Difference between revisions of "1973 USAMO Problems/Problem 3"
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Revision as of 07:26, 16 September 2012
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
![\[p = \frac{ (2n+1){\binom{n}{2}} }{ {\binom{2n+1}{3} } } = \frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } = \frac{ 3(n)(n-1) }{ (2n)(2n-1) }\]](http://latex.artofproblemsolving.com/7/f/9/7f94c400f035b098fcc4a0bef1f76a0a5be343c3.png)
And then the probability we seek is
![\[1-p = \frac{ (2n)(2n-1) - 3(n)(n-1) }{ (2n)(2n-1) } = \frac{ n^2+n }{ 4n^2 - 2n } = \boxed{\frac{n+1}{4n-2}}\]](http://latex.artofproblemsolving.com/3/4/0/34066a7471cbca87f07575f6ac6215708f89ad61.png)
See Also
| 1973 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
