Problem 12 – February 8th, 2019

Is It Really One-seventh?

I learned of this problem in a session on mathematics and art at the 2003 joint AMS/MAA meeting.

Given any triangle ABC, construct a line from each vertex to the opposite side as indicated in the figure below. If point D is one-third of the way from point B to C, point E is one-third of the way from C to A, and point F is one-third of the way from A to B, show that the triangle GHI has area one-seventh of the area of triangle ABC.

Submit your answers to mathpotw@acu.edu.  Details for submissions can be found here.

Correct solutions were submitted by Yunxi Wei.

Solution to Is It Really One-seventh?

We get a solution by tiling the plane using copies of the triangle GHI. The solution becomes apparent by looking at three parallelograms in the figure below, each of which has one of the sides of the triangle ABC as a diagonal.

In particular, look at the parallelogram BJCH. The area of the parallelogram is four times the area of triangle GHI, and by the apparent symmetry, the part of the parallelogram inside triangle ABC has area two times the area of triangle GHI. Similar arguments hold for parallelogram AKBG and parallelogram AICL.

Thus the area inside of triangle ABC is seven times the area of triangle GHI, and the problem is solved.

Hmmm! I wonder what would happen if we went 1/4 of the way along each side? Or how about 1/n th of the way?