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Difference between revisions of "2011 AIME II Problems/Problem 15"

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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>.
 
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==
+
==Solution 1==
  
 
Table of values of <math>P(x)</math>:
 
Table of values of <math>P(x)</math>:
Line 71: Line 71:
  
 
P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.
 
P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.
 +
 +
== 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 right 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>AIMEifying the answer gives <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>
 +
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<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==
 
==See also==
−
{{AIME box | year = 2011 | n = II | num-b=13 | num-a=15}}
+
{{AIME box | year = 2011 | n = II | num-b=14 | after=Last Problem}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:07, 16 June 2020

Problem

Let $P(x) = x^2 - 3x - 9$. A real number $x$ is chosen at random from the interval $5 \le x \le 15$. The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$, $b$, $c$, $d$, and $e$ are positive integers. Find $a + b + c + d + e$.

Solution 1

Table of values of $P(x)$:

\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*}

In order for $\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}$ to hold, $\sqrt{P(\lfloor x \rfloor)}$ must be an integer and hence $P(\lfloor x \rfloor)$ must be a perfect square. This limits $x$ to $5 \le x < 6$ or $6 \le x < 7$ or $13 \le x < 14$ since, from the table above, those are the only values of $x$ for which $P(\lfloor x \rfloor)$ is an perfect square. However, in order for $\sqrt{P(x)}$ to be rounded down to $P(\lfloor x \rfloor)$, $P(x)$ must be less than the next perfect square after $P(\lfloor x \rfloor)$ (for the said intervals). Now, we consider the three cases:


Case $5 \le x < 6$:

$P(x)$ must be less than the first perfect square after $1$, which is $4$, i.e.:

$1 \le P(x) < 4$ (because $\lfloor \sqrt{P(x)} \rfloor = 1$ implies $1 \le \sqrt{P(x)} < 2$)

Since $P(x)$ is increasing for $x \ge 5$, we just need to find the value $v \ge 5$ where $P(v) = 4$, which will give us the working range $5 \le x < v$.

\begin{align*} v^2 - 3v - 9 &= 4 \\ v &= \frac{3 + \sqrt{61}}{2} \end{align*}

So in this case, the only values that will work are $5 \le x < \frac{3 + \sqrt{61}}{2}$.

Case $6 \le x < 7$:

$P(x)$ must be less than the first perfect square after $9$, which is $16$.

\begin{align*} v^2 - 3v - 9 &= 16 \\ v &= \frac{3 + \sqrt{109}}{2} \end{align*}

So in this case, the only values that will work are $6 \le x < \frac{3 + \sqrt{109}}{2}$.

Case $13 \le x < 14$:

$P(x)$ must be less than the first perfect square after $121$, which is $144$.

\begin{align*} v^2 - 3v - 9 &= 144 \\ v &= \frac{3 + \sqrt{621}}{2} \end{align*}

So in this case, the only values that will work are $13 \le x < \frac{3 + \sqrt{621}}{2}$.

Now, we find the length of the working intervals and divide it by the length of the total interval, $15 - 5 = 10$:

\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*}

Thus, the answer is $61 + 109 + 621 + 39 + 20 = \fbox{850}$.

P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.

Solution 2

Make the substitution $y=2x-3$, so $P(x)=\frac{y^2-45}{4}.$ We're looking for solutions to \[\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}\]with the new bounds $y\in{[7,27]}$. Since the right side is an integer, it must be that $\frac{\lfloor{y\rfloor}^2-45}{4}$ is a perfect square. For simplicity, write $\lfloor{y\rfloor}=a$ and \[a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.\]Since $a-2b<a+2b$, it must be that $(a-2b,a+2b)=(1,45),(3,15),(5,9)$, which gives solutions $(23,11),(9,3),(7,1)$, respectively. But this gives us three cases to check:

Case 1: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11$.

In this case, we have \[11\leq{\sqrt{\frac{y^2-45}{4}}}<12\implies{y\in{[23,\sqrt{621})}}.\] Case 2: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3$.

In this case, we have \[3\leq{\sqrt{\frac{y^2-45}{4}}}<4\implies{y\in{[9,\sqrt{109})}}.\] Case 3: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1$

In this case, we have \[1\leq{\sqrt{\frac{y^2-45}{4}}}<2\implies{y\in{[7,\sqrt{61})}}.\] To finish, the total length of the interval from which we choose $y$ is $27-7=20$. The total length of the success intervals is \[(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,\]which means the probability is \[\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.\]AIMEifying the answer gives $\boxed{850}$.

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 $A = \lfloor\sqrt{P(x)}\rfloor$ and $B = \sqrt{P(\lfloor x \rfloor)}$. The graph of $A$ and $B$ will look like this, with $A$ having only integral y-values and $B$ having only integral x-values:

2011 AIME II Problem 15 Graph 1.png

As both $A$ and $B$ consist of a bunch of line segments, the probability that $A = B$ is the "length" of the overlap between the segments of $A$ and $B$ divided by the total length of the segments of $B$.

Looking at the graph, we see that $A$ and $B$ will overlap only when $B$ is an integer. Specifically, each region of overlap will begin when $\sqrt{P(x)}\ = k, 5 \le x \le 15$ has solutions for integral $k$ in the range of $A$, which consists of the integers $1-13$, and end when $A$ 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 $\frac{3 + \sqrt{45 + 4k^2}}{2}$ for $k$ in the specified range, meaning $45 + 4k^2$ must be a perfect square. Plugging in all the possible values of $k$, we get $k = 1, 3, 11$, corresponding to start points of $x = 5, 6, 13$. As already stated, the endpoints will occur when $A$ jumps up to the next integer $k+1$ at each of these segments, at which point the x-value will be $\frac{3 + \sqrt{45 + 4(k+1)^2}}{2}$. On the graph, the overlapping segments of $A$ and $B$ would be represented by the highlighted green segments below:

2011 AIME II Problem 15 Graph 2.png


Taking the difference between this second x-value and the start point for each of our start points $x = 5, 6, 13$ 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 $B$) to give us the probability of overlap between $A$ and $B$.

Doing so gives us:

$\frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} = \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}$

$\implies{61 + 109 + 621 + 39 + 20 = \fbox{850}}$.

~ 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 $\frac{3*3+\sqrt{9+4(4+9)}-10+\sqrt{9+4(16+9)}-12+\sqrt{9+4(144+9)}}{20} \implies \boxed{850}$

~Lcz

See also

2011 AIME II (Problems • Answer Key • Resources)
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