2006 AIME I Problem 7

Below is the professionally curated solution for Problem 7 of the 2006 AIME I, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:similarityarea ratiodifference of squares

Difficulty rating: 2500

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}.

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+j1)2=2x+2j1.(x + j)^2 - (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.

← Problem 6Full ExamProblem 8

Problem 7 in Other Years