Difference between revisions of "2011 AIME II Problems/Problem 15"
Baijiangchen (talk | contribs) (→Solution) |
Baijiangchen (talk | contribs) (→Solution: Strict vs. Not Strict) |
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Case <math>5 \le 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 \ge 5</math>, we just need to find where <math>P(x) = 4</math> and the values that will work will be <math>5 | + | <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 \ge 5</math>, we just need to find where <math>P(x) = 4</math> and the values that will work will be <math>5 \le x < \text{root}</math>. |
<math>\begin{array*} | <math>\begin{array*} | ||
Line 44: | Line 44: | ||
\end{array*}</math> | \end{array*}</math> | ||
− | So in this case, the only values that will work are <math>6 \le | + | So in this case, the only values that will work are <math>6 \le x < \frac{3 + \sqrt{109}}{2}</math>. |
Case <math>13 \le x < 14</math>: | Case <math>13 \le x < 14</math>: |
Revision as of 18:57, 26 August 2012
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 not be greater than the next perfect square after
(for the said intervals). Note that in all the cases the next value of
always passes the next perfect square after
, so in no cases will all values of
in the said intervals work. Now, we consider the three difference cases.
Case :
must not be greater than the first perfect square after
, which is
. Since
is increasing for
, we just need to find where
and the values that will work will be
.
$\begin{array*} x^2 - 3x - 9 = 4 \\ x = \frac{3 + \sqrt{61}}{2} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)
So in this case, the only values that will work are .
Case :
must not be greater than the first perfect square after
, which is
.
$\begin{array*} x^2 - 3x - 9 = 16 \\ x = \frac{3 + \sqrt{109}}{2} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)
So in this case, the only values that will work are .
Case :
must not be greater than the first perfect square after
, which is
.
$\begin{array*} x^2 - 3x - 9 = 144 \\ x = \frac{3 + \sqrt{621}}{2} \end{array*}$ (Error compiling LaTeX. Unknown error_msg)
So in this case, the only values that will work are .
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 .