Difference between revisions of "2011 AIME II Problems/Problem 15"
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− | Problem | + | ==Problem== |
− | Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} | + | Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>. |
− | + | ==Solution 1== | |
− | Solution | ||
− | + | Table of values of <math>P(x)</math>: | |
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 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{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: | ||
+ | |||
+ | |||
+ | Case <math>5 \le x < 6</math>: | ||
+ | |||
+ | <math>P(x)</math> must be less than the first perfect square after <math>1</math>, which is <math>4</math>, ''i.e.'': | ||
+ | |||
+ | <math>1 \le P(x) < 4</math> (because <math>\lfloor \sqrt{P(x)} \rfloor = 1</math> implies <math>1 \le \sqrt{P(x)} < 2</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 &= \frac{3 + \sqrt{61}}{2} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | So in this case, the only values that will work are <math>5 \le x < \frac{3 + \sqrt{61}}{2}</math>. | ||
+ | |||
+ | Case <math>6 \le x < 7</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 &= \frac{3 + \sqrt{109}}{2} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | 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>: | ||
+ | |||
+ | <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 &= \frac{3 + \sqrt{621}}{2} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | So in this case, the only values that will work are <math>13 \le x < \frac{3 + \sqrt{621}}{2}</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{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Thus, the answer is <math>61 + 109 + 621 + 39 + 20 = \fbox{850}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Make the substitution <math>y=2x-3</math>, so <math>P(x)=\frac{y^2-45}{4}.</math> We're looking for solutions to <cmath>\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}</cmath>with the new bounds <math>y\in{[7,27]}</math>. Since the left side is an integer, it must be that <math>\frac{\lfloor{y\rfloor}^2-45}{4}</math> is a perfect square. For simplicity, write <math>\lfloor{y\rfloor}=a</math> and <cmath>a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.</cmath>Since <math>a-2b<a+2b</math>, it must be that <math>(a-2b,a+2b)=(1,45),(3,15),(5,9)</math>, which gives solutions <math>(23,11),(9,3),(7,1)</math>, respectively. But this gives us three cases to check: | ||
+ | |||
+ | Case 1: <math>\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11</math>. | ||
+ | |||
+ | In this case, we have <cmath>11\leq{\sqrt{\frac{y^2-45}{4}}}<12\implies{y\in{[23,\sqrt{621})}}.</cmath> | ||
+ | Case 2: <math>\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3</math>. | ||
+ | |||
+ | In this case, we have <cmath>3\leq{\sqrt{\frac{y^2-45}{4}}}<4\implies{y\in{[9,\sqrt{109})}}.</cmath> | ||
+ | Case 3: <math>\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1</math> | ||
+ | |||
+ | In this case, we have <cmath>1\leq{\sqrt{\frac{y^2-45}{4}}}<2\implies{y\in{[7,\sqrt{61})}}.</cmath> | ||
+ | To finish, the total length of the interval from which we choose <math>y</math> is <math>27-7=20</math>. The total length of the success intervals is <cmath>(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,</cmath>which means the probability is <cmath>\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.</cmath>The requested sum is <math>\boxed{850}</math>. | ||
+ | |||
+ | == Solution 3 (Graphing) == | ||
+ | |||
+ | It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let <math>A = \lfloor\sqrt{P(x)}\rfloor</math> and <math>B = \sqrt{P(\lfloor x \rfloor)}</math>. The graph of <math>A</math> and <math>B</math> will look like this, with <math>A</math> having only integral y-values and <math>B</math> having only integral x-values: | ||
+ | |||
+ | [[File:2011 AIME II Problem 15 Graph 1.png|400px]] | ||
+ | |||
+ | As both <math>A</math> and <math>B</math> consist of a bunch of line segments, the probability that <math>A = B</math> is the "length" of the overlap between the segments of <math>A</math> and <math>B</math> divided by the total length of the segments of <math>B</math>. | ||
+ | |||
+ | Looking at the graph, we see that <math>A</math> and <math>B</math> will overlap only when <math>B</math> is an integer. Specifically, each region of overlap will begin when <math>\sqrt{P(x)}\ = k, 5 \le x \le 15</math> has solutions for integral <math>k</math> in the range of <math>A</math>, which consists of the integers <math>1-13</math>, and end when <math>A</math> jumps up to its next y-value. | ||
+ | |||
+ | Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of <math>\frac{3 + \sqrt{45 + 4k^2}}{2}</math> for <math>k</math> in the specified range, meaning <math>45 + 4k^2</math> must be a perfect square. Plugging in all the possible values of <math>k</math>, we get <math>k = 1, 3, 11</math>, corresponding to start points of <math>x = 5, 6, 13</math>. As already stated, the endpoints will occur when <math>A</math> jumps up to the next integer <math>k+1</math> at each of these segments, at which point the x-value will be <math>\frac{3 + \sqrt{45 + 4(k+1)^2}}{2}</math>. On the graph, the overlapping segments of <math>A</math> and <math>B</math> would be represented by the highlighted green segments below: | ||
+ | |||
+ | [[File:2011 AIME II Problem 15 Graph 2.png|400px]] | ||
+ | |||
+ | |||
+ | Taking the difference between this second x-value and the start point for each of our start points <math>x = 5, 6, 13</math> and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of <math>B</math>) to give us the probability of overlap between <math>A</math> and <math>B</math>. | ||
+ | |||
+ | Doing so gives us: | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | <math>\implies{61 + 109 + 621 + 39 + 20 = \fbox{850}}</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
+ | |||
+ | ==Solution 4== | ||
+ | Note that all the "bounds" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is <math>\frac{3*3+\sqrt{9+4(4+9)}-10+\sqrt{9+4(16+9)}-12+\sqrt{9+4(144+9)}}{20} \implies \boxed{850}</math> | ||
+ | |||
+ | ~Lcz | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box | year = 2011 | n = II | num-b=14 | after=Last Problem}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:03, 7 January 2023
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 1
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 .
Solution 2
Make the substitution , so
We're looking for solutions to
with the new bounds
. Since the left side is an integer, it must be that
is a perfect square. For simplicity, write
and
Since
, it must be that
, which gives solutions
, respectively. But this gives us three cases to check:
Case 1: .
In this case, we have
Case 2:
.
In this case, we have
Case 3:
In this case, we have
To finish, the total length of the interval from which we choose
is
. The total length of the success intervals is
which means the probability is
The requested sum is
.
Solution 3 (Graphing)
It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let and
. The graph of
and
will look like this, with
having only integral y-values and
having only integral x-values:
As both and
consist of a bunch of line segments, the probability that
is the "length" of the overlap between the segments of
and
divided by the total length of the segments of
.
Looking at the graph, we see that and
will overlap only when
is an integer. Specifically, each region of overlap will begin when
has solutions for integral
in the range of
, which consists of the integers
, and end when
jumps up to its next y-value.
Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of for
in the specified range, meaning
must be a perfect square. Plugging in all the possible values of
, we get
, corresponding to start points of
. As already stated, the endpoints will occur when
jumps up to the next integer
at each of these segments, at which point the x-value will be
. On the graph, the overlapping segments of
and
would be represented by the highlighted green segments below:
Taking the difference between this second x-value and the start point for each of our start points and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of
) to give us the probability of overlap between
and
.
Doing so gives us:
.
~ anellipticcurveoverq
Solution 4
Note that all the "bounds" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is
~Lcz
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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