Here's a fun little puzzle that came from an AIME exam paper in 2006:
Problem: In the diagram below, a triangle is drawn over a series of, equally-spaced, parallel lines. If the ratio of the regions C:B is 11:5, what is the ratio of the regions D:A?
Note: The apex of the triangle is not on any of the lines.
What is the ratio D:A?
Answer
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The answer is 408.
Solution
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Because the regions are divided by sub-triangles that are similar, we can distort them without loss of generality.
We can skew the triangles to make a horizontal base. (Or, if you will, slicing the original triangle in half, which will not change the ratio of the areas).
The parallel lines are equally spaced, so let's label them starting at zero.
We'll define the distance from the first line to the apex of the triangle to be x.
Because changing the scale factor of the vertical axis does not affect the ratio of the areas, we can scale this to be something convenient to make the math simpler. We'll scale so that the base of the triangle in region A is (1-x).
We can now look at the ratio of the areas. We know C:B is 11:5. The area of a triangle is: ½ base × height.
We can work out the areas of each region by subtracting the appropriate similar triangles.
This gives us a value for x in our geometry. We can now plug this into the desired ratio to get the answer.
The ratio D:A is 408.