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Difference between revisions of "2001 AIME I Problems/Problem 10"

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(solution (written by JesusFreak197))
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== Problem ==
 
== Problem ==
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Let <math>S</math> be the set of points whose coordinates <math>x,</math> <math>y,</math> and <math>z</math> are integers that satisfy <math>0\le x\le2,</math> <math>0\le y\le3,</math> and <math>0\le z\le4.</math>  Two distinct points are randomly chosen from <math>S.</math>  The probability that the midpoint of the segment they determine also belongs to <math>S</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
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Let <math>S</math> be the [[set]] of points whose [[coordinate]]s <math>x,</math> <math>y,</math> and <math>z</math> are integers that satisfy <math>0\le x\le2,</math> <math>0\le y\le3,</math> and <math>0\le z\le4.</math>  Two distinct points are randomly chosen from <math>S.</math>  The [[probability]] that the [[midpoint]] of the segment they determine also belongs to <math>S</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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The distance between the <math>x</math>, <math>y</math>, and <math>z</math> coordinates must be even so that the midpoint can have integer coordinates. Therefore,
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*For <math>x</math>, we have the possibilities <math>(0,0)</math>, <math>(1,1)</math>, <math>(2,2)</math>, <math>(0,2)</math>, and <math>(2,0)</math>, <math>5</math> possibilities.
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*For <math>y</math>, we have the possibilities <math>(0,0)</math>, <math>(1,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, <math>(0,2)</math>, <math>(2,0)</math>, <math>(1,3)</math>, and <math>(3,1)</math>, <math>8</math> possibilities.
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*For <math>z</math>, we have the possibilities <math>(0,0)</math>, <math>(1,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, <math>(4,4)</math>, <math>(0,2)</math>, <math>(0,4)</math>, <math>(2,0)</math>, <math>(4,0)</math>, <math>(2,4)</math>, <math>(4,2)</math>, <math>(1,3)</math>, and <math>(3,1)</math>, <math>13</math> possibilities.
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However, we have <math>3\cdot 4\cdot 5 = 60</math> cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is <math>\frac {5\cdot 8\cdot 13 - 60}{60\cdot 59} = \frac {23}{177}\Longrightarrow m+n = \boxed{200}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2001|n=I|num-b=9|num-a=11}}
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 +
[[Category:Intermediate Combinatorics Problems]]

Revision as of 17:04, 12 June 2008

Problem

Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

The distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore,

  • For $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities.
  • For $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(0,2)$, $(2,0)$, $(1,3)$, and $(3,1)$, $8$ possibilities.
  • For $z$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(0,2)$, $(0,4)$, $(2,0)$, $(4,0)$, $(2,4)$, $(4,2)$, $(1,3)$, and $(3,1)$, $13$ possibilities.

However, we have $3\cdot 4\cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\frac {5\cdot 8\cdot 13 - 60}{60\cdot 59} = \frac {23}{177}\Longrightarrow m+n = \boxed{200}$.

See also

2001 AIME I (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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