2006 AIME I Problem 7

Attempt Problem 7 of the 2006 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME I solutions, or check the answer key.

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7.

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region C\mathcal{C} to the area of shaded region B\mathcal{B} is 115.\frac{11}{5}. Find the ratio of the area of shaded region D\mathcal{D} to the area of shaded region A.\mathcal{A}.

Answer: 408
Concepts:similarityarea ratiodifference of squares
Difficulty rating: 2500
Solution:

Take the spacing between consecutive lines as the unit, and let xx be the distance from the vertex of the angle to the first line. The triangle cut off by the jjth line is similar to triangle A\mathcal{A} with ratio x+j1x,\frac{x + j - 1}{x}, so its area is proportional to (x+j1)2,(x + j - 1)^2, and the strip between lines jj and j+1j + 1 has area proportional to (x+j)2(x + j)^2 (x+j1)2=2x+2j1.- (x + j - 1)^2 = 2x + 2j - 1.

Regions B,\mathcal{B}, C,\mathcal{C}, and D\mathcal{D} are the strips beginning at lines 2,2, 4,4, and 6,6, with areas proportional to 2x+3,2x + 3, 2x+7,2x + 7, and 2x+11.2x + 11. The given ratio yields 2x+72x+3=115,\frac{2x+7}{2x+3} = \frac{11}{5}, so 10x+35=22x+3310x + 35 = 22x + 33 and x=16.x = \frac{1}{6}.

Since A\mathcal{A} has area proportional to x2,x^2, the requested ratio is 2x+11x2=34/31/36=408.\frac{2x + 11}{x^2} = \frac{34/3}{1/36} = 408.

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