1999 IMO Problems/Problem 1

Problem

Determine all finite sets $S$ of at least three points in the plane which satisfy the following condition:

For any two distinct points $A$ and $B$ in $S$ , the perpendicular bisector of the line segment $AB$ is an axis of symmetry of $S$ .

Solution

Upon reading this problem and drawing some points, one quickly realizes that the set $S$ consists of all the vertices of any regular polygon.

Now to prove it with some numbers:

Let $S=\left\{ P_{0},P_{1},P_{2},...,P_{n-1} \right\}$ , with $n\ge 3$ , where $P_{i}$ is a vertex of a polygon which we can define their $xy$ coordinates as: $P_{i}=\left\langle Rcos\left( \frac{2\pi}{n}i \right),Rsin\left( \frac{2\pi}{n}i \right) \right\rangle$ for $i=0,1,2,...,(n-1)$ .

That defines the vertices of any regular polygon with $R$ being the radius of the circumcircle of the regular $n$ -sided polygon.

Now we can pick any points $A$ and $B$ of the set as:

$A=P_{a}$ and $B=P_{b}$ , where $a=0,1,2,...,(n-1)$ ; $b=0,1,2,...,(n-1)$ ; and $a\ne b$

Then,

$A=\left\langle Rcos\left( \frac{2\pi}{n}a \right),Rsin\left( \frac{2\pi}{n}a \right) \right\rangle$

and $B=\left\langle Rcos\left( \frac{2\pi}{n}b \right),Rsin\left( \frac{2\pi}{n}b \right) \right\rangle$

Let $O$ be point $(0,0)$ which is not part of $S$

Then, $\angle P_{0}OA = \frac{2\pi}{n}a$ , and $\angle P_{0}OB = \frac{2\pi}{n}b$

The perpendicular bisector of $AB$ passes through $O$ .

Let point $M_{AB}$ , not in $S$ be a point that passes through the perpendicular bisector of $AB$ at a distance $R$ from $O$

Then, $\angle P_{0}OM_{AB} =\frac{2\pi}{n}\frac{a+b}{2}$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1999 IMO Problems/Problem 1

Problem

Solution