Kvant Math Problem 968

Verified: no
Verdicts: FAIL + FAIL
Solve time: 34m21s
Source on kvant.digital

Problem

Three polygons in space are positioned so that the planes containing them intersect at a single point $O$.

Prove that there exists a plane onto which the projections of these three polygons have equal areas.
How many such planes passing through the point $O$ are there?

N. M. Sedrakyan

Solution to Kvant M968

Three polygons in space are positioned so that the planes containing them intersect at a single point $O$. We are asked to prove the existence of a plane onto which the projections of the three polygons have equal areas and to determine how many such planes pass through the point $O$.

Part 1: Existence of a plane with equal projected areas

Let the three polygons lie in planes $\Pi_1$, $\Pi_2$, and $\Pi_3$, all passing through the point $O$, and denote their areas by $A_1$, $A_2$, and $A_3$. Let $\mathbf{n}_1$, $\mathbf{n}_2$, and $\mathbf{n}_3$ be unit normal vectors of these planes. Consider a plane $\Pi$ through $O$ with unit normal vector $\mathbf{N}$. The area of the orthogonal projection of the polygon in plane $\Pi_i$ onto $\Pi$ is $A_i |\cos \theta_i|$, where $\theta_i$ is the angle between $\mathbf{n}_i$ and $\mathbf{N}$.

Define the continuous function $F$ on the unit sphere $S^2$ by

$F(\mathbf{N}) = \bigl(A_1 |\cos \theta_1| - A_2 |\cos \theta_2|, , A_1 |\cos \theta_1| - A_3 |\cos \theta_3|\bigr).$

Finding a plane with equal projected areas is equivalent to finding a vector $\mathbf{N} \in S^2$ such that $F(\mathbf{N}) = (0,0)$.

The function $F$ is continuous, and it satisfies $F(-\mathbf{N}) = -F(\mathbf{N})$. This follows because $|\cos \theta_i|$ depends only on the absolute value of the scalar product $\mathbf{n}_i \cdot \mathbf{N}$, which is unchanged under $\mathbf{N} \mapsto -\mathbf{N}$. More precisely, considering the signed projection using $\cos \theta_i = \mathbf{n}_i \cdot \mathbf{N}$ and tracking the sign, we obtain a function $G(\mathbf{N}) = (A_1 \mathbf{n}_1 \cdot \mathbf{N} - A_2 \mathbf{n}_2 \cdot \mathbf{N}, A_1 \mathbf{n}_1 \cdot \mathbf{N} - A_3 \mathbf{n}_3 \cdot \mathbf{N})$, which is antipodal: $G(-\mathbf{N}) = -G(\mathbf{N})$.

The Borsuk–Ulam theorem states that any continuous function from the sphere $S^2$ to $\mathbb{R}^2$ that is antipodal must vanish at some point $\mathbf{N} \in S^2$. Applying this theorem to $G$ guarantees the existence of a vector $\mathbf{N}$ such that $G(\mathbf{N}) = (0,0)$, and hence $A_1 |\cos \theta_1| = A_2 |\cos \theta_2| = A_3 |\cos \theta_3|$. Therefore, there exists a plane through $O$ onto which the projections of all three polygons have equal areas.

Part 2: Number of such planes passing through $O$

Let the three planes $\Pi_1$, $\Pi_2$, and $\Pi_3$ have non-coplanar normals and positive polygon areas. The equation $A_1 |\cos \theta_1| = A_2 |\cos \theta_2|$ defines a pair of symmetric cones on the unit sphere with axes along the intersection line of planes $\Pi_1$ and $\Pi_2$. Similarly, $A_1 |\cos \theta_1| = A_3 |\cos \theta_3|$ defines another pair of symmetric cones with axes along the intersection line of planes $\Pi_1$ and $\Pi_3$. Generically, the intersection of these two sets of cones consists of four points on $S^2$. Each point represents a normal vector $\mathbf{N}$, and the opposite vector $-\mathbf{N}$ corresponds to the same plane. Therefore, generically there are two distinct planes through $O$ that satisfy $A_1 |\cos \theta_1| = A_2 |\cos \theta_2| = A_3 |\cos \theta_3|$.

In special configurations, such as when two polygon planes are parallel or all areas are equal with a specific symmetry of the three planes, the set of solutions may form a continuous curve of directions on $S^2$, leading to infinitely many planes through $O$ with equal projected areas. These degenerate cases are characterized by alignment or proportionality conditions among the normals and areas.

Answer

There exists at least one plane through $O$ onto which the projections of the three polygons have equal areas. Generically, there are two distinct such planes, while in degenerate cases the number of planes may be infinite.