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Difference between revisions of "2011 AIME II Problems/Problem 15"

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\end{array*}</math>
 
\end{array*}</math>
  
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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 < x < 6</math> or <math>6 < x < 7</math> or <math>13 < 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 not be greater than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Note that in all the cases the next value of <math>P(x)</math> always passes the next perfect square after <math>P(\lfloor x \rfloor)</math>, so in no cases will all values of <math>x</math> in the said intervals work. Now, we consider the three difference 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 not be greater than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Note that in all the cases the next value of <math>P(x)</math> always passes the next perfect square after <math>P(\lfloor x \rfloor)</math>, so in no cases will all values of <math>x</math> in the said intervals work. Now, we consider the three difference cases.
  
  
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Case <math>5 < x < 6</math>:
+
Case <math>5 \le x < 6</math>:
  
 
<math>P(x)</math> must not be greater than the first perfect square after <math>1</math>, which is <math>4</math>. Since <math>P(x)</math> is increasing for <math>x > 5</math>, we just need to find where <math>P(x) = 4</math> and the values that will work will be <math>5 < x < \text{root}</math>.
 
<math>P(x)</math> must not be greater than the first perfect square after <math>1</math>, which is <math>4</math>. Since <math>P(x)</math> is increasing for <math>x > 5</math>, we just need to find where <math>P(x) = 4</math> and the values that will work will be <math>5 < x < \text{root}</math>.
Line 35: Line 35:
 
So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>.
 
So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>.
  
−
Case <math>6 < x < 7</math>:
+
Case <math>6 \le x < 7</math>:
  
 
<math>P(x)</math> must not be greater than the first perfect square after <math>9</math>, which is <math>16</math>.
 
<math>P(x)</math> must not be greater than the first perfect square after <math>9</math>, which is <math>16</math>.
Line 46: Line 46:
 
So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>.
 
So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>.
  
−
Case <math>13 < x < 14</math>:
+
Case <math>13 \le x < 14</math>:
  
 
<math>P(x)</math> must not be greater than the first perfect square after <math>121</math>, which is <math>144</math>.
 
<math>P(x)</math> must not be greater than the first perfect square after <math>121</math>, which is <math>144</math>.

Revision as of 17:54, 26 August 2012

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 $P(x)$:

$\begin{array*} P(5) = 1 \\ P(6) = 9 \\ P(7) = 19 \\ P(8) = 31 \\ P(9) = 45 \\ P(10) = 61 \\ P(11) = 79 \\ P(12) = 99 \\ P(13) = 121 \\ P(14) = 145 \\ P(15) = 171 \\ \end{array*}$ (Error compiling LaTeX. Unknown error_msg)

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 not be greater than the next perfect square after $P(\lfloor x \rfloor)$ (for the said intervals). Note that in all the cases the next value of $P(x)$ always passes the next perfect square after $P(\lfloor x \rfloor)$, so in no cases will all values of $x$ in the said intervals work. Now, we consider the three difference cases.


Case $5 \le x < 6$:

$P(x)$ must not be greater than the first perfect square after $1$, which is $4$. Since $P(x)$ is increasing for $x > 5$, we just need to find where $P(x) = 4$ and the values that will work will be $5 < x < \text{root}$.

$\begin{array*} x^2 - 3x - 9 = 4 \\ x = \frac{3 + \sqrt{61}}{2} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)

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

Case $6 \le x < 7$:

$P(x)$ must not be greater than the first perfect square after $9$, which is $16$.

$\begin{array*} x^2 - 3x - 9 = 16 \\ x = \frac{3 + \sqrt{109}}{2} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)

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

Case $13 \le x < 14$:

$P(x)$ must not be greater than the first perfect square after $121$, which is $144$.

$\begin{array*} x^2 - 3x - 9 = 144 \\ x = \frac{3 + \sqrt{621}}{2} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)

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

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

$\begin{array*} \frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} \\ = \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)

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

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