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Difference between revisions of "1992 AIME Problems/Problem 11"

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== Solution ==
 
== Solution ==
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{{solution}}
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Let <math>l</math> be a line that makes an angle of <math>\theta</math> with the positive <math>x</math>-axis. Let <math>l'</math> be the reflection of <math>l</math> in <math>l_1</math>, and let <math>l''</math> be the reflection of <math>l'</math> in <math>l_2</math>.
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The angle between <math>l</math> and <math>l_1</math> is <math>\theta - \frac{\pi}{70}</math>, so the angle between <math>l_1</math> and <math>l'</math> must also be <math>\theta - \frac{\pi}{70}</math>. Thus,  <math>l'</math> makes an angle of <math>\frac{\pi}{70}-\left(\theta-\frac{\pi}{70}\right) = \frac{\pi}{35}-\theta</math> with the positive <math>x</math>-axis. 
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Similarly, since the angle between <math>l'</math> and <math>l_2</math> is <math>\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}</math>, the angle between <math>l''</math> and the positive <math>x</math>-axis is <math>\frac{\pi}{54}-\left(\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}\right) = \frac{\pi}{27}-\frac{\pi}{35}+\theta = \frac{8\pi}{945} + \theta</math>.
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Thus, <math>R(l)</math> makes an <math>\frac{8\pi}{945} + \theta</math> angle with the positive <math>x</math>-axis. So <math>R^{(n)}(l)</math> makes an <math>\frac{8n\pi}{945} + \theta</math> angle with the positive <math>x</math>-axis.
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Therefore, <math>R^{(m)}(l)=l</math> iff <math>\frac{8m\pi}{945}</math> is an integral multiple of <math>\pi</math>. Thus, <math>8m \equiv 0\pmod{945}</math>. Since <math>\gcd(8,945)=1</math>, <math>m \equiv 0 \pmod{945}</math>, so the smallest positive integer <math>m</math> is <math>\boxed{945}</math>.
  
 
== See also ==
 
== See also ==
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* [[1992 AIME Problems/Problem 10 | Previous Problem]]
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{{AIME box|year=1992|num-b=10|num-a=12}}
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* [[1992 AIME Problems/Problem 12 | Next Problem]]
 
  
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* [[1992 AIME Problems]]
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 18:24, 4 July 2013

Problem

Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$, the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$, and the resulting line is reflected in $l_2^{}$. Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)$. Given that $l^{}_{}$ is the line $y=\frac{19}{92}x^{}_{}$, find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$.

Solution

Let $l$ be a line that makes an angle of $\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$.

The angle between $l$ and $l_1$ is $\theta - \frac{\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\theta - \frac{\pi}{70}$. Thus, $l'$ makes an angle of $\frac{\pi}{70}-\left(\theta-\frac{\pi}{70}\right) = \frac{\pi}{35}-\theta$ with the positive $x$-axis.

Similarly, since the angle between $l'$ and $l_2$ is $\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}$, the angle between $l''$ and the positive $x$-axis is $\frac{\pi}{54}-\left(\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}\right) = \frac{\pi}{27}-\frac{\pi}{35}+\theta = \frac{8\pi}{945} + \theta$.

Thus, $R(l)$ makes an $\frac{8\pi}{945} + \theta$ angle with the positive $x$-axis. So $R^{(n)}(l)$ makes an $\frac{8n\pi}{945} + \theta$ angle with the positive $x$-axis.

Therefore, $R^{(m)}(l)=l$ iff $\frac{8m\pi}{945}$ is an integral multiple of $\pi$. Thus, $8m \equiv 0\pmod{945}$. Since $\gcd(8,945)=1$, $m \equiv 0 \pmod{945}$, so the smallest positive integer $m$ is $\boxed{945}$.

See also

1992 AIME (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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