2006 AIME I Problems/Problem 7

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Problem

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A$.

2006AimeA7.PNG

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 $0, 1, 2 \ldots$. The base of region $\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\mathcal{D}$ is on the line $x = 7$. Let the top side of the angle be $y = x - s$ 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 $\frac{1}{2}bh$,

Region $C$/Region $B$ = \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 $s = \frac{5}{6}$.

Using the same reasoning as above, we get Region $D$/Region $A$ = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$, which is$\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)x$and the area of it be$x^2$. Also, let all sections of the line on the same side as the side with length$x$on a trapezoid be equal to$1$.

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$. Multiplying, we get$(x+1)^2$as the area of the triangle, so the area of the trapezoid is$2x+1$. Repeating this process, we get that the area of B is$2x+3$, the area of C is$2x+7$, and the area of D is$2x+11$.

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}$gives us$x = \frac{1}{6}$So now we compute the ratio of D and A, which is$\frac{2(\frac{1}{6} + 11)}{(\frac{1}{6})^2} = \boxed{408.}$

See also

2006 AIME I (ProblemsAnswer KeyResources)
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

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