2006 AIME I Problem 7

With unit spacing, every region is a difference of similar triangles, so areas scale as squares of distances from the vertex

If $x$ is the vertex’s distance to the first line, then $\frac{\mathcal{C}}{\mathcal{B}} = \frac{2x+7}{2x+3} = \frac{11}{5}$ determines $x$

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

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

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