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Difference between revisions of "1999 AIME Problems/Problem 10"

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== Problem ==
 
== Problem ==
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Ten [[point]]s in the plane are given, with no three [[collinear]].  Four distinct [[segment]]s 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 <math>\displaystyle m/n,</math> where <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>\displaystyle m+n.</math>
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Ten [[point]]s in the plane are given, with no three [[collinear]].  Four distinct [[segment]]s 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 <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s.  Find <math>m+n.</math>
  
 
== Solution ==
 
== Solution ==
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== See also ==
 
== See also ==
 
{{AIME box|year=1999|num-b=9|num-a=11}}
 
{{AIME box|year=1999|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 18:40, 4 July 2013

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 $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$

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 ${10\choose3}$ 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 $45 - 3 = 42$ segments remaining.

The total number of ways of picking four distinct segments is ${45\choose4}$. Thus, the requested probability is $\frac{{10\choose3} \cdot 42}{{45\choose4}} = \frac{10 \cdot 9 \cdot 8 \cdot 42 \cdot 4!}{45 \cdot 44 \cdot 43 \cdot 42 \cdot 3!} = \frac{16}{473}$. The solution is $m + n = 489$.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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