2006 AIME I Problems/Problem 7: Difference between revisions
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== Solution == | == Solution == | ||
Let | Apex of the angle is not on the parallel lines. | ||
The set of parallel lines | |||
Let... | |||
One side of the angle be | *The set of parallel lines be | ||
The other side be y = x-h | :perpendicular to x-axis | ||
:& cross x-axis at 0, 1, 2... | |||
*Base of area A be at x = 1; Lower base of Area D at x = 7. | |||
*One side of the angle be x-axis. | |||
*The other side be <math>y = x - h</math> | |||
<br> | |||
Then... | Then... | ||
<br><br> | |||
As area of triangle = .5 base x height... | |||
<br><br> | |||
<math> | <math> | ||
Area C | \frac{Area C}{Area B} = \frac{11}{5} | ||
= | = \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2} | ||
</math> | </math> | ||
By similar method, D/ | <br><br> | ||
h = <math>\frac{5}{6}</math> | |||
By similar method, <math>\frac{Area D}{Area A}</math> seems to be 408. | |||
== See also == | == See also == | ||
Revision as of 18:37, 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 area A be at x = 1; Lower base of Area D at x = 7.
- One side of the angle be x-axis.
- The other side be
Then...
As area of triangle = .5 base x height...
h = 
By similar method, 
seems to be 408.
