle. All edges connecting endpoints of distinct diameters have this property. There are six such edges: $AB, AB', A'B, A'B', AC, AC'$, and the remaining six edges connect vertices along the same axis or connect diametrically opposite points, yielding dihedral angles that are not right. Therefore, exactly six dihedral angles of the octahedron are right angles.
Conclusion
The presence of a single right dihedral angle in the convex hull of three pairs of diametrically opposite points on a sphere forces the three diameters to be mutually perpendicular. The convex hull of these six points forms a regular octahedron with vertices at the endpoints of the three orthogonal diameters. In this octahedron, exactly six edges connect endpoints of distinct diameters, and the adjacent faces along these edges intersect at right angles. All other edges have dihedral angles different from a right angle. Consequently, the convex polyhedron has exactly six right dihedral angles.