1992 AIME Problem 11

Attempt Problem 11 of the 1992 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AIME solutions, or check the answer key.

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11.

Lines l1l_1 and l2l_2 both pass through the origin and make first-quadrant angles of π70\frac{\pi}{70} and π54\frac{\pi}{54} radians, respectively, with the positive xx-axis. For any line l,l, the transformation R(l)R(l) produces another line as follows: ll is reflected in l1,l_1, and the resulting line is reflected in l2.l_2. Let R(1)(l)=R(l)R^{(1)}(l)=R(l) and R(n)(l)=R(R(n1)(l)).R^{(n)}(l)=R(R^{(n-1)}(l)). Given that ll is the line y=(1992)x,y=(\frac{19}{92})x, find the smallest positive integer mm for which R(m)(l)=l.R^{(m)}(l)=l.

Answer: 945
Concepts:reflection (geometry)transformationmodular arithmetic
Difficulty rating: 2230
Small Hint:

Two reflections in intersecting lines compose to a rotation through twice the angle between the lines

Big Hint:

An unoriented line returns to itself when its accumulated rotation is a multiple of π\pi

Solution:

The composition is rotation through 2(π54π70)=8π945.2\left(\frac{\pi}{54}-\frac{\pi}{70}\right)=\frac{8\pi}{945}. A line through the origin is unchanged by a rotation exactly when the rotation angle is a multiple of π.\pi. Thus mm must satisfy 8m945Z.\frac{8m}{945}\in\mathbb Z. Since gcd(8,945)=1,\gcd(8,945)=1, the least such mm is 945.945.

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