Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
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. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below <math>a</math>, which occurs in <math>\dbinom{a-1}{2}</math> ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above <math>a+9</math>, which occurs in <math>\dbinom{43-a}{2}</math> ways. Thus, <cmath>p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.</cmath> Simplifying, we get <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>. If <math>a=22+b</math>, then <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) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}</cmath> 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>. Thus, <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>, and the answer is <math>88+175 = \boxed{263}</math>.
−
 
 
−
Once the two cards are drawn, there are <math>\dbinom{50}{2} = 1225</math> ways for the other two people to draw.  
 
−
 
 
−
Alex and Dylan are the team with higher numbers if Blair and Corey both draw below <math>a</math>, which occurs in <math>\dbinom{a-1}{2}</math> ways.
 
−
 
 
−
Alex and Dylan are the team with lower numbers if Blair and Corey both draw above <math>a+9</math>, which occurs in <math>\dbinom{43-a}{2}</math> ways.  
 
−
 
 
−
Thus, <math>p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}</math>
 
−
 
 
−
<br/>
 
−
 
 
−
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>
 
−
 
 
−
<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) \\
 
−
b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}</cmath>
 
−
 
 
−
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>.
 
−
 
 
−
Thus, <math>p(8) = \frac{616}{1225} = \frac{88}{175}</math>, and the answer is <math>\boxed{263}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 14:37, 26 February 2016

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

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, \[p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.\] Simplifying, we get $p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}$, so we need $(43-a)(42-a)+(a-1)(a-2)\ge (1225)$. If $a=22+b$, then \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 $88+175 = \boxed{263}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_13&oldid=77066"