The power of origami

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The power of origami

December 2009

Doubling the cube using origami — proof

Figure 1
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 B at the point (0,0). See figure 1. Now the line from A to C has equation y=sx, where s is the side length of the square. The line from B to E has equation y=1/2x. So the two lines meet at a point with x-coordinate satisfying 1/2x=sx, so x=2/3s. The y-coordinate of the intersection point is therefore y=s2/3s=1/3s, as required.

The construction claims that the side length s2 of the new cube is ACCB times the side length s1 of the given cube. Now since s2=23s1, we have 23s1=ACCBs1, so we need to show that ACCB=23.

Figure 2
Figure 2

Firstly, we can assume that the length CB is equal to 1, as we can scale the image as we like without affecting the proportions. We'll write X for the length of the line segment AC, so the side length of our square is equal to X+1. See figure 2. The line segment BF corresponds to two thirds of a side of the square and therefore has length 2X+13. This implies that the length of the line segment CF is 2X+131=2X13. We write d for the length BJ. Since the line segment CJ is part of the bottom edge, its length is X+1d. There is a right angle at B, so by Pythagoras' theorem we have d2+1=(X+1d)2. This means that d2+1=X2+2X+12d(X+1)+d2, and therefore d=X2+2X2(X+1). Now consider the two right-angled triangles CFI and CBJ. Considering the 180 degree angle formed at C by the vertical edge of the square, we get α=180δ90=90δ. The angles of the triangle CBJ add up to 180 degrees, so γ=18090δ=90δ=α. So the triangles CFI and CBJ have two equal angles: the right angle and the angle α. This means that they are similar. Therefore, the ratios of corresponding sides of the two triagles are equal. Observing that the side CI of the triangle CFI is a third of the side of the square and therefore has length X+13, we get dX+1d=2X13X+13=2X1X+1. Substituting the expression for d above gives X2+2XX2+2X+2=2X1X+1, so X3+3X2+2X=2X3+3X2+2X2, so X3=2. This proves the result.

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