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Difference between revisions of "2006 AIME A Problems/Problem 10"

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== Problem ==
 
== Problem ==
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Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a <math> 50\% </math> chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team <math> A </math> beats team <math> B. </math> The probability that team <math> A </math> finishes with more points than team <math> B </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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Eight circles of diameter 1 are packed in the first quadrant of the coordinte plane as shown. Let region <math> \mathcal{R} </math> be the union of the eight circular regions. Line <math> l, </math> with slope 3, divides <math> \mathcal{R} </math> into two regions of equal area. Line <math> l </math>'s equation can be expressed in the form <math> ax=by+c, </math> where <math> a, b, </math> and <math> c </math> are positive integers whose greatest common divisor is 1. Find <math> a^2+b^2+c^2. </math>  
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[[Image:2006AimeI10.PNG]]
  
 
== Solution ==
 
== Solution ==

Revision as of 14:41, 25 September 2007

Problem

Eight circles of diameter 1 are packed in the first quadrant of the coordinte plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$

2006AimeI10.PNG

Solution

You can break this into cases based on how many rounds A wins out of the remaining 5 games.

If A wins 0 games, then B must win 0 games and the probability of this is $\frac{{0 \choose 5}}{2^5} \frac{{0 \choose 5}}{2^5} = \frac{1}{1024}$.

If A wins 1 games, then B must win 1 or less games and the probability of this is $\frac{{1 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}}{2^5} = \frac{5}{1024}$.

If A wins 2 games, then B must win 2 or less games and the probability of this is $\frac{{2 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}}{2^5} = \frac{160}{1024}$.

If A wins 3 games, then B must win 3 or less games and the probability of this is $\frac{{3 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}}{2^5} = \frac{260}{1024}$.

If A wins 4 games, then B must win 4 or less games and the probability of this is $\frac{{4 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}+{4 \choose 5}}{2^5} = \frac{155}{1024}$.

If A wins 5 games, then B must win 5 or less games and the probability of this is $\frac{{5 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}+{4 \choose 5}+{5 \choose 5}}{2^5} = \frac{32}{1024}$.

Summing these 6 cases, we get $\frac{638}{1024}$, which simplifies to $\frac{319}{512}$, so out answer is $319 + 512 = 831$.

See also

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