2006 AIME I Problems/Problem 7: Difference between revisions
| Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
Let: | 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... | 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 
to the area of shaded region 
is 11/5. Find the ratio of shaded region 
to the area of shaded region 
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...
![$Area C / Area B = 11 / 5 = [.5(5-h)^2 - .5(4-h)^2] / [.5(3-h)^2 - .5(2-h)^2] Thus h = 5/6$](http://latex.artofproblemsolving.com/2/1/3/21319d787c1bb14478c1d28b517db8fcaf74d6de.png)
By similar method, D/A seems to be 408.
