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, $\angle P_{0}OM_{AB} =\frac{2\pi}{n}\frac{a+b}{2}$ and $M_{AB}=\left\langle Rcos\left( \frac{2\pi}{n}\frac{a+b}{2} \right),Rsin\left( \frac{2\pi}{n}\frac{a+b}{2} \right) \right\rangle$

CASE I: $a+b$ is even

$k=\frac{a+b}{2}$ and $k$ is integer

Then $M_{AB}=\left\langle Rcos\left( \frac{2\pi}{n}k \right),Rsin\left( \frac{2\pi}{n}k \right) \right\rangle=P_{k}$

This means that the perpendicular bisector also passes through a point $P_{k}$ of $S$

Let $c$ be any positive integer

$\angle P_{k}OP_{(k+c)\; mod\; n}=\frac{2\pi}{n}\left( (k+c-k)\; mod\; n \right)=\frac{2\pi}{n}\left( c\; mod\; n \right)$

and

$\angle P_{k}OP_{(k-c)\; mod\; n}=\frac{2\pi}{n}\left( (k-(k-c))\; mod\; n \right)=\frac{2\pi}{n}\left( c\; mod\; n \right)$

Therefore, $\angle P_{k}OP_{(k+c)\; mod\; n}=\angle P_{k}OP_{(k-c)\; mod\; n}$ for any integer $c$ .

Also, since $\left| OP_{(k+c)\; mod\; n} \right|=\left| OP_{(k-c)\; mod\; n} \right|=R$ for any integer $c$

then this proves that the bisector of any points $A$ and $B$ is an axis of symmetry for this case.

CASE II: $a+b$ is odd

$k=\frac{a+b+1}{2}$ and $k$ is integer

$m=\frac{a+b-1}{2}$ and $m$ is integer

Then $M_{AB}=\left\langle Rcos\left( \frac{2\pi}{n}\frac{a+b}{2} \right),Rsin\left( \frac{2\pi}{n}\frac{a+b}{2} \right) \right\rangle$

This means that the perpendicular bisector does not pass through any point of $S$ , but their closest points are $P_{k}$ and $P_{m}$ and that $\angle MOP_{k}=\angle MOP_{m}=\frac{\pi}{n}$

Let $c$ be any positive integer

$\angle P_{k}OP_{(k+c)\; mod\; n}=\frac{2\pi}{n}\left( (k+c-k)\; mod\; n \right)=\frac{2\pi}{n}\left( c\; mod\; n \right)$

$\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{k}+\angle P_{k}OP_{(k+c)\; mod\; n}=\frac{\pi}{n}+\frac{2\pi}{n}\left( c\; mod\; n \right)$

and

$\angle P_{m}OP_{(m-c)\; mod\; n}=\frac{2\pi}{n}\left( (m-(m-c))\; mod\; n \right)=\frac{2\pi}{n}\left( c\; mod\; n \right)$

$\angle MOP_{(m-c)\; mod\; n}=\angle MOP_{m}+\angle P_{m}OP_{(m-c)\; mod\; n}=\frac{\pi}{n}+\frac{2\pi}{n}\left( c\; mod\; n \right)$

Therefore, $\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{(m-c)\; mod\; n}$ for any integer $c$ .

Since $m=k-1$ , $\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{(k-1-c)\; mod\; n}$

Also, since $\left| OP_{(k+c)\; mod\; n} \right|=\left| OP_{(m-c)\; mod\; n} \right|=R$ for any integer $c$

then this proves that the bisector of any points $A$ and $B$ is an axis of symmetry for this case.

Having proven both cases, then the set $S$ of points that comply with the given condition is the set of the vertices of any regular polygon of any number of sides.

~Tomas Diaz. orders@tomasdiaz.com

Solution 2

First we prove no $3$ points can lie on a line. Say $A_1, A_2, A_3$ were sequential points on a line. Considering the axis of symmetry between $A_2$ and $A_3$ one finds there lies a point $A_4$ on the right side of $A_3$ . Then considering the axis of symmetry between $A_3$ and $A_4$ one finds sequential points $A_5$ and $A_6$ that lie on the right side of $A_4$ . One can continue this process ad infinitum to show $S$ must have infinite points. A contradiction.

We define the subset $O \subset S$ of outer points of $S$ as follows: two points $S_1,S_2$ of $S$ are in $O$ iff all other points of $S$ lie on one side of the line the passes through $S_1$ and $S_2$ . $O$ must contain at least $3$ points.

The outer points can clearly be put in a sequence $S_1,...,S_n$ such that as a line "rolls/rotates around" $O$ , keeping all points of $S$ on one side of the line at all times, the line will sequentially touch $S_1,S_2,...,S_n,S_1$ . We will now prove $S_1,...,S_n$ form a regular polygon.

Identify $S_0$ with $S_n$ , $S_{n+1}$ with $S_1$ , and $S_{n+2}$ with $S_2$ .

Considering the axis of symmetry between $S_2$ and $S_3$ , note $\angle S_1S_2S_3$ is greater than $\angle S_kS_2S_3$ for any $k$ (by how $S_1,...,S_n$ were defined). Similarly $\angle S_2S_3S_4$ is greater than $\angle S_2S_3S_k$ for any $k$ . It follows that $S_1,S_4$ must be symmetric to each other about the axis of symmetry between $S_2$ and $S_3$ . So $\angle S_1S_2S_3 = \angle S_2S_3S_4$ . Repeat for all other sequential $4$ points of $O$ to get $\angle S_{k-1}S_kS_{k+1} = \angle S_kS_{k+1}S_{k+2}$ for all $k$ . Now if $S_2$ didn't lie on the axis of symmetry between $S_1$ and $S_3$ there would exist another point $S_2'$ symmetric to $S_2$ about that axis and $S_2'$ would be a point of $O$ between $S_1$ and $S_3$ that isn't $S_2$ . A contradiction to how $S_1,S_2,...S_n$ were ordered. So $S_2$ lies on the axis of symmetry, thus $S_1S_2 = S_2S_3$ . Repeat for all other sequential $3$ points of $O$ to get $S_{k-1}S_k = S_kS_{k+1}$ for all $k$ . We have shown the outer points $S_1,...,S_n$ of $S$ form a regular polygon.

Now, the regular polygon $O$ has a well-defined center. Therefore any axis of symmetry about which we can reflect $S$ must reflect the center of $O$ into itself. i.e. the axis of symmetry must intersect the center of $O$ .

Let $A$ be a point of $S$ . We will show $A$ is a point of $O$ , hence that $S=O$ . Let $O'$ be the center of $O$ . The axis of symmetry between $S_1$ and $A$ intersects $O'$ by the previous, hence $\angle O'S_1A= \angle S_1AO'$ . So $S_1O'A$ is an isosceles triangle with $O'S_1 = O'A$ . Therefore $A$ lies on the circle with center $O'$ and radius $O'S_1$ . If $A$ were not one of $S_1,...,S_n$ , we could suppose $A$ lies between $S_k$ and $S_{k+1}$ on this circle in which case all points of $S$ would lie on one side of $S_kA$ , making $A$ an outer point. A contradiction. So $A$ is in $O$ . So $S=O$ as desired.

~not_detriti

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

1999 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1999 IMO Problems/Problem 1

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Problem

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Solution 2

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