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

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
 
Let:
 
Let:
  The set of parallel lines...
+
The set of parallel lines...perpendicular to x-axis & cross x-axis at 0, 1, 2...
      perpendicular to x-axis &
+
The base of area A is at x = 1.
      cross x-axis at 0, 1, 2...
+
One side of the angle be the x-axis.
 
+
The other side be y = x-h...
  The base of area A is at x = 1.
+
as point of the angle isn't on parallel lines
 
 
  One side of the angle be the x-axis.
 
 
 
  The other side be y = x-h...
 
      as point of the angle isn't on
 
      parallel lines
 
 
 
 
Then...
 
Then...
 
+
<math>
 
Area C / Area B = 11 / 5
 
Area C / Area B = 11 / 5
 
= [.5(5-h)^2 - .5(4-h)^2] / [.5(3-h)^2 - .5(2-h)^2]
 
= [.5(5-h)^2 - .5(4-h)^2] / [.5(3-h)^2 - .5(2-h)^2]
 
Thus h = 5/6
 
Thus h = 5/6
 
+
</math>
 
By similar method, D/A seems to be 408.
 
By similar method, D/A seems to be 408.
  

Revision as of 18:22, 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 $\mathcal{A}.$



Solution

Let: The set of parallel lines...perpendicular to x-axis & cross x-axis at 0, 1, 2... The base of area A is at x = 1. One side of the angle be the x-axis. The other side be y = x-h... as point of the angle isn't on parallel lines Then... $Area C / Area B = 11 / 5 = [.5(5-h)^2 - .5(4-h)^2] / [.5(3-h)^2 - .5(2-h)^2] Thus h = 5/6$ By similar method, D/A seems to be 408.

See also