2006 AIME I Problems/Problem 7
Contents
[hide]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 1
Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof.
Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at . The base of region is on the line . The bigger base of region is on the line . Let the top side of the angle be and the bottom side be x-axis, as dividing the angle doesn't change the problem.
Since the area of the triangle is equal to ,
Solve this to find that .
Using the same reasoning as above, we get , which is .
Solution 2
Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be and the area of it be . Also, let all sections of the line on the same side as the side with length on a trapezoid be equal to .
Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is . Multiplying, we get as the area of the triangle, so the area of the trapezoid is . Repeating this process, we get that the area of B is , the area of C is , and the area of D is .
We can now use the given condition that the ratio of C and B is .
gives us
So now we compute the ratio of D and A, which is
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.