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"

(Remove extra problem section)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
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>.
+
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 form 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>.
  
 
== Solution ==
 
== Solution ==
 
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, <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>.
 +
 +
== Solution 2 ==
 +
 +
Given that Alex and Dylan hold the cards <math>a</math> and <math>a+9</math>, we need to calculate the probability that they end up on the same team. This happens in two scenarios:
 +
 +
1. Both on the Lower Team: This occurs if the other two cards drawn are both greater than <math>a+9</math>. 
 +
2. Both on the Higher Team: This occurs if the other two cards drawn are both less than <math>a</math>.
 +
 +
The total number of ways to choose the other two cards from the remaining 50 cards is <math>\binom{50}{2} = 1225</math>.
 +
 +
The number of favorable outcomes is the sum of: 
 +
The number of ways to choose 2 cards greater than <math>a+9</math>: <math>\binom{43-a}{2}</math> 
 +
The number of ways to choose 2 cards less than <math>a</math>: <math>\binom{a-1}{2}</math> 
 +
 +
Thus, the probability <math>p(a)</math> is:
 +
<cmath>
 +
p(a) = \frac{\binom{a-1}{2} + \binom{43-a}{2}}{\binom{50}{2}} = \frac{\frac{(a-1)(a-2)}{2} + \frac{(43-a)(42-a)}{2}}{1225}.
 +
</cmath>
 +
 +
Finding the Minimum <math>a</math> for <math>p(a) \geq \frac{1}{2}</math>
 +
 +
Solving the inequality:
 +
<cmath>
 +
\frac{\binom{a-1}{2} + \binom{43-a}{2}}{1225} \geq \frac{1}{2}
 +
</cmath>
 +
leads to:
 +
<cmath>
 +
2(a^2 - 44a + 291) \geq 0.
 +
</cmath>
 +
 +
This quadratic inequality is satisfied when <math>a \leq 8</math> or <math>a \geq 36</math>.
 +
 +
The minimum value of <math>p(a)</math> that satisfies <math>p(a) \geq \frac{1}{2}</math> occurs at <math>a = 8</math> (or symmetrically at <math>a = 36</math>), giving:
 +
<cmath>
 +
p(8) = \frac{616}{1225} = \frac{88}{175}
 +
</cmath>
 +
where <math>88</math> and <math>175</math> are coprime.
 +
 +
Final Answer
 +
<cmath>
 +
88 + 175 = 263
 +
</cmath>
 +
 +
Answer: <math>263</math>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
  
 
== See also ==
 
== See also ==

Latest revision as of 02:07, 15 November 2024

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 form 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 $88+175 = \boxed{263}$.

Solution 2

Given that Alex and Dylan hold the cards $a$ and $a+9$, we need to calculate the probability that they end up on the same team. This happens in two scenarios:

1. Both on the Lower Team: This occurs if the other two cards drawn are both greater than $a+9$. 2. Both on the Higher Team: This occurs if the other two cards drawn are both less than $a$.

The total number of ways to choose the other two cards from the remaining 50 cards is $\binom{50}{2} = 1225$.

The number of favorable outcomes is the sum of: The number of ways to choose 2 cards greater than $a+9$: $\binom{43-a}{2}$ The number of ways to choose 2 cards less than $a$: $\binom{a-1}{2}$

Thus, the probability $p(a)$ is: \[p(a) = \frac{\binom{a-1}{2} + \binom{43-a}{2}}{\binom{50}{2}} = \frac{\frac{(a-1)(a-2)}{2} + \frac{(43-a)(42-a)}{2}}{1225}.\]

Finding the Minimum $a$ for $p(a) \geq \frac{1}{2}$

Solving the inequality: \[\frac{\binom{a-1}{2} + \binom{43-a}{2}}{1225} \geq \frac{1}{2}\] leads to: \[2(a^2 - 44a + 291) \geq 0.\]

This quadratic inequality is satisfied when $a \leq 8$ or $a \geq 36$.

The minimum value of $p(a)$ that satisfies $p(a) \geq \frac{1}{2}$ occurs at $a = 8$ (or symmetrically at $a = 36$), giving: \[p(8) = \frac{616}{1225} = \frac{88}{175}\] where $88$ and $175$ are coprime.

Final Answer \[88 + 175 = 263\]

Answer: $263$

~Athmyx

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=234317"