This post was inspired by a problem I came across when studying for a math contest, namely problem 23 in the 2007 Cayley Contest.
I think this is an interesting problem because the solution is not so obvious, and there are several incorrect approaches that I’ve tried (and given up on). I’ve modified the problem slightly to make it more interesting.
Here we have a rectangle, . It’s rotated to the right by degrees, using as the pivot. This forms as the new rectangle. We are given and , we are asked to find the area of the shaded region.
Can you come up with a general formula for the area of the shaded region, using the given variables , , and ?
I will use the notation to denote the area of triangle .
The solution is to draw a line parallel to through , the blue line. This splits the lower shaded area into two triangles: and , and a rectangle: .
Since is the same as , the top shaded area is the same as the bottom shaded area.
Let’s find the area of triangle .
The angle is equal to . Of course, is equal to . Using some basic trigonometry, , , and . Now we know the area of triangle :
Next we find the area of . , and using trigonometry again, .
Finally, for the rectangle :
Now we add all this up and simplify ( is total):
It’s rather tedious to get from the second to the last step, but Wolfram Alpha could do it for me.
Notice that we assume that intersects . If is large enough, then would intersect , and this formula would no longer work.
Of course it would also fail if is greater than , so my formula works for only a rather limited range of values.