Difference between revisions of "2006 AIME I Problems/Problem 7"
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m (→Solution) |
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| Line 13: | Line 13: | ||
:perpendicular to x-axis | :perpendicular to x-axis | ||
:& cross x-axis at 0, 1, 2... | :& cross x-axis at 0, 1, 2... | ||
| − | *Base of | + | *Base of Region <math>\mathcal{A}</math> be at <math>x = 1</math>; Lower base of Region <math>\mathcal{B}</math> at <math>x = 7</math> |
*One side of the angle be x-axis. | *One side of the angle be x-axis. | ||
*The other side be <math>y = x - h</math> | *The other side be <math>y = x - h</math> | ||
| Line 22: | Line 22: | ||
<br><br> | <br><br> | ||
<math> | <math> | ||
| − | \frac{ | + | \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} | = \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2} | ||
</math> | </math> | ||
<br><br> | <br><br> | ||
| − | + | <math>h = \frac{5}{6}</math> | |
| − | By similar method, <math>\frac{ | + | By similar method, <math>\frac{Region \mathcal{D}}{Region \mathcal{A}}</math> seems to be 408. |
== See also == | == See also == | ||
Revision as of 18:41, 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
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
be at 
; Lower base of Region 
at 
- One side of the angle be x-axis.
- The other side be
be at 
; Lower base of Region 
Then...
As area of triangle = .5 base x height...
By similar method, 
seems to be 408.
