Areas in Square by Dissection

The July 15 issue of the NCTM Newsletter contained the following problem:

In the figure below, quadrilateral ABCD is a square, and E is the midpoint of the side AD. How do the areas of regions I, II, III, and IV compare? Another way to think about this is to consider the question, What are the ratios of the areas of the four subdivisions, I : II : III : IV?

The August issue offered a very nice solution by dissection.

The applet below presents a generalization of the problem and its solution.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Buy this applet
What if applet does not run?

Let DE = AD/N, N ≥ 2. Then

I : II : III : IV = 1 : N : N² : N² + N - 1.

The key to these ratios is the fact that, if points are marked on the sides of a triangle that divide the sides into N parts then the lines parallel to the sides joining the points split the triangle into N² equal triangles. This follows from the simple dependency of the area on the linear dimensions of a triangle (but is true for any other shape.) If the sides of a triangle are increased by the factor of N, then its area increases by the factor of N². This is the same dependency that was used by Euclid in his second and less known proof of the Pythagorean theorem.

49552088