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

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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>.
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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.
  
 
==See also==
 
==See also==

Revision as of 16:49, 3 September 2017

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

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.

See also

2011 AIME II (ProblemsAnswer KeyResources)
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

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