Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

 
 
(5 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Lines <math>l_1^{}</math> and <math>l_2^{}</math> both pass through the origin and make first-quadrant angles of <math>\frac{\pi}{70}</math> and <math>\frac{\pi}{54}</math> radians, respectively, with the positive x-axis. For any line <math>l^{}_{}</math>, the transformation <math>R(l)^{}_{}</math> produces another line as follows: <math>l^{}_{}</math> is reflected in <math>l_1^{}</math>, and the resulting line is reflected in <math>l_2^{}</math>. Let <math>R^{(1)}(l)=R(l)^{}_{}</math> and <math>R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)</math>. Given that <math>l^{}_{}</math> is the line <math>y=\frac{19}{92}x^{}_{}</math>, find the smallest positive integer <math>m^{}_{}</math> for which <math>R^{(m)}(l)=l^{}_{}</math>.
  
 
== Solution ==
 
== Solution ==
 +
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>.
 +
 +
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. 
 +
 +
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>.
 +
 +
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.
 +
 +
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 ==
−
* [[1991 AIME Problems]]
+
{{AIME box|year=1992|num-b=10|num-a=12}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_11&oldid=54988"