Kvant Math Problem 726

Verified: yes
Verdicts: PASS + PASS
Solve time: 23m55s
Source on kvant.digital

Problem

A point inside a regular $2n$-gon is joined to all the vertices. The resulting $2n$ triangles are colored alternately blue and red. Prove that the sum of the areas of the blue triangles is equal to the sum of the areas of the red triangles

for $n=4$,
for $n=3$,
for an arbitrary natural $n$.

V. V. Prasolov

Exploration

Let the regular $2n$-gon have vertices $A_1,A_2,\dots,A_{2n}$ in cyclic order, and let $P$ be a point inside the polygon. Denote by $S_i$ the area of triangle $PA_iA_{i+1}$, where indices are taken modulo $2n$. The triangles are colored alternately blue and red. The problem reduces to proving

$S_1+S_3+\cdots+S_{2n-1} = S_2+S_4+\cdots+S_{2n}.$

Each triangle has a base $A_iA_{i+1}$ of length $s$ and an altitude equal to the perpendicular distance from $P$ to the line containing $A_iA_{i+1}$. Let $l_i$ be the line containing side $A_iA_{i+1}$ and $d_i$ the perpendicular distance from $P$ to $l_i$. Then

$S_i = \frac12 s, d_i.$

This converts the area identity into the linear identity

$\sum_{i=1}^{2n}(-1)^{i-1} d_i = 0.$

For the octagon, the sides $A_iA_{i+1}$ and $A_{i+4}A_{i+5}$ are parallel. Let $h_i$ denote the distance between these two parallel lines. The sum $d_i+d_{i+4}$ equals $h_i$, independent of the position of $P$. Pairing opposite sides and summing the alternating distances produces zero, which proves the identity for $n=4$. A similar pairing of opposite sides works for $n=3$.

To generalize, consider the inward unit normal $u_i$ to each side line $l_i$. Let the incenter of the polygon be the origin, and let $r$ denote the inradius. Then each side line satisfies $u_i\cdot x = r$. For a point $P$ with position vector $p$, the perpendicular distance is

$d_i = r - u_i \cdot p.$

Hence

$\sum_{i=1}^{2n}(-1)^{i-1} d_i = r \sum_{i=1}^{2n} (-1)^{i-1} - p \cdot \sum_{i=1}^{2n} (-1)^{i-1} u_i.$

The first sum vanishes because it contains equal numbers of $+1$ and $-1$ terms. Therefore it suffices to prove

$\sum_{i=1}^{2n} (-1)^{i-1} u_i = 0.$

Problem Understanding

The problem asks for the equality of the sums of areas of alternately colored triangles in a regular $2n$-gon for $n=3$, $n=4$, and general $n$. The essential idea is that the linear combination of distances from an interior point to the side lines, taken with alternating signs, vanishes for any point inside the polygon. This converts the geometric problem into a linear algebraic statement about the normals to the sides. The validity of the formula for the signed distances and the sum of normals is central.

Proof Architecture

Let $s$ be the common side length of the polygon. For each side $A_iA_{i+1}$, let $l_i$ be the line containing the side, and let $u_i$ be the inward unit normal vector. With the origin at the polygon center, the line has equation $u_i \cdot x = r$, where $r$ is the inradius. For a point $P$ with position vector $p$, the signed perpendicular distance to the line is $d_i = r - u_i \cdot p$. Then

$S_i = \frac12 s, d_i,$

and

$\sum_{i=1}^{2n}(-1)^{i-1} S_i = \frac{s}{2} \sum_{i=1}^{2n}(-1)^{i-1} d_i.$

Hence it suffices to prove $\sum_{i=1}^{2n}(-1)^{i-1} d_i = 0$.

To justify this, represent the plane as the complex plane. Each unit normal vector $u_i$ corresponds to $e^{i(\theta + (i-1)\pi/n)}$ for some angle $\theta$ depending on the orientation of the polygon. Then

$\sum_{i=1}^{2n} (-1)^{i-1} u_i = \sum_{i=1}^{2n} (-1)^{i-1} e^{i(\theta + (i-1)\pi/n)} = e^{i\theta} \sum_{k=0}^{2n-1} (-1)^k e^{i k \pi/n} = e^{i\theta} \sum_{k=0}^{2n-1} (-e^{i\pi/n})^k.$

The ratio of the geometric series is $q = -e^{i\pi/n}$. Since $q \neq 1$ and $q^{2n} = (-e^{i\pi/n})^{2n} = e^{2\pi i} = 1$, the sum of the series is

$\sum_{k=0}^{2n-1} q^k = \frac{1 - q^{2n}}{1 - q} = 0.$

Therefore $\sum_{i=1}^{2n} (-1)^{i-1} u_i = 0$, which implies $\sum_{i=1}^{2n} (-1)^{i-1} d_i = 0$, and consequently $\sum_{i=1}^{2n} (-1)^{i-1} S_i = 0$. This proves the equality of the sums of areas of blue and red triangles.

Solution

Label the vertices of the regular $2n$-gon as $A_1,A_2,\dots,A_{2n}$ in cyclic order. Let $P$ be any interior point. Define $S_i = [PA_iA_{i+1}]$ for $i=1,\dots,2n$. Let $s$ be the common side length. Let $u_i$ denote the inward unit normal to the line $l_i$ containing side $A_iA_{i+1}$. Choose the origin at the center of the polygon, with inradius $r$. Then $l_i$ satisfies $u_i \cdot x = r$, so the perpendicular distance from $P$ to $l_i$ is $d_i = r - u_i \cdot p$, and $S_i = \frac12 s, d_i$.

Compute the alternating sum:

$\sum_{i=1}^{2n} (-1)^{i-1} S_i = \frac{s}{2} \sum_{i=1}^{2n} (-1)^{i-1} d_i = -\frac{s}{2} p \cdot \sum_{i=1}^{2n} (-1)^{i-1} u_i.$

Express the normals as complex numbers $u_i = e^{i(\theta + (i-1)\pi/n)}$. Then

$\sum_{i=1}^{2n} (-1)^{i-1} u_i = \sum_{k=0}^{2n-1} (-1)^k e^{i k \pi/n} = \sum_{k=0}^{2n-1} (-e^{i\pi/n})^k = 0.$

Hence $\sum_{i=1}^{2n} (-1)^{i-1} d_i = 0$ and $\sum_{i=1}^{2n} (-1)^{i-1} S_i = 0$. Consequently

$S_1+S_3+\cdots+S_{2n-1} = S_2+S_4+\cdots+S_{2n}.$

This establishes the statement for $n=3$, $n=4$, and all natural $n$.

∎

Verification of Key Steps

The formula $S_i = \frac12 s, d_i$ is justified because the area of a triangle equals one half the base times its height. The perpendicular distance $d_i$ from an interior point to the side line is positive. The formula $d_i = r - u_i \cdot p$ is valid because $u_i$ is a unit vector, and $r$ is the distance from the origin to the side line. Representing the normals as $u_i = e^{i(\theta + (i-1)\pi/n)}$ is exact because the polygon is regular, which ensures equal angular spacing. The alternating sum forms a geometric series with ratio $q = -e^{i\pi/n}$, whose $2n$-th power equals 1. Since $q \neq 1$, the sum of the series is zero, which rigorously proves the identity.

Alternative Approaches

For $n=3$ and $n=4$, one can pair opposite sides of the hexagon or octagon, which are parallel. Let $d_i$ denote the distance from $P$ to side $i$. Then $d_i + d_{i+n} = h_i$ is constant, where $h_i$ is the distance between parallel lines. The alternating sum of areas becomes the sum over these pairs with