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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 10]] |
| − | 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]]
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| − | == Solution ==
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| − | You can break this into cases based on how many rounds A wins out of the remaining 5 games.
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| − | If A wins 0 games, then B must win 0 games and the probability of this is <math> \frac{{0 \choose 5}}{2^5} \frac{{0 \choose 5}}{2^5} = \frac{1}{1024} </math>.
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| − | If A wins 1 games, then B must win 1 or less games and the probability of this is <math> \frac{{1 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}}{2^5} = \frac{5}{1024} </math>.
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| − | If A wins 2 games, then B must win 2 or less games and the probability of this is <math> \frac{{2 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}}{2^5} = \frac{160}{1024} </math>.
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| − | If A wins 3 games, then B must win 3 or less games and the probability of this is <math> \frac{{3 \choose 5}}{2^5} \frac{{0 \choose 5}+{1 \choose 5}+{2 \choose 5}+{3 \choose 5}}{2^5} = \frac{260}{1024} </math>.
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| − | If A wins 4 games, then B must win 4 or less games and the probability of this is <math> \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} </math>.
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| − | If A wins 5 games, then B must win 5 or less games and the probability of this is <math> \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} </math>.
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| − | Summing these 6 cases, we get <math> \frac{638}{1024} </math>, which simplifies to <math> \frac{319}{512} </math>, so out answer is <math>319 + 512 = 831</math>.
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| − | == See also ==
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| − | *[[2006 AIME II Problems/Problem 9 | Previous problem]]
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| − | *[[2006 AIME II Problems/Problem 11 | Next problem]]
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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Combinatorics Problems]]
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