AoPSWiki

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

Personal tools

Search

Toolbox

2001 AIME I Problems/Problem 10

From AoPSWiki

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 • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2001_AIME_I_Problems/Problem_10"

Category: Intermediate Combinatorics Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us