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Difference between revisions of "2006 AIME I Problems/Problem 9"

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== Problem ==
 
== Problem ==
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The sequence <math> a_1, a_2, \ldots </math> is geometric with <math> a_1=a </math> and common ratio <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
  
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Circles <math> \mathcal{C}_1, \mathcal{C}_2, </math> and <math> \mathcal{C}_3 </math> have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line <math> t_1 </math> is a common internal tangent to <math>\mathcal{C}_1</math> and <math>\mathcal{C}_2</math> and has a positive slope, and line <math>t_2</math> is a common internal tangent to <math>\mathcal{C}_2</math> and <math>\mathcal{C}_3</math> and has a negative slope. Given that lines <math>t_1</math> and <math>t_2</math> intersect at <math>(x,y),</math> and that <math>x=p-q\sqrt{r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p+q+r.</math>
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== Solution ==
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<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) = </math>
  
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== Solution ==
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<math> = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8  (a^{12}r^{66}) </math>
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[[Image:2006_I_AIME-9.png]]
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So our question is equivalent to solving <math>\log_8 (a^{12}r^{66})=2006</math> for <math>a, r</math> [[positive integer]]s. 
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<math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so
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<math>a^{2}r^{11}=2^{1003}</math>
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The product of <math>a^2</math> and <math>r^{11}</math> is a power of 2.  Since both numbers have to be integers, this means that <math>a</math> and <math>r</math> are themselves powers of 2.  Now, let <math>a=2^x</math> and <math>r=2^y</math>:
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<math>(2^x)^2\cdot(2^y)^{11}=2^{1003}</math>
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<math>2^{2x}\cdot 2^{11y}=2^{1003}</math>
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<math>2^{2x+11y}=2^{1003}</math>
  
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Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[x axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have [[vertical angle]]s, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 8</math> and <math>s_1s = 4\sqrt{3}</math>.
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<math>2x+11y=1003</math>
  
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The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>.
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<math>11y=1003-2x</math>
  
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From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{3}}</math>.
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<math>y=\frac{1003-2x}{11}</math>
  
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The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} \Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1} = \frac{76 - 12\sqrt{5}}{4} = 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow 027</math>.
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For <math>y</math> to be an integer, the [[numerator]] must be [[divisible]] by 11.  This occurs when <math>x=1</math> because <math>1001=91*11</math>. Because only [[even integer]]s are being subtracted from 1003, the numerator never equals an even [[multiple]] of 11.  Therefore, the numerator takes on the value of every [[odd integer | odd]] multiple of 11 from 11 to 1001.  Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>. (We must add 1 because both endpoints are being included.) So the answer is <math>46</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:20, 25 September 2007

Problem

The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$

Solution

$\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) =$

$= \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8 (a^{12}r^{66})$

So our question is equivalent to solving $\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers.

$a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so

$a^{2}r^{11}=2^{1003}$

The product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that $a$ and $r$ are themselves powers of 2. Now, let $a=2^x$ and $r=2^y$:

$(2^x)^2\cdot(2^y)^{11}=2^{1003}$

$2^{2x}\cdot 2^{11y}=2^{1003}$

$2^{2x+11y}=2^{1003}$

$2x+11y=1003$

$11y=1003-2x$

$y=\frac{1003-2x}{11}$

For $y$ to be an integer, the numerator must be divisible by 11. This occurs when $x=1$ because $1001=91*11$. Because only even integers are being subtracted from 1003, the numerator never equals an even multiple of 11. Therefore, the numerator takes on the value of every odd multiple of 11 from 11 to 1001. Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is $1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46$. (We must add 1 because both endpoints are being included.) So the answer is $46$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions
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