EDIT: I have a better answer, but had too much help to claim it as my own! See the first comment.

I have a sort-of answer to this week’s Riddler Classic. Probably not the exact answer sought, but it’s further than I usually get!

Say you have an “L” shape formed by two rectangles touching each other. These two rectangles could have any dimensions and they don’t have to be equal to each other in any way. (A few examples are shown below.)

Using only a straightedge and a pencil (no rulers, protractors or compasses), how can you draw a single straight line that cuts the L into two halves of exactly equal area, no matter what the dimensions of the L are? You can draw as many lines as you want to get to the solution, but the bisector itself can only be one single straight line.

I suggest doing this:

- Draw a line that divides the shape back up into its two original rectangles.
- Draw both diagonals in both original rectangles (four lines in total). The intersections of the diagonals are the centres of the rectangles
- Draw a straight line that connects the two centres and continues in both directions until it has passed through as much of the shape as possible.
*This is the bisector.*

This line *does* divide the shape up into “two halves”, in the sense that both sides of the lines have equal area. However, it doesn’t always divide the original shape up into two continuous parts: for lots of possible shapes, the line passes through two parts of the shape, and divides one side of the line into two parts, not just one.

So these sections *are* halves, in the sense of being exactly 50% of the original area, but they are *not* halves in the sense of each being only one part. It would feel like a bit of a trick if this was the real answer!

I got diverted into a discussion on r/askmath about the centres of shapes:

https://www.reddit.com/r/askmath/comments/904xhk/centres_of_2d_shapes/

But from that, with some hints from u/wijwijwij , I got the better answer that I figured must exist.