Difference between revisions of "2012 USAMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
| + | If you rotate the red points 431 times, they will overlap with blue points <math>108\times 108</math> times, for an average of <math>\frac{108\times 108}{431}</math> per rotation. Note that this average is slightly greater than 27. Therefore at some point 28 red points overlap with blue points. In other words, there exist 28 red and blue points such that the convex 28-gons formed by them are congruent. | ||
| − | ==See | + | Rotate these 28 red points 431 times. They will overlap with green points <math>108\times 28</math> times, for an average of <math>\frac{108\times 28}{431}</math> per rotation. This average is slightly greater than 7, so at some point 8 of those red points overlap with green points. In other words, there exist 8 red points, 8 blue points, and 8 green points such that the convex octagons formed by them are congruent. |
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| + | Rotate these 8 red points 431 times. They will overlap with yellow points <math>108\times 8</math> times, for an average of <math>\frac{108\cdot 8}{431}</math> per rotation. This average is slightly greater than 2, so at some point 3 of those red points will overlap with yellow points. In other words, there exist 3 red points, 3 blue points, 3 green points, and 3 yellow points such that the triangles formed by them are congruent. | ||
| + | |||
| + | ==Note== | ||
| + | Huge bash of Pigeonhole Principle: | ||
| + | |||
| + | <cmath>108\longrightarrow 28\longrightarrow 8\longrightarrow 3</cmath> | ||
| + | |||
| + | <math>\left \lceil{\left \lceil{\left \lceil{108\cdot \frac{108}{431}}\right \rceil \cdot \frac{108}{431}}\right \rceil \cdot \frac{108}{431}}\right \rceil =\left \lceil{\left \lceil{28\cdot \frac{108}{431}}\right \rceil \cdot \frac{108}{431}}\right \rceil =\left \lceil{8\cdot \frac{108}{431}}\right \rceil =3</math> | ||
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| + | ==See Also== | ||
*[[USAMO Problems and Solutions]] | *[[USAMO Problems and Solutions]] | ||
{{USAMO newbox|year=2012|num-b=1|num-a=3}} | {{USAMO newbox|year=2012|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Olympiad Combinatorics Problems]] | ||
Latest revision as of 19:53, 17 April 2016
Contents
Problem
A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors such that some 108 points are colored Red, some 108 points are colored Green, some 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.
Solution
If you rotate the red points 431 times, they will overlap with blue points 
times, for an average of 
per rotation. Note that this average is slightly greater than 27. Therefore at some point 28 red points overlap with blue points. In other words, there exist 28 red and blue points such that the convex 28-gons formed by them are congruent.
Rotate these 28 red points 431 times. They will overlap with green points 
times, for an average of 
per rotation. This average is slightly greater than 7, so at some point 8 of those red points overlap with green points. In other words, there exist 8 red points, 8 blue points, and 8 green points such that the convex octagons formed by them are congruent.
Rotate these 8 red points 431 times. They will overlap with yellow points 
times, for an average of 
per rotation. This average is slightly greater than 2, so at some point 3 of those red points will overlap with yellow points. In other words, there exist 3 red points, 3 blue points, 3 green points, and 3 yellow points such that the triangles formed by them are congruent.
Note
Huge bash of Pigeonhole Principle:
![\[108\longrightarrow 28\longrightarrow 8\longrightarrow 3\]](http://latex.artofproblemsolving.com/4/f/b/4fb909cdf09dd8c9d7370ffadd702a74c7d52191.png)
See Also
| 2012 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
