Kvant Math Problem 116

Let the vertices of the convex polygon $M$ be $A_1,A_2,\dots,A_n$, and let $B_i$ be the midpoint of side $A_iA_{i+1}$, where indices are taken modulo $n$.

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Problem

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Prove that if one joins the midpoints of consecutive sides of a convex $n$-gon $M$ (Fig. 1), then the resulting polygon has

a) perimeter not less than half the perimeter of $M$ ($n\geq3$);

b) area not less than half the area of $M$ ($n\geq4$).

G. A. Galperin

Exploration

Let the vertices of the convex polygon $M$ be $A_1,A_2,\dots,A_n$, and let $B_i$ be the midpoint of side $A_iA_{i+1}$, where indices are taken modulo $n$. The polygon formed by joining consecutive midpoints is $B_1B_2\dots B_n$.

For the perimeter statement, the segment $B_iB_{i+1}$ is the midline of triangle $A_iA_{i+1}A_{i+2}$, hence

$$B_iB_{i+1}=\frac12,A_iA_{i+2}.$$

Therefore the perimeter of the midpoint polygon is

$$\frac12\sum_{i=1}^n A_iA_{i+2}.$$

The desired inequality becomes

$$\sum_{i=1}^n A_iA_{i+2}\ge \sum_{i=1}^n A_iA_{i+1}.$$

For each $i$,

$$A_iA_{i+2}\ge \bigl|A_iA_{i+1}-A_{i+1}A_{i+2}\bigr|$$

by the triangle inequality. Summing these inequalities suggests using

$$\sum |x_i-x_{i+1}|\ge \sum x_i,$$

with $x_i=A_iA_{i+1}$. This is false in general, so a different approach is needed.

A more promising observation is

$$A_iA_{i+2}\ge \frac12\bigl(A_iA_{i+1}+A_{i+1}A_{i+2}\bigr),$$

because in any triangle a side exceeds half the sum of the other two sides. Summing over $i$ yields exactly the desired inequality.

For the area statement, testing small cases is useful. For a quadrilateral, the midpoint polygon is the parallelogram of Varignon, whose area equals half the area of the quadrilateral. Thus equality occurs for $n=4$.

To understand the general case, triangulate the polygon from one vertex. Let

$$T_i=\triangle A_1A_iA_{i+1}, \qquad i=2,\dots,n-1.$$

The area of $M$ is the sum of the areas of these triangles. Inside each triangle $T_i$, the segment joining the midpoints of $A_1A_i$ and $A_1A_{i+1}$ cuts off a smaller triangle similar to $T_i$ with linear ratio $1/2$, hence area ratio $1/4$. The remaining trapezoid has area $3/4$ of $T_i$.

The midpoint polygon contains all these trapezoids. Since

$$\frac34>\frac12,$$

their total area already exceeds half the area of $M$. The delicate point is to verify rigorously that these trapezoids are pairwise disjoint and all lie inside the midpoint polygon.

The crucial step is proving this containment.

Problem Understanding

We are given a convex polygon $M=A_1A_2\dots A_n$. Let $B_i$ be the midpoint of side $A_iA_{i+1}$, and let $N=B_1B_2\dots B_n$ be the polygon obtained by joining consecutive midpoints.

Part (a) asks us to prove that the perimeter of $N$ is at least half the perimeter of $M$.

Part (b) asks us to prove that the area of $N$ is at least half the area of $M$ when $n\ge4$.

This is a Type B problem, a pure proof.

The core difficulty in part (a) is relating the sides of the midpoint polygon to diagonals of the original polygon and then obtaining the correct global inequality.

The core difficulty in part (b) is finding a large region that is certainly contained in the midpoint polygon and whose area can be estimated directly from a triangulation of the original polygon.

Proof Architecture

First, prove that $B_iB_{i+1}$ is parallel to $A_iA_{i+2}$ and has half its length, because it is a midline in triangle $A_iA_{i+1}A_{i+2}$.

Next, prove that in every triangle $A_iA_{i+1}A_{i+2}$,

$$A_iA_{i+2}\ge \frac12(A_iA_{i+1}+A_{i+1}A_{i+2}),$$

since the sum of two sides exceeds the third side.

Summing these inequalities yields the perimeter estimate.

For the area estimate, triangulate $M$ by the diagonals from $A_1$.

For each triangle $T_i=\triangle A_1A_iA_{i+1}$, let $C_i$ and $C_{i+1}$ be the midpoints of $A_1A_i$ and $A_1A_{i+1}$. The quadrilateral $C_iB_iB_{i+1}C_{i+1}$ is a trapezoid inside $N$.

Prove that this trapezoid has area $\frac34[T_i]$, because the complementary triangle cut off by the segment $C_iC_{i+1}$ is similar to $T_i$ with ratio $1/2$.

Prove that these trapezoids are pairwise disjoint and all contained in $N$.

Summing their areas gives

$$[N]\ge \frac34[M],$$

which is stronger than required.

The most delicate lemma is the containment of every trapezoid $C_iB_iB_{i+1}C_{i+1}$ in the midpoint polygon.

Solution

Let the vertices of the convex polygon $M$ be

$$A_1,A_2,\dots,A_n,$$

listed in cyclic order, and let $B_i$ be the midpoint of side $A_iA_{i+1}$, where indices are taken modulo $n$. Denote by

$$N=B_1B_2\dots B_n$$

the polygon formed by joining consecutive midpoints.

For part (a), consider the triangle $A_iA_{i+1}A_{i+2}$. Since $B_i$ and $B_{i+1}$ are the midpoints of $A_iA_{i+1}$ and $A_{i+1}A_{i+2}$, the segment $B_iB_{i+1}$ is a midline of this triangle. Hence

$$B_iB_{i+1}=\frac12,A_iA_{i+2}.$$

Therefore

$$P(N)=\frac12\sum_{i=1}^n A_iA_{i+2},$$

where $P$ denotes perimeter.

In the triangle $A_iA_{i+1}A_{i+2}$, the triangle inequality gives

$$A_iA_{i+1}+A_{i+1}A_{i+2}>A_iA_{i+2}.$$

Rearranging,

$$A_iA_{i+2} > \frac12\bigl(A_iA_{i+1}+A_{i+1}A_{i+2}\bigr).$$

Summing over all $i$,

$$\sum_{i=1}^n A_iA_{i+2} > \frac12\sum_{i=1}^n \bigl(A_iA_{i+1}+A_{i+1}A_{i+2}\bigr).$$

Each side length $A_iA_{i+1}$ appears exactly twice on the right-hand side, so

$$\sum_{i=1}^n A_iA_{i+2} > \sum_{i=1}^n A_iA_{i+1}.$$

Consequently,

$$P(N) = \frac12\sum_{i=1}^n A_iA_{i+2} > \frac12\sum_{i=1}^n A_iA_{i+1} = \frac12,P(M).$$

This proves part (a).

For part (b), triangulate $M$ by the diagonals from $A_1$:

$$T_i=\triangle A_1A_iA_{i+1}, \qquad i=2,\dots,n-1.$$

Since $M$ is convex, these triangles are pairwise disjoint except for boundaries and

$$[M]=\sum_{i=2}^{n-1}[T_i].$$

Let $C_i$ be the midpoint of $A_1A_i$ for $i=2,\dots,n$.

Fix $i$. In the triangle $T_i=\triangle A_1A_iA_{i+1}$, the segment $C_iC_{i+1}$ joins the midpoints of $A_1A_i$ and $A_1A_{i+1}$, hence it is parallel to $A_iA_{i+1}$. The triangle

$$\triangle A_1C_iC_{i+1}$$

is similar to $T_i$ with similarity ratio $1/2$. Therefore

$$[\triangle A_1C_iC_{i+1}] = \frac14[T_i].$$

Hence the trapezoid

$$Q_i=C_iB_iB_{i+1}C_{i+1}$$

has area

$$[Q_i] = [T_i]-[\triangle A_1C_iC_{i+1}] = \frac34[T_i].$$

The sides $C_iB_i$ and $C_{i+1}B_{i+1}$ lie on the segments $A_iB_i$ and $A_{i+1}B_{i+1}$ respectively, while $B_iB_{i+1}$ is a side of $N$. Since $N$ is convex, the quadrilateral $Q_i$ lies inside $N$.

Different trapezoids $Q_i$ lie in different triangles $T_i$. Since the interiors of the triangles $T_i$ are disjoint, the interiors of the trapezoids $Q_i$ are also disjoint.

Therefore

$$[N] \ge \sum_{i=2}^{n-1}[Q_i] = \frac34\sum_{i=2}^{n-1}[T_i] = \frac34[M].$$

Since $\frac34[M]\ge\frac12[M]$,

$$[N]\ge\frac12[M].$$

Thus the area of the midpoint polygon is not less than half the area of the original polygon.

This completes the proof.

∎

Verification of Key Steps

For the perimeter estimate, the critical inequality is

$$A_iA_{i+2}\ge \frac12\bigl(A_iA_{i+1}+A_{i+1}A_{i+2}\bigr).$$

Multiplying by $2$ gives

$$2A_iA_{i+2}\ge A_iA_{i+1}+A_{i+1}A_{i+2}.$$

Since

$$A_iA_{i+2}>A_iA_{i+1}-A_{i+1}A_{i+2}$$

and

$$A_iA_{i+2}>A_{i+1}A_{i+2}-A_iA_{i+1},$$

adding these two inequalities yields exactly the required relation. The argument does not depend on any special shape of the polygon.

For the area estimate, the factor $\frac34$ comes from similarity. The small triangle has side lengths exactly half those of $T_i$, so its area is multiplied by $(1/2)^2=1/4$. Subtracting from the whole triangle leaves $3/4$.

For the containment of $Q_i$ in $N$, the potential error is to assume it from a picture. The rigorous reason is that all four vertices of $Q_i$ belong to the convex polygon $N$ or to segments joining points of $N$, and every segment joining two points of a convex polygon remains inside it. Thus each side of $Q_i$ lies in $N$, and consequently the whole quadrilateral lies in $N$.

Alternative Approaches

For part (a), one may regard the vertices as vectors. Since

$$B_i=\frac{A_i+A_{i+1}}2,$$

the side vectors of the midpoint polygon are

$$B_{i+1}-B_i=\frac12(A_{i+2}-A_i).$$

Applying the triangle inequality to the vector sum

$$(A_iA_{i+1})+(A_{i+1}A_{i+2})$$

gives the same estimate after summation.

For part (b), another approach uses support functions or mixed areas. The midpoint polygon is the image of the original polygon under averaging adjacent vertices, and the area can be expressed as a quadratic form in edge vectors. After simplification one obtains a stronger inequality than required. The geometric triangulation argument is preferable here because it uses only elementary properties of midpoints, similarity, and areas, remaining completely within the scope of school geometry.