Difference between revisions of "1973 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | There are <math>\binom{2n+1}{3}</math> ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some <math>n</math> consecutive vertices of the polygon. | + | There are <math>\binom{2n+1}{3}</math> ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some <math>n+1</math> consecutive vertices of the polygon. |
− | We will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily (<math>2n+1</math> possibilities). Once we pick it, we have to pick <math>2</math> out of the next <math>n | + | We will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily (<math>2n+1</math> possibilities). Once we pick it, we have to pick <math>2</math> out of the next <math>n</math> vertices (<math>\binom{n}{2}</math> possibilities). |
Then the probability that our triangle does NOT contain the center is | Then the probability that our triangle does NOT contain the center is | ||
Line 10: | Line 10: | ||
p | p | ||
= | = | ||
− | \frac{ (2n+1){\binom{n | + | \frac{ (2n+1){\binom{n}{2}} }{ {\binom{2n+1}{3} } } |
= | = | ||
− | \frac{ (1/2)(2n+1)(n | + | \frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } |
= | = | ||
− | \frac{ 3(n | + | \frac{ 3(n)(n-1) }{ (2n)(2n-1) } |
</cmath> | </cmath> | ||
Line 21: | Line 21: | ||
1-p | 1-p | ||
= | = | ||
− | \frac{ (2n)(2n-1) - 3(n | + | \frac{ (2n)(2n-1) - 3(n)(n-1) }{ (2n)(2n-1) } |
= | = | ||
− | + | \frac{ n^2+n }{ 4n^2 - 2n } | |
+ | = | ||
+ | \boxed{\frac{n+1}{4n-2}} | ||
</cmath> | </cmath> | ||
− | |||
− | ==See | + | |
+ | |||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | ==See Also== | ||
{{USAMO box|year=1973|num-b=2|num-a=4}} | {{USAMO box|year=1973|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Combinatorics Problems]] | [[Category:Olympiad Combinatorics Problems]] |
Latest revision as of 18:54, 3 July 2013
Problem
Three distinct vertices are chosen at random from the vertices of a given regular polygon of sides. If all such choices are equally likely, what is the probability that the center of the given polygon lies in the interior of the triangle determined by the three chosen random points?
Solution
There are ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some consecutive vertices of the polygon. We will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily ( possibilities). Once we pick it, we have to pick out of the next vertices ( possibilities).
Then the probability that our triangle does NOT contain the center is
And then the probability we seek is
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1973 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.