2010 AIME II Problems/Problem 13

Problem

The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards from a team, and the two persons with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$ , and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$ . where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Clearly $p(a)$ is a quadratic centered at $a=22$ .

Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw.

Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$ , which occurs in $\dbinom{a-1}{2}$ ways.

Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$ , which occurs in $\dbinom{43-a}{2}$ ways.

Thus, $p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}$

We can look at $p(a)$ as $p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}$

So we need $(43-a)(42-a)+(a-1)(a-2)\ge (1225)$

Let $a=22+b$

$\begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}$

So $b> 13$ or $b< -13$ , and $a=22+b<9$ or $a>35$ , so $a=8$ or $a=36$ .

Thus, $p(8) = \frac{616}{1225} = \frac{88}{175}$ , and the answer is $\boxed{263}$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2010 AIME II Problems/Problem 13

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