1963 IMO Problems/Problem 3

Problem

In an $n$ -gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation

$a_1\ge a_2\ge \cdots \ge a_n$ .

Prove that $a_1=a_2=\cdots = a_n$ .

Solution

Let $a_1 = p_1p_2$ , $a_2 = p_2p_3$ , etc.

Plot the $n$ -gon on the cartesian plane such that $p_1p_2$ is on the $x$ -axis and the entire shape is above the $x$ -axis. There are two cases: the number of sides is even, and the number of sides is odd:

$\textbf{Case 1: Even}$

In this case, the side with the topmost points will be $p_{\frac{n}{2}+1}p_{\frac{n}{2}+2}$ . To obtain the $y$ -coordinate of this top side, we can multiply the lengths of the sides $a_1$ , $a_2$ , ... $a_{\frac{n}{2}}$ by the sine of the angle they make with the $x$ -axis:

$y\textrm{-coordinate} = \sum_{k = 1}^{\frac{n}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (1)}$

We can obtain the $y$ -coordinate of the top side in a different way by multiplying the lengths of the sides $a_{\frac{n}{2}+1}$ , $a_{\frac{n}{2}+2}$ , ... $a_n$ by the sine of the angle they make with the $x$ -axis to get the $\emph{negated}$ $y$ -coordinate of the top side:

$-y\textrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}$ $y\textrm{-coordinate} = \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot -\sin \frac{2\pi(k-1)}{n}$ $= \sum_{k = \frac{n}{2}+1}^{n}a_k \cdot \sin \frac{2\pi(k-\frac{n}{2}-1)}{n}$ $= \sum_{k = 1}^{\frac{n}{2}}a_{k+\frac{n}{2}} \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (2)}$

It must be true that $\textbf{(1)} = \textbf{(2)}$ . This implies that $a_k = a_{k+\frac{n}{2}}$ for all $1 \leq k \leq \frac{n}{2}$ , and therefore $a_1=a_2=\cdots = a_n$ .

$\textbf{Case 2: Odd}$

This case is very similar to before. We will compute the $y$ -coordinate of the top point $p_{frac{n+3}{2}}$ two ways:

$y\textrm{-coordinate} = \sum_{k = 2}^{\frac{n+1}{2}}a_k \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (3)}$ $-y\textrm{-coordinate} = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(k-1)}{n}$ $y\textrm{-coordinate} = \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot -\sin \frac{2\pi(k-1)}{n}$ $= \sum_{k = \frac{n+3}{2}}^{n}a_k \cdot \sin \frac{2\pi(n - k + 1)}{n}$ $= \sum_{k = 2}^{\frac{n+1}{2}}a_{n-k+2} \cdot \sin \frac{2\pi(k-1)}{n}.\textbf{ (4)}$

It must be true that $\textbf{(3)} = \textbf{(4)}$ . Then, we get $a_k = a_{n-k+2}$ for all $2 \leq k \leq \frac{n+1}{2}$ . Therefore, $a_2=a_3=\cdots = a_n$ . It is trivial that $a_1$ is then equal to the other values, so $a_1=a_2=\cdots = a_n$ . This completes the proof. $\square$

~mathboy100

Solution 2

Define the vector $\vec{v_i}$ to equal $\cos{\left(\frac{2\pi}{n}i\right)}\vec{i}+\sin{\left(\frac{2\pi}{n}i\right)}\vec{j}$ . Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length $a_i$ is parallel to $\vec{v_i}$ . We then have that

$\sum_{i=1}^{n} a_i\vec{v_i}=\vec{0}\Rightarrow \sum_{i=1}^{n} a_i\cos{\left(\frac{2\pi}{n}i\right)} = \sum_{i=1}^{n} a_i\sin{\left(\frac{2\pi}{n}i\right)} =0$

But $a_i\geq a_{n-i}$ for all $i\leq \lfloor \frac{n}{2}\rfloor$ , so

$a_i \sin{\left(\frac{2\pi}{n}i\right)} = -a_i\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq -a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}$

for all $i\leq \lfloor \frac{n}{2}\rfloor$ . This shows that $a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0$ , with equality when $a_i=a_{n-i}$ . Therefore

$\sum_{i=1}^{n} a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}\rfloor} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0$

There is equality only when $a_i=a_{n-i}$ for all $i$ . This implies that $a_1=a_{n-1}$ and $a_2=a_n$ , so we have that $a_1=a_2=\cdots =a_n$ . $\blacksquare$

1963 IMO (Problems) • Resources
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1963 IMO Problems/Problem 3

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Problem

Solution

Solution 2

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