Difference between revisions of "2006 AIME I Problems/Problem 7"
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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 19:37, 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.