Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1999 IMO Problems/Problem 1"
m (remove a word solution 2)
 
(14 intermediate revisions by the same user not shown)
Line 91: Line 91:
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
  
 +
==Solution 2==
 +
First we prove no <math>3</math> points can lie on a line. Say <math>A_1, A_2, A_3</math> were sequential points on a line. Considering the axis of symmetry between <math>A_2</math> and <math>A_3</math> one finds there lies a point <math>A_4</math> on the right side of <math>A_3</math>. Then considering the axis of symmetry between <math>A_3</math> and <math>A_4</math> one finds sequential points <math>A_5</math> and <math>A_6</math> that lie on the right side of <math>A_4</math>. One can continue this process ad infinitum to show <math>S</math> must have infinite points. A contradiction.
 +
 +
Let a line roll/rotate around the perimeter of <math>S</math> such that at any time all points of <math>S</math> are on one side of the line and the line is always touching one point of <math>S</math>, but may touch two points of <math>S</math> at a time. Say the line sequentially touches the points <math>S_1,S_2,...,S_n,S_1</math>. We will now prove <math>S_1,...,S_n</math> form a regular polygon.
 +
 +
Identify <math>S_0</math> with <math>S_n</math>, <math>S_{n+1}</math> with <math>S_1</math>, and <math>S_{n+2}</math> with <math>S_2</math>.
 +
 +
Considering the axis of symmetry between <math>S_2</math> and <math>S_3</math>, a line which rolls counterclockwise around <math>S</math> with starting position passing through <math>S_2</math> and <math>S_3</math> will roll symmetrically to a line which rolls clockwise around <math>S</math> with starting position passing through <math>S_2</math> and <math>S_3</math>. Let <math>l</math> be the perpendicular bisector of <math>S_2S_3</math>. There are two possibilities: <math>S_1</math> lies on the same side of <math>l</math> as <math>S_2</math> or <math>S_1</math> lies on <math>l</math>. In the latter case apparently <math>S_4 = S_1</math>. In any event, <math>S_1</math> has the property all points of <math>S</math> are on one side of the line passing through <math>S_1</math> and <math>S_2</math>, and this is not true for any other point on the same side of <math>l</math> as <math>S_2</math>. A similar statement holds for <math>S_4</math> and <math>S_3</math>. It follows <math>S_1</math> must be symmetric to <math>S_4</math> about <math>l</math>. So <math>\angle S_1S_2S_3 = \angle S_2S_3S_4</math>. Repeat for all other sequential <math>4</math> points of <math>O</math> to get
 +
<cmath>
 +
\angle S_{k-1}S_kS_{k+1} = \angle S_kS_{k+1}S_{k+2}
 +
</cmath>
 +
for all <math>k</math>. Now, if <math>S_2</math> didn't lie on the axis of symmetry between <math>S_1</math> and <math>S_3</math> there would exist another point <math>S_2'</math> symmetric to <math>S_2</math> about that axis, and as the line rolled around <math>S</math> one would find it sequentially touched <math>S_1,S_2',S_2,S_3</math> or <math>S_1,S_2,S_2',S_3</math>. A contradiction to show <math>S_1,...,S_n</math> were defined. So <math>S_2</math> lies on the axis of symmetry, thus <math>S_1S_2 = S_2S_3</math>. Repeat for all other sequential <math>3</math> points of <math>S_1,...,S_n</math> to get
 +
<cmath>
 +
S_{k-1}S_k = S_kS_{k+1}
 +
</cmath>
 +
for all <math>k</math>. We have shown <math>S_1,...,S_n</math> form a regular polygon.
 +
 +
Now, the regular polygon <math>S_1...S_n</math> has a well-defined center <math>O</math> and there is certainly not another isometric polygon among <math>S</math>. Therefore any axis of symmetry about which we can reflect <math>S</math> must reflect <math>O</math> into itself. i.e. the axis of symmetry must intersect <math>O</math>.
 +
 +
Let <math>A</math> be a point of <math>S</math>. We will show <math>A</math> is one of <math>S_1,...,S_n</math>. The axis of symmetry between <math>S_1</math> and <math>A</math> intersects <math>O</math> by the previous. So <math>S_1OA</math> is an isosceles triangle with <math>OS_1 = OA</math>. Therefore <math>A</math> lies on the circle with center <math>O</math> and radius <math>OS_1</math>. Note all points <math>S_1,...,S_n</math> lie on this circle. If <math>A</math> were not one of <math>S_1,...,S_n</math>, we could suppose <math>A</math> lies between <math>S_k</math> and <math>S_{k+1}</math> on this circle, whereas all points of <math>S</math> lie on one side of the line passing through <math>S_k</math> and <math>S_{k+1}</math>. A contradiction. So <math>A</math> is one of <math>S_1,...,S_n</math>. So <math>S</math> is the regular polygon <math>S_1...S_n</math>.
 +
 +
~not_detriti
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 16:34, 21 July 2025

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.

Let a line roll/rotate around the perimeter of $S$ such that at any time all points of $S$ are on one side of the line and the line is always touching one point of $S$, but may touch two points of $S$ at a time. Say the line sequentially touches the points $S_1,S_2,...,S_n,S_1$. 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 $S_3$, a line which rolls counterclockwise around $S$ with starting position passing through $S_2$ and $S_3$ will roll symmetrically to a line which rolls clockwise around $S$ with starting position passing through $S_2$ and $S_3$. Let $l$ be the perpendicular bisector of $S_2S_3$. There are two possibilities: $S_1$ lies on the same side of $l$ as $S_2$ or $S_1$ lies on $l$. In the latter case apparently $S_4 = S_1$. In any event, $S_1$ has the property all points of $S$ are on one side of the line passing through $S_1$ and $S_2$, and this is not true for any other point on the same side of $l$ as $S_2$. A similar statement holds for $S_4$ and $S_3$. It follows $S_1$ must be symmetric to $S_4$ about $l$. 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 as the line rolled around $S$ one would find it sequentially touched $S_1,S_2',S_2,S_3$ or $S_1,S_2,S_2',S_3$. A contradiction to show $S_1,...,S_n$ were defined. So $S_2$ lies on the axis of symmetry, thus $S_1S_2 = S_2S_3$. Repeat for all other sequential $3$ points of $S_1,...,S_n$ to get \[S_{k-1}S_k = S_kS_{k+1}\] for all $k$. We have shown $S_1,...,S_n$ form a regular polygon.

Now, the regular polygon $S_1...S_n$ has a well-defined center $O$ and there is certainly not another isometric polygon among $S$. Therefore any axis of symmetry about which we can reflect $S$ must reflect $O$ into itself. i.e. the axis of symmetry must intersect $O$.

Let $A$ be a point of $S$. We will show $A$ is one of $S_1,...,S_n$. The axis of symmetry between $S_1$ and $A$ intersects $O$ by the previous. So $S_1OA$ is an isosceles triangle with $OS_1 = OA$. Therefore $A$ lies on the circle with center $O$ and radius $OS_1$. Note all points $S_1,...,S_n$ lie on this circle. 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, whereas all points of $S$ lie on one side of the line passing through $S_k$ and $S_{k+1}$. A contradiction. So $A$ is one of $S_1,...,S_n$. So $S$ is the regular polygon $S_1...S_n$.

~not_detriti

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


See Also

1999 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_IMO_Problems/Problem_1&oldid=254928"