Difference between revisions of "2011 AIME II Problems/Problem 15"
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\end{array*}</math> | \end{array*}</math> | ||
− | 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 | + | 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 \le x < 6</math> or <math>6 \le x < 7</math> or <math>13 \le 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. |
− | Case <math>5 | + | 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 > 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>. | <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>. | ||
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So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>. | So in this case, the only values that will work are <math>5 < x < \frac{3 + \sqrt{61}}{2}</math>. | ||
− | Case <math>6 | + | Case <math>6 \le 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>. | <math>P(x)</math> must not be greater than the first perfect square after <math>9</math>, which is <math>16</math>. | ||
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So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>. | So in this case, the only values that will work are <math>6 < x < \frac{3 + \sqrt{109}}{2}</math>. | ||
− | Case <math>13 | + | Case <math>13 \le 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>. | <math>P(x)</math> must not be greater than the first perfect square after <math>121</math>, which is <math>144</math>. |
Revision as of 17:54, 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 :
$
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 .
$
So in this case, the only values that will work are .
Case :
must not be greater than the first perfect square after , which is .
$
So in this case, the only values that will work are .
Case :
must not be greater than the first perfect square after , which is .
$
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, :
$
So the answer is .