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)
So the answer is .