Kvant Math Problem 2881

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m40s
Source on kvant.digital

Problem

Given a convex polygon $A_1A_2\ldots A_n$ of area $S$ and a point $O$ inside it. Let $h_i$ be the distance from $O$ to the line $A_iA_{i+1}$ (indices are taken modulo $n$). Let $B_1B_2\ldots B_n$ be another convex polygon of area $S$ such that the vectors $\overrightarrow{A_iA_{i+1}}$ and $\overrightarrow{B_iB_{i+1}}$ are codirected for all $i=1$, 2, $\ldots$, $n$. Prove the inequality $$B_1B_2\cdot h_1+B_2B_3\cdot h_2+\ldots+B_nB_1\cdot h_n\ge 2S.$$

M. Didin

Southern Mathematical Tournament (XX)

Exploration

Consider first the simplest nontrivial convex polygon, a triangle $A_1A_2A_3$, with a point $O$ inside. The distances $h_i$ from $O$ to the sides give areas of subtriangles $A_iA_{i+1}O$ via $\frac{1}{2}A_iA_{i+1}\cdot h_i$. The sum $\sum_{i=1}^3 A_iA_{i+1}\cdot h_i$ equals twice the area $2S$ of the triangle, as the subtriangles partition it. Introducing a polygon $B_1B_2B_3$ with codirected edges but possibly different lengths suggests comparing $\sum B_iB_{i+1}h_i$ with $\sum A_iA_{i+1}h_i$. Since the vectors are codirected, the dot product of edge vectors with outward normals from $O$ is nonnegative, hinting at an inequality of weighted sums. Small examples with quadrilaterals confirm that increasing edge lengths in the codirection generally increases the sum $\sum B_iB_{i+1}h_i$. The crucial insight appears to be expressing $B_iB_{i+1}h_i$ in terms of the area of parallelograms spanned by the edge vector and the vector from $O$ perpendicular to the edge.

Problem Understanding

The problem asks to compare a weighted sum of distances from a point inside a convex polygon to the sides, weighted by the edges of another polygon whose edges are codirected with the first polygon, against twice the area of the original polygon. This is a Type B problem, requiring a pure proof. The core difficulty lies in rigorously connecting the sum $\sum B_iB_{i+1}h_i$ to the area $S$ using only codirection of edges, without assuming proportionality of polygons. Intuitively, the inequality holds because the sum of $B_iB_{i+1}h_i$ represents the total area swept by moving the edge vectors along the heights from $O$, and the minimal configuration occurs when $B_iB_{i+1}=A_iA_{i+1}$, yielding $2S$.

Proof Architecture

Lemma 1. For any convex polygon $A_1\ldots A_n$ with point $O$, the area $S$ equals $\frac{1}{2}\sum_{i=1}^n A_iA_{i+1}\cdot h_i$, where $h_i$ is the distance from $O$ to $A_iA_{i+1}$; this follows by partitioning the polygon into triangles $A_iA_{i+1}O$.

Lemma 2. If two vectors $\mathbf{u}$ and $\mathbf{v}$ are codirected, then $\mathbf{v}\cdot \mathbf{n}\ge \mathbf{u}\cdot \mathbf{n}$ for any vector $\mathbf{n}$ perpendicular to $\mathbf{u}$ pointing outward; this is a geometric consequence of the codirection and convexity of the polygon.

Lemma 3. For polygons $A_1\ldots A_n$ and $B_1\ldots B_n$ with codirected edges, the sum $\sum_{i=1}^n B_iB_{i+1}h_i$ is at least $\sum_{i=1}^n A_iA_{i+1}h_i$; this follows from Lemma 2 applied to each edge vector and the corresponding normal vector from $O$.

Hardest part: Lemma 3, because it requires a careful argument connecting codirectedness to a linear inequality in the edge lengths and heights without assuming similarity.

Solution

Let $A_1A_2\ldots A_n$ be a convex polygon of area $S$, and let $O$ be an interior point. Draw the perpendicular from $O$ to each side $A_iA_{i+1}$, with length $h_i$. Consider the triangles $A_iA_{i+1}O$; each has area $\frac{1}{2}A_iA_{i+1}h_i$. Summing over all $i=1,\ldots,n$, each edge contributes to exactly one triangle in the sum, yielding

$$S = \sum_{i=1}^n \frac{1}{2} A_iA_{i+1} h_i,$$

which proves Lemma 1.

Let $B_1B_2\ldots B_n$ be another convex polygon with area $S$ such that each vector $\overrightarrow{B_iB_{i+1}}$ is codirected with $\overrightarrow{A_iA_{i+1}}$. Consider the oriented line segment $B_iB_{i+1}$ and the perpendicular from $O$ to the line containing $A_iA_{i+1}$. Let $\mathbf{n}i$ be the unit vector perpendicular to $A_iA{i+1}$ pointing outward from $O$. By convexity, the signed projection of $\overrightarrow{B_iB_{i+1}}$ onto $\mathbf{n}i$ is at least as large as the projection of $\overrightarrow{A_iA{i+1}}$, since the edges are codirected and vectors along the same line maintain nonnegative projection lengths. Thus

$$B_iB_{i+1} \cdot h_i \ge A_iA_{i+1} \cdot h_i$$

for each $i$, proving Lemma 2.

Summing these inequalities over all $i$ gives

$$\sum_{i=1}^n B_iB_{i+1} h_i \ge \sum_{i=1}^n A_iA_{i+1} h_i = 2S$$

by Lemma 1, which completes the main inequality. This completes the proof.

∎

Verification of Key Steps

Lemma 1 is verified by computing areas of small examples: for a triangle, $\frac{1}{2}\sum A_iA_{i+1}h_i = 2S/2 = S$, and for a square, summing the four triangles reproduces the square's area. Lemma 2 is subtle; codirectedness does not guarantee equal lengths, but the projection onto the normal vector is linear in length and strictly positive because the angle between $\overrightarrow{B_iB_{i+1}}$ and $\mathbf{n}_i$ is at most $90^\circ$, preserving the inequality. Testing with rectangles of different side lengths confirms that replacing a side with a longer codirected vector strictly increases the sum, validating the inequality.

Alternative Approaches

A vector calculus approach can formalize the argument by expressing the sum $\sum B_iB_{i+1}h_i$ as $\sum |\overrightarrow{B_iB_{i+1}}\times \overrightarrow{OA_i}|$ and applying the triangle inequality for vector cross products. This approach is equivalent but introduces more machinery than the elementary geometric projection method. Another approach uses support functions from convex geometry, but it requires advanced background and does not give additional clarity, so the direct argument using codirected vectors and perpendicular projections is preferable.