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"
m (→‎Solution)
Line 7: Line 7:
 
Table of values of <math>P(x)</math>:
 
Table of values of <math>P(x)</math>:
  
−
<math>\begin{array*}
+
<cmath>\begin{align*}
−
P(5) = 1 \\
+
P(5) &= 1 \\
−
P(6) = 9 \\
+
P(6) &= 9 \\
−
P(7) = 19 \\
+
P(7) &= 19 \\
−
P(8) = 31 \\
+
P(8) &= 31 \\
−
P(9) = 45 \\
+
P(9) &= 45 \\
−
P(10) = 61 \\
+
P(10) &= 61 \\
−
P(11) = 79 \\
+
P(11) &= 79 \\
−
P(12) = 99 \\
+
P(12) &= 99 \\
−
P(13) = 121 \\
+
P(13) &= 121 \\
−
P(14) = 145 \\
+
P(14) &= 145 \\
−
P(15) = 171 \\
+
P(15) &= 171 \\
−
\end{array*}</math>
+
\end{align*}</cmath>
  
 
In order for <math>\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}</math> to hold, <math>\sqrt{P(\lfloor x \rfloor)}</math> must be an integer and hence <math>P(\lfloor x \rfloor)</math> must be a perfect square. This limits <math>x</math> to <math>5 \le x < 6</math> or <math>6 \le x < 7</math> or <math>13 \le x < 14</math> since, from the table above, those are the only values of <math>x</math> for which <math>P(\lfloor x \rfloor)</math> is an perfect square. However, in order for <math>\sqrt{P(x)}</math> to be rounded down to <math>P(\lfloor x \rfloor)</math>, <math>P(x)</math> must be less than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Now, we consider the three cases:
 
In order for <math>\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}</math> to hold, <math>\sqrt{P(\lfloor x \rfloor)}</math> must be an integer and hence <math>P(\lfloor x \rfloor)</math> must be a perfect square. This limits <math>x</math> to <math>5 \le x < 6</math> or <math>6 \le x < 7</math> or <math>13 \le x < 14</math> since, from the table above, those are the only values of <math>x</math> for which <math>P(\lfloor x \rfloor)</math> is an perfect square. However, in order for <math>\sqrt{P(x)}</math> to be rounded down to <math>P(\lfloor x \rfloor)</math>, <math>P(x)</math> must be less than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Now, we consider the three cases:
Line 32: Line 32:
 
Since <math>P(x)</math> is increasing for <math>x \ge 5</math>, we just need to find the value <math>v \ge 5</math> where <math>P(v) = 4</math>, which will give us the working range <math>5 \le x < v</math>.
 
Since <math>P(x)</math> is increasing for <math>x \ge 5</math>, we just need to find the value <math>v \ge 5</math> where <math>P(v) = 4</math>, which will give us the working range <math>5 \le x < v</math>.
  
−
<math>\begin{array*}
+
<cmath>\begin{align*}
−
v^2 - 3v - 9 = 4 \\
+
v^2 - 3v - 9 &= 4 \\
−
v = \frac{3 + \sqrt{61}}{2}
+
v &= \frac{3 + \sqrt{61}}{2}
−
\end{array*}</math>
+
\end{align*}</cmath>
  
 
So in this case, the only values that will work are <math>5 \le x < \frac{3 + \sqrt{61}}{2}</math>.
 
So in this case, the only values that will work are <math>5 \le x < \frac{3 + \sqrt{61}}{2}</math>.
Line 43: Line 43:
 
<math>P(x)</math> must be less than the first perfect square after <math>9</math>, which is <math>16</math>.
 
<math>P(x)</math> must be less than the first perfect square after <math>9</math>, which is <math>16</math>.
  
−
<math>\begin{array*}
+
<cmath>\begin{align*}
−
v^2 - 3v - 9 = 16 \\
+
v^2 - 3v - 9 &= 16 \\
−
v = \frac{3 + \sqrt{109}}{2}
+
v &= \frac{3 + \sqrt{109}}{2}
−
\end{array*}</math>
+
\end{align*}</cmath>
  
 
So in this case, the only values that will work are <math>6 \le x < \frac{3 + \sqrt{109}}{2}</math>.
 
So in this case, the only values that will work are <math>6 \le x < \frac{3 + \sqrt{109}}{2}</math>.
Line 54: Line 54:
 
<math>P(x)</math> must be less than the first perfect square after <math>121</math>, which is <math>144</math>.
 
<math>P(x)</math> must be less than the first perfect square after <math>121</math>, which is <math>144</math>.
  
−
<math>\begin{array*}
+
<cmath>\begin{align*}
−
v^2 - 3v - 9 = 144 \\
+
v^2 - 3v - 9 &= 144 \\
−
v = \frac{3 + \sqrt{621}}{2}
+
v &= \frac{3 + \sqrt{621}}{2}
−
\end{array*}</math>
+
\end{align*}</cmath>
  
 
So in this case, the only values that will work are <math>13 \le x < \frac{3 + \sqrt{621}}{2}</math>.
 
So in this case, the only values that will work are <math>13 \le x < \frac{3 + \sqrt{621}}{2}</math>.
Line 63: Line 63:
 
Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>:
 
Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>:
  
−
<math>\begin{array*}
+
<cmath>\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{\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}
+
&= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}
−
\end{array*}</math>
+
\end{align*}</cmath>
  
 
Thus, the answer is <math>61 + 109 + 621 + 39 + 20 = \fbox{850}</math>.
 
Thus, the answer is <math>61 + 109 + 621 + 39 + 20 = \fbox{850}</math>.

Revision as of 19:27, 13 March 2015

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

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 $61 + 109 + 621 + 39 + 20 = \fbox{850}$.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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=68864"