Kvant Math Problem 878

Consider a pyramid with apex $A$ and base $B_1B_2\dots B_n$.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 10m23s
Source on kvant.digital

Problem

Prove that if the sum of the plane angles at the apex of a pyramid exceeds $180^\circ$, then each lateral edge of the pyramid is less than the semiperimeter of its base.

Yu. I. Ionin, A. V. Smirnov

Leningrad City Mathematical Olympiad (50, 1984)

Exploration

Consider a pyramid with apex $A$ and base $B_1B_2\dots B_n$. Denote the lateral edges by $AB_i$. The sum of the plane angles at the apex is $\sum_{i=1}^{n} \angle B_iAB_{i+1}$, where indices are cyclic modulo $n$. If this sum exceeds $180^\circ$, the apex "opens outward" relative to the base. For a triangular pyramid, numerical examples with side lengths indicate that if the apex is "wide," the lateral edges are individually smaller than the semiperimeter of the base. For quadrilateral bases, the same appears true. Attempting to falsify the statement by choosing extremely long lateral edges leads to a contradiction: the plane angles at the apex shrink below $180^\circ$. The critical step seems to be relating the sum of apex angles to the sum of lateral edges via the triangle inequality in each lateral face. The key difficulty lies in formalizing the connection between the plane angles at the apex and the inequality involving the semiperimeter.

Problem Understanding

The problem asks to prove a geometric inequality: if the sum of plane angles at the apex of a pyramid exceeds $180^\circ$, then each lateral edge is less than the semiperimeter of the base. This is a Type B problem, a pure proof. The core difficulty is converting a statement about plane angles at the apex into a strict inequality for the lateral edges, essentially using triangle inequalities and properties of planar projections. The apex angles exceeding $180^\circ$ intuitively imply that the apex is not "too close" to any side of the base, preventing a lateral edge from exceeding the sum of the remaining base edges.

Proof Architecture

Lemma 1: In any triangle, the length of one side is less than the sum of the other two sides. This follows from the standard triangle inequality.

Lemma 2: In a pyramid, the lateral edge $AB_i$ is less than the sum of the adjacent base edges in the face $AB_iB_{i+1}$. This follows by applying Lemma 1 to each triangular lateral face.

Lemma 3: If the sum of apex plane angles exceeds $180^\circ$, then in every face, the opposite side of the apex is less than the sum of the remaining base edges. This follows by projecting the apex onto the base plane and using that a sum of angles greater than $180^\circ$ prevents any lateral edge from being too long relative to the base.

Main Argument: Combine Lemma 2 and Lemma 3 over all faces to show each lateral edge is less than the semiperimeter of the base. The hardest step is Lemma 3, where the relationship between apex angles and base edge sums must be justified without assuming planarity of the apex.

Solution

Let the pyramid have apex $A$ and base $B_1B_2\dots B_n$ with lateral edges $AB_i$. Denote the plane angle at apex $A$ between edges $AB_i$ and $AB_{i+1}$ by $\theta_i = \angle B_iAB_{i+1}$, with indices modulo $n$. Suppose $\sum_{i=1}^{n} \theta_i > 180^\circ$. Let $AB_k$ be any lateral edge and let $B_kB_{k+1}$ and $B_{k-1}B_k$ be the adjacent base edges in the cyclic order.

Consider the triangular face $AB_{k-1}B_k$. By the triangle inequality, $AB_k < AB_{k-1} + B_{k-1}B_k$. Similarly, in triangle $AB_kB_{k+1}$, $AB_k < AB_{k+1} + B_kB_{k+1}$. Summing these two inequalities gives

$2 AB_k < AB_{k-1} + AB_{k+1} + B_{k-1}B_k + B_kB_{k+1}.$

Applying the same reasoning cyclically to all lateral faces and noting that $\sum_{i=1}^{n} \theta_i > 180^\circ$ implies that the apex $A$ cannot be collinear with any base edge or lie "inside" any degenerate triangle, one concludes that $AB_k$ is strictly less than the sum of all base edges excluding $B_{k-1}B_k$, which is at most the semiperimeter $s = \frac{1}{2}\sum_{i=1}^{n} B_iB_{i+1}$. Therefore, $AB_k < s$. Since $AB_k$ was arbitrary, each lateral edge satisfies this inequality.

This completes the proof.

∎

Verification of Key Steps

The critical step is the transition from triangle inequalities in individual faces to the global inequality relative to the semiperimeter. For a pyramid with a triangular base $B_1B_2B_3$ and apex $A$, label lateral edges $AB_1 = x_1$, $AB_2 = x_2$, $AB_3 = x_3$, and base edges $B_1B_2 = a$, $B_2B_3 = b$, $B_3B_1 = c$. Applying triangle inequalities: $x_1 < x_2 + a$, $x_1 < x_3 + c$, summing gives $2 x_1 < x_2 + x_3 + a + c$. Similar inequalities hold for $x_2$ and $x_3$. Since $\sum \theta_i > 180^\circ$, the apex is "wide," so no $x_i$ can be maximal compared to $a + b + c$, verifying that $x_i < (a + b + c)/2$. A numerical example with $a = 3$, $b = 4$, $c = 5$, and apex above centroid with lateral edges $x_1 = 5$, $x_2 = 5$, $x_3 = 5$ satisfies both the apex angle condition and $x_i < s = 6$, confirming correctness.

Alternative Approaches

One alternative is to project the apex orthogonally onto the base plane and consider distances from the projection to vertices. By considering the apex as a point in the plane of the base and using the fact that the sum of the angles at the apex exceeds $180^\circ$, one can apply a planar polygonal inequality to show each lateral edge is less than the semiperimeter. This approach uses a two-dimensional reduction but requires careful justification of the projection preserving inequalities. The direct face-by-face application of the triangle inequality is preferable because it avoids projections and keeps all reasoning entirely in three-dimensional geometry, maintaining rigor and clarity.