Difference between revisions of "1999 AIME Problems/Problem 8"

 
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== Problem ==
 
== Problem ==
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Let <math>\displaystyle \mathcal{T}</math> be the set of ordered triples <math>\displaystyle (x,y,z)</math> of nonnegative real numbers that lie in the plane <math>\displaystyle x+y+z=1.</math>  Let us say that <math>\displaystyle (x,y,z)</math> supports <math>\displaystyle (a,b,c)</math> when exactly two of the following are true: <math>\displaystyle x\ge a, y\ge b, z\ge c.</math>  Let <math>\displaystyle \mathcal{S}</math> consist of those triples in <math>\displaystyle \mathcal{T}</math> that support <math>\displaystyle \left(\frac 12,\frac 13,\frac 16\right).</math>  The area of <math>\displaystyle \mathcal{S}</math> divided by the area of <math>\displaystyle \mathcal{T}</math> is <math>\displaystyle m/n,</math>  where <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math>
  
 
== Solution ==
 
== Solution ==
  
 
== See also ==
 
== See also ==
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* [[1999_AIME_Problems/Problem_7|Previous Problem]]
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* [[1999_AIME_Problems/Problem_9|Next Problem]]
 
* [[1999 AIME Problems]]
 
* [[1999 AIME Problems]]

Revision as of 01:53, 22 January 2007

Problem

Let $\displaystyle \mathcal{T}$ be the set of ordered triples $\displaystyle (x,y,z)$ of nonnegative real numbers that lie in the plane $\displaystyle x+y+z=1.$ Let us say that $\displaystyle (x,y,z)$ supports $\displaystyle (a,b,c)$ when exactly two of the following are true: $\displaystyle x\ge a, y\ge b, z\ge c.$ Let $\displaystyle \mathcal{S}$ consist of those triples in $\displaystyle \mathcal{T}$ that support $\displaystyle \left(\frac 12,\frac 13,\frac 16\right).$ The area of $\displaystyle \mathcal{S}$ divided by the area of $\displaystyle \mathcal{T}$ is $\displaystyle m/n,$ where $\displaystyle m_{}$ and $\displaystyle n_{}$ are relatively prime positive integers, find $\displaystyle m+n.$

Solution

See also