I got this problem from Rustan Leino, who was asked it by Vlad Rusu, who in turn heard it from Grigore Rosu and family.
I solved it and wrote up my solution.
Two arbitrary rectangles are placed to form an "L". That is, the lower left-hand corner of the two rectangles share the same point. (What I'm trying to say is that there's an "L" whose "I" and "_" parts have arbitrary widths and heights.) Using only a (pen and a) straightedge (that is, no measuring device and no compass), figure out a way to, with a single straight cut, divide the "L" into two pieces of equal area.
Let $L$ be the "L" shape. It consists of a rectangle $R$ with a smaller rectangle $r$ cut out of its upper right corner.
Use the straightedge to extend the top edge of $L$ to the right, and then to extend the right edge of $L$ up. In this way, you wind up drawing the remaining edges of $R$ and $r.$
Given any rectangle, you can draw both its diagonals with the straightedge and thereby obtain their intersection, which is the center of the rectangle. Do this to obtain the centers of $R$ and $r.$
Finally, use the straightedge to draw a line passing through the centers of $R$ and $r.$ This line bisects $L,$ by the following reasoning. Any line drawn through the center of a rectangle bisects that rectangle; since our line passes through the centers of both $R$ and $r,$ it bisects both. The area of $L$ is that of $R$ minus $r,$ so bisecting both also bisects $L.$