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

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
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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1963 IMO Problems/Problem 3

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