Difference between revisions of "1973 USAMO Problems/Problem 3"
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\frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } | \frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } | ||
= | = | ||
| − | \frac{ 3 | + | \frac{ 3n-3 }{ 4n-2 } |
</cmath> | </cmath> | ||
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1-p | 1-p | ||
= | = | ||
| − | \frac{ ( | + | \frac{ (4n-2) - (3n-3) }{ (4n-2) } |
| − | |||
| − | |||
= | = | ||
\boxed{\frac{n+1}{4n-2}} | \boxed{\frac{n+1}{4n-2}} | ||
Latest revision as of 01:46, 27 December 2024
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
![\[p = \frac{ (2n+1){\binom{n}{2}} }{ {\binom{2n+1}{3} } } = \frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } = \frac{ 3n-3 }{ 4n-2 }\]](http://latex.artofproblemsolving.com/f/7/3/f7331e9240c0940d80927530d00b9b1986fbc194.png)
And then the probability we seek is
![\[1-p = \frac{ (4n-2) - (3n-3) }{ (4n-2) } = \boxed{\frac{n+1}{4n-2}}\]](http://latex.artofproblemsolving.com/4/1/3/413092f8529e371cf5c9ed28ea7b9dfacdbfc1d1.png)
Perplexing Edge Case
However, for 
, (a rare edge case), the circumcenter of the triangle must lie in the center of the triangle, which is notably difficult to prove if not assumed, resulting in a probability of 
.
Observe that the probability tends to 
as approaches infinity. Why does this occur?
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. 
