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

(Created page with '== Problem 13 == 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…')
 
(→‎Solution)
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Thus, <math>p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}</math>
 
Thus, <math>p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}</math>
  
 +
<br/>
  
−
<br/> After this, we have to "guesstimate" <math>a</math>.
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We can look at <math>p(a)</math> as <math>p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}</math>
 +
 
 +
So we need <math>(43-a)(42-a)+(a-1)(a-2)\ge (1225)</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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<math>b^2\ge \frac{385}{2} = 192.5 >13^2</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>
  
 
And we'll find <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>
 
And we'll find <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>

Revision as of 21:28, 3 April 2010

Problem 13

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 $a=22+b$

$(43-a)(42-a)+(a-1)(a-2)=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225)$

$b^2\ge \frac{385}{2} = 192.5 >13^2$

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

And we'll find $p(8) = \frac{616}{1225} = \frac{88}{175}$

Answer: $\boxed{263}$

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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