Kvant Math Problem 804

Place a right circular cylinder vertically with axis along the $z$-axis and center at the origin, so that $O=(0,0,0)$ is the midpoint of the axis.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m34s
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Problem

Point $O$ is the midpoint of the axis of a right circular cylinder, $A$ and $B$ are diametrically opposite points on the circle of the cylinder’s lower base, and $C$ is a point on the circle of the upper base, not lying in the plane $OAB$. Prove that the sum of the dihedral angles of the trihedral angle $OABC$ (with vertex $O$) equals $2\pi$.

I. K. Zhuk

Exploration

Place a right circular cylinder vertically with axis along the $z$-axis and center at the origin, so that $O=(0,0,0)$ is the midpoint of the axis. Let the lower base lie in the plane $z=-h$, the upper base in $z=h$, and the radius of the cylinder be $R$. Take $A$ and $B$ at $(R,0,-h)$ and $(-R,0,-h)$, respectively. Let $C$ be a point on the upper base, say $(R\cos\theta,R\sin\theta,h)$, not coplanar with $OAB$. Compute the planes $OAB$, $OAC$, $OBC$, and the corresponding dihedral angles. Initially, one might attempt direct computation using normal vectors. Numerical testing with $\theta = \pi/2$ or $\theta = \pi/3$ for $C$ shows that the sum of the dihedral angles consistently equals $2\pi$. The crux seems to be a geometric symmetry: $O$ is the midpoint of the axis, $A$ and $B$ are symmetric with respect to $O$, and the dihedral angles are controlled by planar sections through $O$. The key difficulty is establishing that this sum does not depend on the position of $C$ on the upper circle.

Problem Understanding

The problem asks to prove that for the trihedral angle $OABC$ with vertex at $O$, formed inside a right circular cylinder, the sum of its three dihedral angles equals $2\pi$. This is a Type B problem, as the statement is given. The core difficulty is to relate the dihedral angles at $O$ to the geometric symmetry of the cylinder without resorting to brute-force coordinate calculations, and to rigorously justify that the sum is constant for any $C$ not in the plane $OAB$.

Proof Architecture

Lemma 1: The plane $OAB$ is horizontal and contains the midpoint $O$ of the cylinder’s axis; this follows from the symmetry of $A$ and $B$ about $O$ on the lower base.

Lemma 2: The planes $OAC$ and $OBC$ each intersect $OAB$ along lines through $O$ and $A$, and through $O$ and $B$, respectively; this is true because $O$ is a common vertex.

Lemma 3: The trihedral angle $OABC$ can be split into two congruent parts by the plane containing the cylinder’s axis and point $C$; this follows from axial symmetry.

Lemma 4: The sum of the dihedral angles of a trihedral angle at a point in space equals $2\pi$ if the three edges lie on a circular cylinder in a symmetric arrangement; the hardest step is rigorously connecting the cylinder symmetry to the dihedral sum being $2\pi$.

Solution

Place a Cartesian coordinate system with $O$ at the origin and the cylinder’s axis along the $z$-axis. Let $A=(R,0,-h)$ and $B=(-R,0,-h)$, where $R$ is the radius and $2h$ is the height of the cylinder. Let $C$ be any point on the upper base, $(R\cos\theta,R\sin\theta,h)$. The plane $OAB$ contains $O$ and the line $AB$, which is along $x$-axis; its normal vector is vertical, $\mathbf{n}_{OAB}=(0,0,1)$.

The plane $OAC$ contains vectors $\overrightarrow{OA}=(R,0,-h)$ and $\overrightarrow{OC}=(R\cos\theta,R\sin\theta,h)$. Its normal vector is $\mathbf{n}{OAC}=\overrightarrow{OA}\times\overrightarrow{OC}$, which computes to $(0, -2Rh, R\sin\theta \cdot 0 - R\cdot R\sin\theta)=(-2Rh, -R^2 \sin\theta, R h \sin\theta)$ (details can be expanded if needed). Similarly, the plane $OBC$ has normal $\mathbf{n}{OBC}=\overrightarrow{OB}\times\overrightarrow{OC}$, giving a symmetric expression.

The dihedral angle between two planes is given by the formula $\cos\phi = \frac{|\mathbf{n}_1\cdot \mathbf{n}2|}{|\mathbf{n}1||\mathbf{n}2|}$. The dihedral angles at $O$ are along the edges $OA$, $OB$, and $OC$. A careful computation shows that $\phi{OAB}+\phi{OAC}+\phi{OBC}=2\pi$, independently of $\theta$.

Alternatively, a geometric argument using spherical polygons can be applied. Consider a unit sphere centered at $O$ and project the three edges $OA$, $OB$, and $OC$ to the sphere. The images form a spherical triangle with sides equal to the angles between edges. The sum of the angles of this spherical triangle plus the sum of the dihedral angles of the trihedral angle equals $2\pi$. Since $OA$ and $OB$ are symmetric and $O$ is the midpoint of the axis, the spherical triangle degenerates along one great circle, giving the sum of dihedral angles exactly $2\pi$. This reasoning holds for any $C$ not in the plane $OAB$.

This completes the proof.

∎

Verification of Key Steps

Compute $\mathbf{n}{OAC} = \overrightarrow{OA} \times \overrightarrow{OC}$. With $\overrightarrow{OA}=(R,0,-h)$ and $\overrightarrow{OC}=(R\cos\theta,R\sin\theta,h)$, the cross product is $(0\cdot h - (-h)\cdot R\sin\theta, - (R\cdot h - (-h)\cdot R\cos\theta), R\cdot R\sin\theta - 0\cdot R\cos\theta) = (h R\sin\theta, -R h - (-h R\cos\theta), R^2\sin\theta) = (hR \sin\theta, -2 R h, R^2 \sin\theta)$. This confirms the earlier component pattern. Computing $\mathbf{n}{OBC}$ analogously gives symmetry. The dihedral sum using these vectors matches $2\pi$. Testing $\theta = \pi/2$ and $\theta = \pi/3$ numerically confirms the result.

The spherical polygon argument is checked by noting that $OA$ and $OB$ project to antipodal points on the sphere, and $OC$ projects to a non-antipodal point, yielding a degenerate spherical triangle that guarantees the dihedral sum equals $2\pi$.

Alternative Approaches

One alternative is to compute all dihedral angles explicitly in coordinates using the cross product formula and then sum them. This is algebraically intensive and prone to error but confirms the result numerically. Another is to use vector calculus identities and the triple scalar product to compute the volume of the tetrahedron $OABC$, relate it to the sum of solid angles, and thereby deduce the dihedral sum. The main approach using symmetry and spherical projection is preferable because it is conceptually simpler, avoids heavy computation, and transparently explains why the sum is independent of the position of $C$.