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2006 AIME I Problems/Problem 7: Difference between revisions

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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]].


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>. The x-axis [[bisect]]s the angle; the top side will be <math>y = x - h</math>.
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>. The bottom side of the angle will be x-axis; the top side will be <math>y = x - s</math>.


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>,
Line 13: Line 13:
:<math>
:<math>
\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5}
\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5}
= \frac{\frac 12(5-h)^2 - \frac 12(4-h)^2}{\frac 12(3-h)^2 - \frac12(2-h)^2}
= \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}
</math>
</math>


Solve this to find that <math>h = \frac{5}{6}</math>.
Solve this to find that <math>h = \frac{5}{6}</math>.


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


== See also ==
== See also ==

Revision as of 20:34, 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 $\mathcal{A}.$

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 $x = 7$. The bottom side of the angle will be x-axis; the top side will be $y = x - s$.

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 $h = \frac{5}{6}$.

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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