# The power of origami

December 2009

### Doubling the cube using origami — proof Figure 1

First we show that the first three steps amount to dividing the height of the square into three equal parts. Place the square in a coordinate system with corner at the point . See figure 1. Now the line from to has equation where is the side length of the square. The line from to has equation So the two lines meet at a point with -coordinate satisfying so The -coordinate of the intersection point is therefore as required.

The construction claims that the side length of the new cube is times the side length of the given cube. Now since , we have so we need to show that  Figure 2

Firstly, we can assume that the length is equal to 1, as we can scale the image as we like without affecting the proportions. We'll write for the length of the line segment , so the side length of our square is equal to . See figure 2. The line segment corresponds to two thirds of a side of the square and therefore has length This implies that the length of the line segment is We write for the length . Since the line segment is part of the bottom edge, its length is . There is a right angle at , so by Pythagoras' theorem we have This means that and therefore Now consider the two right-angled triangles and . Considering the 180 degree angle formed at by the vertical edge of the square, we get The angles of the triangle add up to 180 degrees, so So the triangles and have two equal angles: the right angle and the angle This means that they are similar. Therefore, the ratios of corresponding sides of the two triagles are equal. Observing that the side of the triangle is a third of the side of the square and therefore has length , we get Substituting the expression for above gives so so This proves the result.