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Difference between revisions of "2007 AIME II Problems/Problem 11"

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== Problem ==
 
== Problem ==
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The increasing [[geometric sequence]] <math>x_{0},x_{1},x_{2},\ldots</math> consists entirely of [[integer|integral]] powers of <math>3.</math> Given that
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Two long [[cylinder|cylindrical]] tubes of the same length but different [[diameter]]s lie [[parallel]] to each other on a [[plane|flat surface]]. The larger tube has [[radius]] <math>72</math> and rolls along the surface toward the smaller tube, which has radius <math>24</math>. It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its [[circumference]] as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a [[distance]] <math>x</math> from where it starts. The distance <math>x</math> can be expressed in the form <math>a\pi+b\sqrt{c},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[integer]]s and <math>c</math> is not divisible by the [[square]] of any [[prime]]. Find <math>a+b+c.</math>
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<math>\sum_{n=0}^{7}\log_{3}(x_{n}) = 308</math> and <math>56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,</math>
 
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find <math>\displaystyle \log_{3}(x_{14}).</math>
 
  
 
== Solution ==
 
== Solution ==
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Suppose that <math>\displaystyle x_0 = a</math>, and that the common [[ratio]] between the terms is <math>r</math>.
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[[Image:2007 AIME II-11.png]]
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The first conditions tells us that <math>\displaystyle \log_3 a + \log_3 ar \ldots \log_3 ar^7 = 308</math>. Using the rules of [[logarithm]]s, we can simplify that to <math>\displaystyle \log_3 a^8r^{1 + 2 + \ldots + 7} = 308</math>. Thus, <math>\displaystyle a^8r^{28} = 3^{308}</math>. Since all of the terms of the geometric sequence are integral powers of <math>3</math>, we know that both <math>a</math> and <math>r</math> must be powers of 3. Denote <math>\displaystyle 3^x = a</math> and <math>\displaystyle 3^y = r</math>. We find that <math>8x + 28y = 308</math>. The possible positive integral pairs of <math>(x,y)</math> are <math>(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)</math>.
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If it weren’t for the small tube, the larger tube would travel <math>144\pi</math>. Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.
  
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Drawing the radii as shown in the diagram, notice that the [[hypotenuse]] of the [[right triangle]] in the diagram has a length of <math>72 + 24 = 96</math>. The horizontal line divides the radius of the larger circle into <math>72 - 24 = 48</math> on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a <math>30-60-90 \triangle</math>.
  
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The second condition tells us that <math>56 \le \log_3 (a + ar + \ldots ar^7) \le 57</math>. Using the sum formula for a [[geometric series]] and substituting <math>x</math> and <math>y</math>, this simplifies to <math>3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}</math>. The fractional part <math>\approx \frac{3^{8y}}{3^y} = 3^{7y}</math>. Thus, we need <math>\approx 56 \le x + 7y \le 57</math>. Checking the pairs above, only <math>\displaystyle (21,5)</math> is close.
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Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree [[central angle]] indicates to take <math>\frac{60}{360} = \frac 16</math> of the circumference of the larger circle (twice), while the <math>180 - 2(30) = 120^{\circ}</math> central angle in the smaller circle indicates to take <math>\frac{120}{360} = \frac 13</math> of the circumference. This adds up to <math>2 \cdot \frac 16 144\pi + \frac 13 48\pi = 64\pi</math>.  
  
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The actual horizontal distance it takes can be found by using the <math>30-60-90 \triangle</math>s. The missing leg is equal in length to <math>48\sqrt{3}</math>. Thus, the total horizontal distance covered is <math>96\sqrt{3}</math>.
  
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Our solution is therefore <math>\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = 091 \displaystyle</math>.
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Thus, we get <math>144\pi - 64\pi + 96\sqrt{3} = 80\pi + 96\sqrt{3}</math>, and our answer is <math>\boxed{179}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 15:31, 29 February 2020

Problem

Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$. It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c.$

Solution

2007 AIME II-11.png

If it weren’t for the small tube, the larger tube would travel $144\pi$. Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.

Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of $72 + 24 = 96$. The horizontal line divides the radius of the larger circle into $72 - 24 = 48$ on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a $30-60-90 \triangle$.

Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take $\frac{60}{360} = \frac 16$ of the circumference of the larger circle (twice), while the $180 - 2(30) = 120^{\circ}$ central angle in the smaller circle indicates to take $\frac{120}{360} = \frac 13$ of the circumference. This adds up to $2 \cdot \frac 16 144\pi + \frac 13 48\pi = 64\pi$.

The actual horizontal distance it takes can be found by using the $30-60-90 \triangle$s. The missing leg is equal in length to $48\sqrt{3}$. Thus, the total horizontal distance covered is $96\sqrt{3}$.

Thus, we get $144\pi - 64\pi + 96\sqrt{3} = 80\pi + 96\sqrt{3}$, and our answer is $\boxed{179}$.

See also

2007 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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