Difference between revisions of "2011 AIME II Problems/Problem 15"
Michelangelo (talk | contribs) |
Mathgeek2006 (talk | contribs) m (→Solution) |
||
Line 7: | Line 7: | ||
Table of values of <math>P(x)</math>: | Table of values of <math>P(x)</math>: | ||
− | < | + | <cmath>\begin{align*} |
− | P(5) = 1 \\ | + | P(5) &= 1 \\ |
− | P(6) = 9 \\ | + | P(6) &= 9 \\ |
− | P(7) = 19 \\ | + | P(7) &= 19 \\ |
− | P(8) = 31 \\ | + | P(8) &= 31 \\ |
− | P(9) = 45 \\ | + | P(9) &= 45 \\ |
− | P(10) = 61 \\ | + | P(10) &= 61 \\ |
− | P(11) = 79 \\ | + | P(11) &= 79 \\ |
− | P(12) = 99 \\ | + | P(12) &= 99 \\ |
− | P(13) = 121 \\ | + | P(13) &= 121 \\ |
− | P(14) = 145 \\ | + | P(14) &= 145 \\ |
− | P(15) = 171 \\ | + | P(15) &= 171 \\ |
− | \end{ | + | \end{align*}</cmath> |
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 be less than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Now, we consider the three cases: | 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 be less than the next perfect square after <math>P(\lfloor x \rfloor)</math> (for the said intervals). Now, we consider the three cases: | ||
Line 32: | Line 32: | ||
Since <math>P(x)</math> is increasing for <math>x \ge 5</math>, we just need to find the value <math>v \ge 5</math> where <math>P(v) = 4</math>, which will give us the working range <math>5 \le x < v</math>. | Since <math>P(x)</math> is increasing for <math>x \ge 5</math>, we just need to find the value <math>v \ge 5</math> where <math>P(v) = 4</math>, which will give us the working range <math>5 \le x < v</math>. | ||
− | < | + | <cmath>\begin{align*} |
− | v^2 - 3v - 9 = 4 \\ | + | v^2 - 3v - 9 &= 4 \\ |
− | v = \frac{3 + \sqrt{61}}{2} | + | v &= \frac{3 + \sqrt{61}}{2} |
− | \end{ | + | \end{align*}</cmath> |
So in this case, the only values that will work are <math>5 \le x < \frac{3 + \sqrt{61}}{2}</math>. | So in this case, the only values that will work are <math>5 \le x < \frac{3 + \sqrt{61}}{2}</math>. | ||
Line 43: | Line 43: | ||
<math>P(x)</math> must be less than the first perfect square after <math>9</math>, which is <math>16</math>. | <math>P(x)</math> must be less than the first perfect square after <math>9</math>, which is <math>16</math>. | ||
− | < | + | <cmath>\begin{align*} |
− | v^2 - 3v - 9 = 16 \\ | + | v^2 - 3v - 9 &= 16 \\ |
− | v = \frac{3 + \sqrt{109}}{2} | + | v &= \frac{3 + \sqrt{109}}{2} |
− | \end{ | + | \end{align*}</cmath> |
So in this case, the only values that will work are <math>6 \le x < \frac{3 + \sqrt{109}}{2}</math>. | So in this case, the only values that will work are <math>6 \le x < \frac{3 + \sqrt{109}}{2}</math>. | ||
Line 54: | Line 54: | ||
<math>P(x)</math> must be less than the first perfect square after <math>121</math>, which is <math>144</math>. | <math>P(x)</math> must be less than the first perfect square after <math>121</math>, which is <math>144</math>. | ||
− | < | + | <cmath>\begin{align*} |
− | v^2 - 3v - 9 = 144 \\ | + | v^2 - 3v - 9 &= 144 \\ |
− | v = \frac{3 + \sqrt{621}}{2} | + | v &= \frac{3 + \sqrt{621}}{2} |
− | \end{ | + | \end{align*}</cmath> |
So in this case, the only values that will work are <math>13 \le x < \frac{3 + \sqrt{621}}{2}</math>. | So in this case, the only values that will work are <math>13 \le x < \frac{3 + \sqrt{621}}{2}</math>. | ||
Line 63: | Line 63: | ||
Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>: | Now, we find the length of the working intervals and divide it by the length of the total interval, <math>15 - 5 = 10</math>: | ||
− | < | + | <cmath>\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{\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} | + | &= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20} |
− | \end{ | + | \end{align*}</cmath> |
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>. |
Revision as of 20:27, 13 March 2015
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 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 .
So in this case, the only values that will work are .
Case :
must be less than the first perfect square after , which is .
So in this case, the only values that will work are .
Case :
must be less 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, :
Thus, the answer is .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.