2006 AIME A Problems/Problem 10

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 $a^2+b^2+c^2.$

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 $319 + 512 = 831$ .

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2006 AIME A Problems/Problem 10

Problem

Solution

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