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 "2011 AIME II Problems/Problem 15"

(→‎Solution)
m (→‎See also)
Line 88: Line 88:
  
 
==See also==
 
==See also==
−
{{AIME box | year = 2011 | n = II | num-b=13 | num-a=15}}
+
{{AIME box | year = 2011 | n = II | num-b=14 | Last problem}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:15, 17 March 2019

Problem

Let $P(x) = x^2 - 3x - 9$. A real number $x$ is chosen at random from the interval $5 \le x \le 15$. The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$, $b$, $c$, $d$, and $e$ are positive integers. Find $a + b + c + d + e$.

Solution 1

Table of values of $P(x)$:

\begin{align*} P(5) &= 1 \\ P(6) &= 9 \\ P(7) &= 19 \\ P(8) &= 31 \\ P(9) &= 45 \\ P(10) &= 61 \\ P(11) &= 79 \\ P(12) &= 99 \\ P(13) &= 121 \\ P(14) &= 145 \\ P(15) &= 171 \\ \end{align*}

In order for $\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}$ to hold, $\sqrt{P(\lfloor x \rfloor)}$ must be an integer and hence $P(\lfloor x \rfloor)$ must be a perfect square. This limits $x$ to $5 \le x < 6$ or $6 \le x < 7$ or $13 \le x < 14$ since, from the table above, those are the only values of $x$ for which $P(\lfloor x \rfloor)$ is an perfect square. However, in order for $\sqrt{P(x)}$ to be rounded down to $P(\lfloor x \rfloor)$, $P(x)$ must be less than the next perfect square after $P(\lfloor x \rfloor)$ (for the said intervals). Now, we consider the three cases:


Case $5 \le x < 6$:

$P(x)$ must be less than the first perfect square after $1$, which is $4$, i.e.:

$1 \le P(x) < 4$ (because $\lfloor \sqrt{P(x)} \rfloor = 1$ implies $1 \le \sqrt{P(x)} < 2$)

Since $P(x)$ is increasing for $x \ge 5$, we just need to find the value $v \ge 5$ where $P(v) = 4$, which will give us the working range $5 \le x < v$.

\begin{align*} v^2 - 3v - 9 &= 4 \\ v &= \frac{3 + \sqrt{61}}{2} \end{align*}

So in this case, the only values that will work are $5 \le x < \frac{3 + \sqrt{61}}{2}$.

Case $6 \le x < 7$:

$P(x)$ must be less than the first perfect square after $9$, which is $16$.

\begin{align*} v^2 - 3v - 9 &= 16 \\ v &= \frac{3 + \sqrt{109}}{2} \end{align*}

So in this case, the only values that will work are $6 \le x < \frac{3 + \sqrt{109}}{2}$.

Case $13 \le x < 14$:

$P(x)$ must be less than the first perfect square after $121$, which is $144$.

\begin{align*} v^2 - 3v - 9 &= 144 \\ v &= \frac{3 + \sqrt{621}}{2} \end{align*}

So in this case, the only values that will work are $13 \le x < \frac{3 + \sqrt{621}}{2}$.

Now, we find the length of the working intervals and divide it by the length of the total interval, $15 - 5 = 10$:

\begin{align*} \frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} \\ &= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20} \end{align*}

Thus, the answer is $61 + 109 + 621 + 39 + 20 = \fbox{850}$.

P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.

Solution 2

Make the substitution $y=2x-3$, so $P(x)=\frac{y^2-45}{4}.$ We're looking for solutions to \[\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}\]with the new bounds $y\in{[7,27]}$. Since the left side is an integer, it must be that $\frac{\lfloor{y\rfloor}^2-45}{4}$ is a perfect square. For simplicity, write $\lfloor{y\rfloor}=a$ and \[a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.\]Since $a-2b<a+2b$, it must be that $(a-2b,a+2b)=(1,45),(3,15),(5,9)$, which gives solutions $(23,11),(9,3),(7,1)$, respectively. But this gives us three cases to check:

Case 1: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11$.

In this case, we have \[11\leq{\sqrt{\frac{y^2-45}{4}}}<12\implies{y\in{[23,\sqrt{621})}}.\] Case 2: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3$.

In this case, we have \[3\leq{\sqrt{\frac{y^2-45}{4}}}<4\implies{y\in{[9,\sqrt{109})}}.\] Case 3: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1$

In this case, we have \[1\leq{\sqrt{\frac{y^2-45}{4}}}<2\implies{y\in{[7,\sqrt{61})}}.\] To finish, the total length of the interval from which we choose $y$ is $27-7=20$. The total length of the success intervals is \[(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,\]which means the probability is \[\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.\]AIMEifying the answer gives $\boxed{850}$.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
[[2011 AIME II Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]
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=2011_AIME_II_Problems/Problem_15&oldid=104642"