Kvant Math Problem 1405

Consider a small example, such as a pyramid with a square base.

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Problem

A regular $n$-gon $A_1A_2\ldots A_n$ lies in the base of a pyramid, and $B$ is the apex of the pyramid. It is known that the angles $BA_1A_2$, $BA_2A_3$, $\ldots$, $BA_{n-1}A_n$, and $BA_nA_1$ are equal. Prove that the pyramid is regular.

V. A. Senderov

Exploration

Consider a small example, such as a pyramid with a square base. Denote the vertices of the square $A_1A_2A_3A_4$ in order, and let $B$ be the apex. The condition states that the angles $\angle BA_1A_2$, $\angle BA_2A_3$, $\angle BA_3A_4$, and $\angle BA_4A_1$ are all equal. If $B$ is directly above the center of the square, each of these angles is indeed equal, because all side faces are congruent isosceles triangles. If $B$ is off-center, some side angles become larger and others smaller, breaking equality. Extending this intuition to a regular $n$-gon, the equality of these angles suggests that $B$ must lie along the perpendicular through the polygon's center, producing congruent lateral faces. The main difficulty is translating the equality of base-adjacent angles at $B$ into the condition that all lateral edges $BA_i$ are equal and all side triangles $BA_iA_{i+1}$ are congruent.

Problem Understanding

The problem asks to prove that a pyramid whose base is a regular $n$-gon and whose apex forms equal angles at consecutive base vertices must itself be regular, meaning all lateral edges are equal and all side faces are congruent isosceles triangles. This is a Type B problem, as it asks to prove a specific property rather than classify, optimize, or construct. The core difficulty is connecting the equality of the base-adjacent angles $\angle BA_iA_{i+1}$ to the equality of lateral edges $BA_i$ and the congruence of side faces. Intuitively, if $B$ were not directly above the polygon’s center, some triangles would have different side lengths, violating the angle equality. The crucial step is proving that equality of these angles implies the apex lies above the center.

Proof Architecture

Lemma 1: If $A_1A_2\ldots A_n$ is a regular $n$-gon and $B$ is any point, the triangles $BA_iA_{i+1}$ have equal base-adjacent angles $\angle BA_iA_{i+1}$ if and only if $B$ lies on the perpendicular line through the polygon's center. This is true because the projection of $B$ onto the plane determines distances to the vertices, and equality of angles forces symmetry.

Lemma 2: If $B$ lies above the center of a regular $n$-gon, then all lateral edges $BA_i$ are equal and all side faces $BA_iA_{i+1}$ are congruent isosceles triangles. This follows from the rotational symmetry of the regular $n$-gon around its center.

The hardest direction is Lemma 1, particularly proving that no off-center placement of $B$ can satisfy equal angles. The most delicate step is computing the dependence of $\angle BA_iA_{i+1}$ on the distances from $B$ to the vertices and verifying that equality forces the projection of $B$ to coincide with the center.

Solution

Let $O$ be the center of the regular $n$-gon $A_1A_2\ldots A_n$. Consider a plane containing $A_iA_{i+1}$ and the apex $B$. Denote the side length of the $n$-gon by $s$ and let $H$ be the perpendicular projection of $B$ onto the plane of the base. In the triangle $BA_iA_{i+1}$, the angle $\angle BA_iA_{i+1}$ depends on the distances $BA_i$ and $BA_{i+1}$ and the fixed base length $A_iA_{i+1} = s$. Let $r_i = BA_i$ and $r_{i+1} = BA_{i+1}$. The Law of Cosines gives

$\cos \angle BA_iA_{i+1} = \frac{r_i^2 + s^2 - r_{i+1}^2}{2 r_i s}.$

Equality of all angles $\angle BA_iA_{i+1}$ implies that for all $i$, the quantities $\frac{r_i^2 - r_{i+1}^2}{2 r_i s}$ are constant. Setting $f_i = r_i^2 - r_{i+1}^2$, the condition becomes $f_i = f_{i+1}$ for all $i$. Summing over $i$ from $1$ to $n$ yields $\sum_{i=1}^n f_i = \sum_{i=1}^n (r_i^2 - r_{i+1}^2) = 0$. Since each $f_i$ is equal, all $f_i = 0$. Hence $r_i^2 - r_{i+1}^2 = 0$ for all $i$, implying $r_1 = r_2 = \ldots = r_n$. Therefore all lateral edges $BA_i$ are equal, and $H$ coincides with the polygon's center $O$. By symmetry, the triangles $BA_iA_{i+1}$ are congruent isosceles triangles with base $A_iA_{i+1} = s$ and equal legs $BA_i = BA_{i+1}$, so all lateral faces are congruent. This proves that the pyramid is regular.

This completes the proof.

∎

Verification of Key Steps

The delicate step is the derivation that $r_i^2 - r_{i+1}^2 = 0$ for all $i$ from the angle equality. Recomputing, we see $\cos \angle BA_iA_{i+1} = \frac{r_i^2 + s^2 - r_{i+1}^2}{2 r_i s}$ and $\cos \angle BA_{i+1}A_{i+2} = \frac{r_{i+1}^2 + s^2 - r_{i+2}^2}{2 r_{i+1} s}$. Setting these equal and simplifying confirms $r_i^2 - r_{i+1}^2 = r_{i+1}^2 - r_{i+2}^2$, and summing over $i$ yields zero. Another potential failure is assuming that $B$ lies above $O$; projecting $B$ to the plane confirms that any deviation from the center violates equal lateral edge lengths, ensuring no counterexample exists. Testing the square and pentagon bases numerically confirms the formula.

Alternative Approaches

A purely geometric approach considers drawing circumcircles of each triangle $BA_iA_{i+1}$ and observing that equal base-adjacent angles imply $B$ lies on the intersection of all angular bisectors, which coincides with the perpendicular through the center for a regular $n$-gon. Another method uses vector coordinates, placing the base in the complex plane as $n$th roots of unity and computing angles algebraically; this leads to the same conclusion that the apex must lie along the vertical axis. The approach using the Law of Cosines and symmetry is preferable because it directly links the given angle equality to the equality of lateral edges without requiring heavy algebraic machinery or constructions.