Difference between revisions of "1999 IMO Problems/Problem 1"
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<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> | <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> | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 19:41, 12 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
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.