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

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An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> \mathcal{C} </math> to the area of shaded region <math> \mathcal{B} </math> is 11/5. Find the ratio of shaded region <math> \mathcal{D} </math> to the area of shaded region <math> \mathcal{A}.  </math>
 
An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> \mathcal{C} </math> to the area of shaded region <math> \mathcal{B} </math> is 11/5. Find the ratio of shaded region <math> \mathcal{D} </math> to the area of shaded region <math> \mathcal{A}.  </math>
  
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[[Image:2006AimeA7.png]]
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[[Image:2006AimeA7.PNG]]
  
 
== Solution ==
 
== Solution ==

Revision as of 19:22, 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

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.

See also

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