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

Revision as of 19:26, 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

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 $x = 1$ ; bigger base of region $\mathcal{D}$ 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.

@@ Line 11: / Line 11: @@
 :perpendicular to x-axis
 :& cross x-axis at 0, 1, 2...
-*Base of Region <math>\mathcal{A}</math> be at <math>x = 1</math>; bigger base of Region <math>\mathcal{D}</math> at <math>x = 7</math>
+*Base of region <math>\mathcal{A}</math> be at <math>x = 1</math>; bigger base of region <math>\mathcal{D}</math> at <math>x = 7</math>
 *One side of the angle be x-axis.
 *The other side be <math>y = x - h</math>

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

Revision as of 19:26, 11 March 2007

Problem

Solution

See also