2006 AIME I Problems/Problem 7

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 ,

Region /Region = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2} $$ (Error compiling LaTeX. Unknown error_msg)

Solve this to find that .

Using the same reasoning as above, we get Region /Region = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}\boxed{408}$.

== 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$ (Error compiling LaTeX. Unknown error_msg)xx^2x1$.

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$ (Error compiling LaTeX. Unknown error_msg)(\frac{x+1}{x})^2(x+1)^22x+12x+32x+72x+11$.

We can now use the given condition that the ratio of C and B is$ (Error compiling LaTeX. Unknown error_msg)\frac{11}{5}\frac{11}{5} = \frac{2x+7}{2x+3}x = \frac{1}{6}\frac{2(\frac{1}{6} + 11)}{(\frac{1}{6})^2} = \boxed{408.}$

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.