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1992 AIME Problems/Problem 11

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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