Kvant Math Problem 846

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Problem

Prove that the arithmetic mean of the side lengths of a regular polygon is less than the arithmetic mean of the lengths of its diagonals.

V. F. Lev

Exploration

Let the regular polygon have $n$ sides and circumradius $R$. The lengths of all chords are

$$d_k=2R\sin\frac{k\pi}{n}, \qquad k=1,2,\dots,\left\lfloor\frac n2\right\rfloor .$$

Here $d_1$ is the side length. Every diagonal has length $d_k$ with $k\ge2$.

The statement compares the arithmetic mean of all sides with the arithmetic mean of all diagonals. Since every side has the same length, the mean side length is simply $d_1$.

The mean diagonal length is a weighted average of the numbers $d_2,d_3,\dots$. Since every one of these numbers exceeds $d_1$, the result would be immediate provided there is at least one diagonal.

For a regular polygon, a diagonal corresponds to $k\ge2$. Since $\sin x$ is strictly increasing on $[0,\pi/2]$, and

$$\frac{\pi}{n}<\frac{k\pi}{n}\le\frac{\pi}{2},$$

we obtain

$$d_k>d_1.$$

The only point requiring care is the counting of diagonals. We must ensure that every diagonal indeed belongs to one of the classes $k\ge2$, and that no class with $k\ge2$ contains sides. Once this is established, the arithmetic mean of all diagonals is a weighted average of numbers all strictly larger than $d_1$, hence is itself strictly larger than $d_1$.

For a numerical check, consider a regular pentagon. The side length is $2R\sin 36^\circ$. Every diagonal has length $2R\sin72^\circ$, which is larger. For a regular hexagon, side length is $R$, while diagonal lengths are $\sqrt3,R$ and $2R$, again larger than the side length.

The potentially hidden error is assuming without proof that every diagonal is longer than a side. This must be justified through the chord formula and the monotonicity of $\sin x$ on $[0,\pi/2]$.

Problem Understanding

We are given a regular polygon and must prove that the arithmetic mean of the lengths of all its sides is smaller than the arithmetic mean of the lengths of all its diagonals.

This is a Type B problem. A statement is given and must be proved.

The core difficulty is showing that every diagonal of a regular polygon is strictly longer than a side, and then converting that fact into a statement about arithmetic means.

Proof Architecture

The first lemma is that in a regular $n$-gon inscribed in a circle of radius $R$, the chord joining vertices whose indices differ by $k$ has length $2R\sin\frac{k\pi}{n}$. This follows from the standard chord formula.

The second lemma is that for every $k$ with $2\le k\le\lfloor n/2\rfloor$,

$$2R\sin\frac{k\pi}{n}>2R\sin\frac{\pi}{n}.$$

This follows because $\sin x$ is strictly increasing on $[0,\pi/2]$.

The third lemma is that every diagonal belongs to one of the classes $k\ge2$, whereas the sides are exactly the chords with $k=1$.

The final step is that the arithmetic mean of a collection of numbers all strictly greater than $d_1$ is itself strictly greater than $d_1$.

The lemma most likely to fail under scrutiny is the second one, because it uses the monotonicity of the sine function and requires checking that all relevant angles lie in $[0,\pi/2]$.

Solution

Let the regular polygon have $n\ge4$ sides and circumradius $R$. Number its vertices cyclically by

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

For each integer $k$ with

$$1\le k\le \left\lfloor\frac n2\right\rfloor,$$

consider the chord joining $A_i$ and $A_{i+k}$.

The central angle subtending this chord equals

$$\frac{2k\pi}{n}.$$

A chord subtending a central angle $\theta$ in a circle of radius $R$ has length

$$2R\sin\frac{\theta}{2}.$$

Hence the length of every chord of step $k$ is

$$d_k=2R\sin\frac{k\pi}{n}.$$

The sides of the polygon are precisely the chords with $k=1$, so the common side length is

$$d_1=2R\sin\frac{\pi}{n}.$$

Every diagonal corresponds to some integer

$$k\in\left{2,3,\dots,\left\lfloor\frac n2\right\rfloor\right}.$$

For such $k$,

$$\frac{\pi}{n}<\frac{k\pi}{n}\le\frac{\pi}{2}.$$

Since $\sin x$ is strictly increasing on the interval $[0,\pi/2]$,

$$\sin\frac{k\pi}{n}>\sin\frac{\pi}{n}.$$

Multiplying by $2R$ gives

$$d_k>d_1.$$

Thus every diagonal of the regular polygon is strictly longer than a side.

The arithmetic mean of the side lengths equals $d_1$, because all sides have the same length.

The arithmetic mean of the diagonal lengths is a weighted average of the numbers

$$d_2,d_3,\dots,d_{\lfloor n/2\rfloor},$$

each of which is strictly greater than $d_1$. A weighted average of numbers all exceeding $d_1$ must itself exceed $d_1$.

Therefore the arithmetic mean of the lengths of the diagonals is strictly greater than the arithmetic mean of the side lengths.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the formula

$$d_k=2R\sin\frac{k\pi}{n}.$$

The chord joins two vertices whose radii form a central angle $2k\pi/n$. The chord length corresponding to a central angle $\theta$ is obtained from the isosceles triangle formed by the two radii and the chord. Bisecting the triangle gives a right triangle with hypotenuse $R$ and half-chord $R\sin(\theta/2)$. Hence the full chord length is $2R\sin(\theta/2)$, yielding the stated formula.

The second delicate point is the monotonicity argument. For every diagonal we have

$$2\le k\le\left\lfloor\frac n2\right\rfloor.$$

Consequently

$$\frac{\pi}{n}<\frac{k\pi}{n}\le\frac{\pi}{2}.$$

The interval $[0,\pi/2]$ is exactly the interval on which $\sin x$ is strictly increasing. If one overlooked the upper bound $\pi/2$, the argument would not be justified.

The third delicate point is the passage from individual inequalities to an inequality of means. If numbers $x_1,\dots,x_m$ all satisfy $x_i>d_1$, then

$$x_1+\cdots+x_m>md_1.$$

Dividing by $m$ gives

$$\frac{x_1+\cdots+x_m}{m}>d_1.$$

This is precisely the statement needed for the diagonal lengths.

Alternative Approaches

A geometric proof avoids the explicit chord formula. In a fixed circle, larger central angles subtend longer chords. A side of the regular polygon subtends the central angle $2\pi/n$. Every diagonal subtends a central angle $2k\pi/n$ with $k\ge2$, hence a strictly larger central angle not exceeding $\pi$. Therefore every diagonal is longer than a side. The argument about arithmetic means then proceeds exactly as above.

The main approach is preferable because it converts the problem into a simple comparison of the values $\sin(k\pi/n)$ and $\sin(\pi/n)$. The comparison is completely explicit and leaves no ambiguity about why each diagonal exceeds the side length.