2006 AIME I Problems/Problem 7
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
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 .
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 .
So now we compute the ratio of D and A, which is
Solution 3 (Bash)
Let the distances from the apex to the parallel lines be and and the distance between the intersections be We know the area ratio means which simplifying yields The ratio we seek is We know that so the ratio we seed is Finally note that by similar triangles Therefore the ratio we seek is Finally note that so the final ratio is
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