2006 AIME I Problems/Problem 7

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

Apex of the angle is not on the parallel lines.

Let...

The set of parallel lines be

perpendicular to x-axis

& cross x-axis at 0, 1, 2...

Base of Region $\mathcal{A}$ be at $x = 1$ ; Lower base of Region $\mathcal{B}$ at $x = 7$
One side of the angle be x-axis.
The other side be $y = x - h$

Then...

As area of triangle =.5 base x height...

$\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5} = \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2}$

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

By similar method, $\frac{Region \mathcal{D}}{Region \mathcal{A}}$ seems to be 408.

Page

Toolbox

Search

2006 AIME I Problems/Problem 7

Problem

Solution

See also