In the parallelogram below, each vertex has a line from it to the midpoint of the opposite 2 sides. What is the ratio of the area of the shaded region to the area of the entire parallelogram?

Source: Giles Cohen, 50 Mathematical Puzzles and Problems: Red Collection, Key Curriculum Press, 2001.

Level: Senior

Let the center of the parallelogram be the center of 2D plane. Rotate the y-axis to coincide with the closest side of the parallelogram. While preserving the desired ratio, we can treat the parallelogram as a rectangle. Consider the quarter rectangle in the first quadrant. Without loss of generality, scale the axes so that the quarter parallelogram becomes a square with side length 2. We can show that the three lines forming the shaded region have equations
(1) y = x/2+1;
(2) y = -x/2+1;
(3) y = -2x+2.
Solving (1) and (3), we get 5x/2 = 1 or x = 2/5, and so y = 6/5. Solving (2) and (3), we get 3x/2 = 1 or x = 2/3, and so y = 2/3. The leftmost point of the shaded region is clearly at (0, 1), which is the solution to (1) and (2). It follows that the 3 vertices of the shaded region are (0, 1), (2/5, 6/5), and (2/3, 2/3). It follows that the area of this triangle is
|det(1 1 1; 0 2/5 2/3; 1 6/5 2/3)|/2
= |(4/15-12/15)+2/3-2/5|/2
= |(-8+10-6)/15|/2
= 2/15.
We know that the area of the entire square is 4(2*2) = 16. Therefore the ratio of the shaded region to the area of the entire square is 2/15/16 = 1/120, which is also the area of the shaded region to the area of the entire parallelogram.