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

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Then, <math>\angle P_{0}OA = \frac{2\pi}{n}a</math>, and <math>\angle P_{0}OB = \frac{2\pi}{n}b</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>
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The perpendicular bisector of <math>AB</math> passes through <math>O</math>.
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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>
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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>
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'''CASE I:''' <math>a+b</math> is even
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<math>k=\frac{a+b}{2}</math> and <math>k</math> is integer
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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>
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This means that the perpendicular bisector also passes through a point <math>P_{k}</math> of <math>S</math>
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Let <math>c</math> be any positive integer
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<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>
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and
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<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>
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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>.
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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>
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then this proves that the bisector of any points <math>A</math> and <math>B</math> is an axis of symmetry for this case.
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'''CASE II:''' <math>a+b</math> is odd
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<math>k=\frac{a+b+1}{2}</math> and <math>k</math> is integer
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<math>m=\frac{a+b-1}{2}</math> and <math>m</math> is integer
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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>
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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>
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Let <math>c</math> be any positive integer
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<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>
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<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>
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and
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<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>
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<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>
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Therefore, <math>\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{(m-c)\; mod\; n}</math> for any integer <math>c</math>.
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Since <math>m=k-1</math>, <math>\angle MOP_{(k+c)\; mod\; n}=\angle MOP_{(k-1-c)\; mod\; n}</math>
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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>
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then this proves that the bisector of any points <math>A</math> and <math>B</math> is an axis of symmetry for this case.
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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.
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~Tomas Diaz. orders@tomasdiaz.com
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{{alternate solutions}}
 
{{alternate solutions}}
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==See Also==
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{{IMO box|year=1999|before=First Question|num-a=2}}
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[[Category:Olympiad Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Latest revision as of 01:21, 21 November 2023

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


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
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