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

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== Solution ==
 
== Solution ==
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<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}=
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{{image}}
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    \log_8 a+\log_8 (ar)+\ldots+\log_8 (ar^{11}) = </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> = \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8  (a^{12}r^{66}) </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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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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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>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so
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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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<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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<math>2x+11y=1003</math>
 
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<math>11y=1003-2x</math>
 
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<math>y=\frac{1003-2x}{11}</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 15:39, 25 September 2007

Problem

Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope, and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y),$ and that $x=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$

Solution

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Call the centers $O_1, O_2, O_3$, the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$, and $s_2$ on $\mathcal{C}_2$), and the intersection of each common internal tangent to the x axis $r, s$. $\triangle O_1r_1r \sim \triangle O_2r_2r$ since both triangles have a right angle and have vertical angles, and the same goes for $\triangle O_2s_2s \sim \triangle O_3s_1s$. By proportionality, we find that $O_1r = 4$; solving $\triangle O_1r_1r$ by the Pythagorean theorem yields $r_1r = \sqrt{15}$. On $\mathcal{C}_3$, we can do the same thing to get $O_3s_1 = 8$ and $s_1s = 4\sqrt{3}$.

The vertical altitude of each of $\triangle O_1r_1r$ and $\triangle O_3s_1s$ can each by found by the formula $c \cdot h = a \cdot b$ (as both products equal twice of the area of the triangle). Thus, the respective heights are $\frac{\sqrt{15}}{4}$ and $2\sqrt{3}$. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: $\sqrt{15 - \frac{15}{16}} = \frac{15}{4}$, and by 30-60-90: $6$.

From this information, the slope of each tangent can be uncovered. The slope of $t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}$. The slope of $t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{3}}$.

The equation of $t_1$ can be found by substituting the point $r (4,0)$ into $y = \frac{1}{\sqrt{15}}x + b$, so $y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}$. The equation of $t_2$, found by substituting point $s (16,0)$, is $y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}$. Putting these two equations together results in the desired $\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} \Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1} = \frac{76 - 12\sqrt{5}}{4} = 19 - 3\sqrt{5}$. Thus, $p + q + r = 19 + 3 + 5 \Longrightarrow 027$.

See also

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