1963 IMO Problems/Problem 3


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 $a_1=a_2=\cdots = a_n$.


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. $\square$


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 $a_1=a_2=\cdots =a_n$. $\blacksquare$

See Also

1963 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
All IMO Problems and Solutions
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