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

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
 
==Solution==
 
==Solution==
  
−
Table of values of P([x]):
+
Table of values of <math>P(x)</math>:
  
−
P(5) = 1
+
<math>\begin{array*}
−
P(6) = 9
+
P(5) = 1 \\
−
P(7) = 19
+
P(6) = 9 \\
−
P(8) = 31
+
P(7) = 19 \\
−
P(9) = 45
+
P(8) = 31 \\
−
P(10) = 61
+
P(9) = 45 \\
−
P(11) = 79
+
P(10) = 61 \\
−
P(12) = 99
+
P(11) = 79 \\
−
P(13) = 121
+
P(12) = 99 \\
−
P(14) = 145
+
P(13) = 121 \\
−
P(15) = 171
+
P(14) = 145 \\
 +
P(15) = 171 \\
 +
\end{array*}</math>
  
−
In order for [√P(x)] = √P([x]) to hold, √P([x]) must be an integer and hence P([x]) must be a perfect square. This limits x to 5 < x < 6 or 6 < x < 7 or 13 < x < 14 since, from the table above, those are the only values of x for which P([x]) is an integer. However, in order for √P(x) to be rounded down to √P([x]), P(x) must not be greater than the next perfect square after P([x]) (for 5 < x < 6, etc.). Note that in all the cases the next value of P(x) always passes the next perfect square after P([x]), so in no cases will all values of x 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 < 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.
  
−
5 < 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 < root.
 
  
−
x^2 - 3x - 9 = 4
+
Case <math>1</math>:
−
x = (3 + √61)/2
+
<math>5 < 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>.
  
−
So in this case, the only values that will work are 5 < x < (3 + √61)/2.
+
<math>\begin{array*}
 +
x^2 - 3x - 9 = 4 \\
 +
x = \frac{3 + \sqrt{61}}{2}
 +
\end{array*}</math>
  
−
6 < x < 7:
+
So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>.
−
P(x) must not be greater than the first perfect square after 9, which is 16.
 
  
−
x^2 - 3x - 9 = 16
+
Case <math>2</math>:
−
x = (3 + √109)/2
+
<math>6 < 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>.
  
−
So in this case, the only values that will work are 6 < x < (3 + √109)/2.
+
<math>\begin{array*}
 +
x^2 - 3x - 9 = 16 \\
 +
x = \frac{3 + \sqrt{109}}{2}
 +
\end{array*}</math>
  
−
13 < x < 14:
+
So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>.
−
P(x) must not be greater than the first perfect square after 121, which is 144.
 
  
−
x^2 - 3x - 9 = 144
+
Case <math>3</math>:
−
x = (3 + √721)/2
+
<math>13 < 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>.
  
−
So in this case, the only values that will work are 13 < x < (3 + √721)/2.
+
<math>\begin{array*}
 +
x^2 - 3x - 9 = 144 \\
 +
x = \frac{3 + \sqrt{621}}{2}
 +
\end{array*}</math>
  
−
Now, we find the length of the working intervals and divide it by the length of the total interval, 15 - 5 = 10:
+
So in this case, the only values that will work are <math>13 < x < \frac{3 + \sqrt{621}}{2}</math>.
  
−
(((3 + √61)/2 - 5) + ((3 + √109)/2 - 6) + ((3 + √721)/2 - 13))/10
+
Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>:
−
= (√61 + √109 + √721 - 39)/20
 
  
−
So the answer is 61 + 109 + 721 + 39 + 20 = 950.
+
<math>\begin{array*}
 +
\frac{(\frac{3 + \sqrt{61}}{2} - 5) + (\frac{3 + \sqrt{109}}{2} - 6) + (\frac{3 + \sqrt{621}}{2} - 13)}{10} \\
 +
= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39)}{20}
 +
\end{array*}</math>
 +
 
 +
So the answer is <math>61 + 109 + 621 + 39 + 20 = 850</math>.

Revision as of 19:08, 19 April 2011

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 $5 < x < 6$ or $6 < x < 7$ or $13 < x < 14$ 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 $1$: $5 < x < 6 \$ (Error compiling LaTeX. Unknown error_msg): $P(x)$ 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 $2$: $6 < x < 7 \$ (Error compiling LaTeX. Unknown error_msg): $P(x)$ 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 $3$: $13 < x < 14 \$ (Error compiling LaTeX. Unknown error_msg): $P(x)$ 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, $15 - 5 = 10$:

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

So the answer is $61 + 109 + 621 + 39 + 20 = 850$.

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