Difference between revisions of "1999 IMO Problems/Problem 1"

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.

@@ Line 6: / Line 6: @@
 ==Solution==
-{{solution}}
+Upon reading this problem and drawing some points, one quickly realizes that the set <math>S</math> consists of all the vertices of any regular polygon.
+Now to prove it with some numbers:
+Let <math>S=\left\{ P_{0},P_{1},P_{2},...,P_{n-1} \right\}</math>, with <math>n\ge 3</math>, where <math>P_{i}</math> is a vertex of a polygon which we can define their <math>xy</math> coordinates as: <math>P_{i}=\left\langle Rcos\left( \frac{2\pi}{n}i \right),Rsin\left( \frac{2\pi}{n}i \right) \right\rangle</math> for <math>i=0,1,2,...,(n-1)</math>.
+That defines the vertices of any regular polygon with <math>R</math> being the radius of the circumcircle of the regular <math>n</math>-sided polygon.
+Now we can pick any points <math>A</math> and <math>B</math> of the set as:
+<math>A=P_{a}</math> and <math>B=P_{b}</math>, where <math>a=0,1,2,...,(n-1)</math>; <math>b=0,1,2,...,(n-1)</math>; and <math>a\ne b</math>
+Then,
+<math>A=\left\langle Rcos\left( \frac{2\pi}{n}a \right),Rsin\left( \frac{2\pi}{n}a \right) \right\rangle</math>
+and <math>B=\left\langle Rcos\left( \frac{2\pi}{n}b \right),Rsin\left( \frac{2\pi}{n}b \right) \right\rangle</math>
+Let <math>O</math> be point <math>(0,0)</math> which is not part of <math>S</math>
+Then, <math>\angle P_{0}OA = \frac{2\pi}{n}a</math>, and <math>\angle P_{0}OB = \frac{2\pi}{n}b</math>
+The perpendicular bisector of <math>AB</math> passes through <math>O</math>.
+Let point <math>M_{AB}</math>, not in <math>S</math> be a point that passes through the perpendicular bisector of <math>AB</math> at a distance <math>R</math> from <math>O</math>
+Then, <math>\angle P_{0}OM_{AB} =\frac{2\pi}{n}\frac{a+b}{2}</math> and <math>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</math>
+'''CASE I:''' <math>a+b</math> is even
+<math>k=\frac{a+b}{2}</math> and <math>k</math> is integer
+Then <math>M_{AB}=\left\langle Rcos\left( \frac{2\pi}{n}k \right),Rsin\left( \frac{2\pi}{n}k \right) \right\rangle=P_{k}</math>
+This means that the perpendicular bisector also passes through a point <math>P_{k}</math> of <math>S</math>
+Let <math>c</math> be any positive integer
+<math>\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)</math>
+and
+<math>\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)</math>
+Therefore, <math>\angle P_{k}OP_{(k+c)\; mod\; n}=\angle P_{k}OP_{(k-c)\; mod\; n}</math> for any integer <math>c</math>.
+Also, since <math>\left| OP_{(k+c)\; mod\; n} \right|=\left| OP_{(k-c)\; mod\; n} \right|=R</math> for any integer <math>c</math>
+then this proves that the bisector of any points <math>A</math> and <math>B</math> is an axis of symmetry for this case.
+'''CASE II:''' <math>a+b</math> is odd
+<math>k=\frac{a+b+1}{2}</math> and <math>k</math> is integer
+<math>m=\frac{a+b-1}{2}</math> and <math>m</math> is integer
+Then <math>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</math>
+This means that the perpendicular bisector does not pass through any point of <math>S</math>, but their closest points are <math>P_{k}</math> and <math>P_{m}</math> and that <math>\angle MOP_{k}=\angle MOP_{m}=\frac{\pi}{n}</math>
+Let <math>c</math> be any positive integer
+<math>\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)</math>
+<math>\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)</math>
+and
+<math>\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)</math>
+<math>\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)</math>
+Therefore, <math>\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{(m-c)\; mod\; n}</math> for any integer <math>c</math>.
+Since <math>m=k-1</math>, <math>\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{(k-1-c)\; mod\; n}</math>
+Also, since <math>\left| OP_{(k+c)\; mod\; n} \right|=\left| OP_{(m-c)\; mod\; n} \right|=R</math> for any integer <math>c</math>
+then this proves that the bisector of any points <math>A</math> and <math>B</math> is an axis of symmetry for this case.
+Having proven both cases, then the set <math>S</math> 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 <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}}
+==See Also==
+{{IMO box|year=1999|before=First Question|num-a=2}}
+[[Category:Olympiad Geometry Problems]]
+[[Category:3D Geometry Problems]]

1999 IMO (Problems) • Resources
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Difference between revisions of "1999 IMO Problems/Problem 1"

Latest revision as of 16:34, 21 July 2025

Contents

Problem

Solution

Solution 2

See Also