Originally Posted by
ebaines Similar to the previous post, I have a way to solve this with only two measurements, not three, and without using the 7cm length information. It involves breaking the polygon into five adjacent trapezoids, each with vertical sides, and using the base of the rectangle as its base, and with slanted top edges, and then combining two adjacent trapezoids into a single trapezoid of equal area. This combining of trapezoids requires use of the compass and straightedge - the process is to find the midpoint of the upper edge of both adjacent trapezoids and then draw a straight line through those two points, creating a single trapezoid with combined width of the original two trapezoids, which now has the same area as the original two. Repeat this process four times and you end up with a single trapezoid with base length equal to the original large rectangle. The only trick is that for the right-most trapezoid, which has the triangle area cut out of the bottom, you can move that triangle to an equivalent area at the top - simply copy its height using the compass to extend the upper portion down an equal amount. After all five sections have been converted into one large trapezoid the last step is to convert that trapezoid into a rectangle - simply find the midpoint of its top slanted edge and draw a horizontal line through it. So now the original polygon has been converted into a rectangle with width equal to the large rectangle. Use the ruler to measure the height of the inner rectangle that has been converted from the original polygon (call this height x) and the height of the large rectangle (call this y), and the probability of the bug being in the original polygon is the same as the probability of it being in the equivalent rectangle, which is x/y. So again - only two measurements are needed.