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Difference between revisions of "2010 AIME II Problems/Problem 13"

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== Problem 13 ==
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== Problem ==
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The <math>52</math> cards in a deck are numbered <math>1, 2, \cdots, 52</math>. 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 <math>p(a)</math> be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards <math>a</math> and <math>a+9</math>, and Dylan picks the other of these two cards. The minimum value of <math>p(a)</math> for which <math>p(a)\ge\frac{1}{2}</math> can be written as <math>\frac{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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The <math>52</math> cards in a deck are numbered <math>1, 2, \cdots, 52</math>. 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 <math>p(a)</math> be the [[probability]] that Alex and Dylan are on the same team, given that Alex picks one of the cards <math>a</math> and <math>a+9</math>, and Dylan picks the other of these two cards. The minimum value of <math>p(a)</math> for which <math>p(a)\ge\frac{1}{2}</math> can be written as <math>\frac{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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== Solution ==
 
== Solution ==
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Clearly <math>p(a)</math> is a [[quadratic]] centered at <math>a=22</math>.  
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Clearly <math>p(a)</math> is a quadratic centered at <math>a=22</math>.  
 
  
 
Once the two cards are drawn, there are <math>\dbinom{50}{2} = 1225</math> ways for the other two people to draw.  
 
Once the two cards are drawn, there are <math>\dbinom{50}{2} = 1225</math> ways for the other two people to draw.  
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Let <math>a=22+b</math>
 
Let <math>a=22+b</math>
  
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<math>(43-a)(42-a)+(a-1)(a-2)=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225)</math>
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<cmath>\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) \\
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b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}</cmath>
  
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<math>b^2\ge \frac{385}{2} = 192.5 >13^2</math>
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So <math>b&> 13</math> or <math>b< -13</math>, and <math>a=22+b<9</math> or <math>a>35</math>, so <math>a=8</math> or <math>a=36</math>.
  
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<math>b> 13</math> or <math>b< -13</math>, So <math>a=22+b<9</math> or <math>a>35</math>, so <math>a=8</math> or <math>a=36</math>
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Thus, <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>, and the answer is <math>\boxed{263}</math>.
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And we'll find <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>
 
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Answer: <math>\boxed{263}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=12|num-a=14|n=II}}
 
{{AIME box|year=2010|num-b=12|num-a=14|n=II}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 11:02, 6 April 2010

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 $b&> 13$ (Error compiling LaTeX. Unknown error_msg) 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 $\boxed{263}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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