Difference between revisions of "1999 IMO Problems/Problem 1"
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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> | 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 | + | '''CASE I:''' <math>a+b</math> is even |
<math>k=\frac{a+b}{2}</math> and <math>k</math> is integer | <math>k=\frac{a+b}{2}</math> and <math>k</math> is integer | ||
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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> | 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. | + | 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 | + | '''CASE II:''' <math>a+b</math> is odd |
<math>k=\frac{a+b+1}{2}</math> and <math>k</math> is integer | <math>k=\frac{a+b+1}{2}</math> and <math>k</math> is integer | ||
Line 65: | Line 65: | ||
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> | 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}</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 | 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 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 | and | ||
− | <math>\angle P_{ | + | <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 | |
− | |||
{{alternate solutions}} | {{alternate solutions}} | ||
+ | |||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1999|before=First Question|num-a=2}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 02:21, 21 November 2023
Problem
Determine all finite sets of at least three points in the plane which satisfy the following condition:
For any two distinct points and
in
, the perpendicular bisector of the line segment
is an axis of symmetry of
.
Solution
Upon reading this problem and drawing some points, one quickly realizes that the set consists of all the vertices of any regular polygon.
Now to prove it with some numbers:
Let , with
, where
is a vertex of a polygon which we can define their
coordinates as:
for
.
That defines the vertices of any regular polygon with being the radius of the circumcircle of the regular
-sided polygon.
Now we can pick any points and
of the set as:
and
, where
;
; and
Then,
and
Let be point
which is not part of
Then, , and
The perpendicular bisector of passes through
.
Let point , not in
be a point that passes through the perpendicular bisector of
at a distance
from
Then, and
CASE I: is even
and
is integer
Then
This means that the perpendicular bisector also passes through a point of
Let be any positive integer
and
Therefore, for any integer
.
Also, since for any integer
then this proves that the bisector of any points and
is an axis of symmetry for this case.
CASE II: is odd
and
is integer
and
is integer
Then
This means that the perpendicular bisector does not pass through any point of , but their closest points are
and
and that
Let be any positive integer
and
Therefore, for any integer
.
Since ,
Also, since for any integer
then this proves that the bisector of any points and
is an axis of symmetry for this case.
Having proven both cases, then the set 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 |