Difference between revisions of "2011 AIME II Problems/Problem 15"
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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: | ||
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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>. | ||
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<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>. | ||
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<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>. | ||
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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.