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Difference between revisions of "2006 AIME I Problems/Problem 7"

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Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].
 
Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].
  
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Let the set of parallel lines be [[perpendicular]] to the [[x-axis]], such that they cross it at <math>0, 1, 2 \ldots</math>. The base of region <math>\mathcal{A}</math> is on the line <math>x = 1</math>. The bigger base of region <math>\mathcal{D}</math> is on the line <math>x = 7</math>. x-axis will bisect the angle; the top side will be <math>y = \frac{1}{2}(x - s)</math>; the bottom side will be <math>y = -\frac{1}{2}(x - s)</math>.  
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Let the set of parallel lines be [[perpendicular]] to the [[x-axis]], such that they cross it at <math>0, 1, 2 \ldots</math>. The base of region <math>\mathcal{A}</math> is on the line <math>x = 1</math>. The bigger base of region <math>\mathcal{D}</math> is on the line <math>x = 7</math>.  
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Cutting the angle by half don't change the problem. After the cut, the top side of the angle will be <math>y = x - s</math>; the bottom side will be x-axis.  
  
 
Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,
 
Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,

Revision as of 21:03, 11 March 2007

Problem

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is 11/5. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}.$

2006AimeA7.PNG

Solution

Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof.

Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at $0, 1, 2 \ldots$. The base of region $\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\mathcal{D}$ is on the line $x = 7$. Cutting the angle by half don't change the problem. After the cut, the top side of the angle will be $y = x - s$; the bottom side will be x-axis.

Since the area of the triangle is equal to $\frac{1}{2}bh$,

$\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}$

Solve this to find that $s = \frac{5}{6}$.

By a similar method, $\frac{Region \mathcal{D}}{Region \mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$ is $408$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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