2006 AIME I Problems/Problem 7: Difference between revisions
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== Solution == | == Solution == | ||
{{solution}} | {{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 == | == See also == | ||
Revision as of 18:21, 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
This problem needs a solution. If you have a solution for it, please help us out by adding it. 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.
