2011 AIME II Problems/Problem 15

Problem

Let . A real number is chosen at random from the interval . The probability that is equal to , where , , , , and are positive integers. Find .

Solution

Table of values of :

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

In order for to hold, must be an integer and hence must be a perfect square. This limits to or or since, from the table above, those are the only values of for which is an perfect square. However, in order for to be rounded down to , must be less than the next perfect square after (for the said intervals). Now, we consider the three cases:

Case :

must be less than the first perfect square after , which is , i.e.:

(because implies )

Since is increasing for , we just need to find the value where , which will give us the working range .

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

So in this case, the only values that will work are .

Case :

must be less than the first perfect square after , which is .

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

So in this case, the only values that will work are .

Case :

must be less than the first perfect square after , which is .

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

So in this case, the only values that will work are .

Now, we find the length of the working intervals and divide it by the length of the total interval, :

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

Thus, the answer is .

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.