Difference between revisions of "1999 AIME Problems/Problem 10"
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The total number of ways of picking four distinct segments is <math>{45\choose4}</math>. Thus, the requested probability is <math>\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}</math>. The solution is <math>m + n = 489</math>. | The total number of ways of picking four distinct segments is <math>{45\choose4}</math>. Thus, the requested probability is <math>\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}</math>. The solution is <math>m + n = 489</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Note that 4 points can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the endpoints of the first one, then the one segment that completed the triangle. | ||
+ | Note that there are <math>(9 - 1) \times 2 = 16</math> segments that share an endpoint with the first segment. The answer is then $4 \times \frac{16}{44} \times \frac{1}{43} = \frac{16}{11 \times 43} = \frac{16}{473} \implies m + n = \boxed{489} | ||
+ | -whatRthose | ||
== See also == | == See also == | ||
{{AIME box|year=1999|num-b=9|num-a=11}} | {{AIME box|year=1999|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:34, 13 June 2019
Contents
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
.
Solution 2
Note that 4 points can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the endpoints of the first one, then the one segment that completed the triangle.
Note that there are segments that share an endpoint with the first segment. The answer is then $4 \times \frac{16}{44} \times \frac{1}{43} = \frac{16}{11 \times 43} = \frac{16}{473} \implies m + n = \boxed{489}
-whatRthose
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.