1973 USAMO Problems/Problem 3

Revision as of 02:13, 20 August 2009 by God of Math (talk | contribs) (Solution)


Three distinct vertices are chosen at random from the vertices of a given regular polygon of $(2n+1)$ 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?


There are $\binom{2n+1}{3}$ 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 $n$ 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 ($2n+1$ possibilities). Once we pick it, we have to pick $2$ out of the next $n-1$ vertices ($\binom{n-1}{2}$ possibilities).

Then the probability that our triangle does NOT contain the center is \[p = \frac{ (2n+1){\binom{n-1}{2}} }{ {\binom{2n+1}{3} } } = \frac{ (1/2)(2n+1)(n-1)(n-2) }{ (1/6)(2n+1)(2n)(2n-1) } =  \frac{ 3(n-1)(n-2) }{ (2n)(2n-1) }\]

And then the probability we seek is \[1-p =  \frac{ (2n)(2n-1) - 3(n-1)(n-2) }{ (2n)(2n-1) } = \boxed{\frac{ n^2+7n-6 }{ 4n^2 - 2n }}\] Note: Trying the regular pentagon case, i.e n = 2, we get 12/12, or 1. The real probability is 1/2 for the pentagon, and the correct general formula is (n+1)/(4n-2).

See also

1973 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions
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