1999 AIME Problems/Problem 10
From AoPSWiki
Problem
Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is 
where 
and 
are relatively prime positive integers. Find 
Solution
First, let us find the number of triangles that can be formed from the 10 points. Since none of the points are collinear, it is possible to pick 
sets of 3 points which form triangles. However, a fourth distinct segment must also be picked. Since the triangle accounts for 3 segments, there are 
segments remaining.
The total number of ways of picking four distinct segments is 
. Thus, the requested probability is 
. The solution is 
.
See also
| 1999 AIME (Problems • Resources) | ||
| Preceded by Problem 9 | Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
