# Paper Folding: The proof of the cube root applet

Last week, we have discussed the second part of our Paper Folding series, a fold that extracts the cube root of any number.   In this post, we are going to discuss its proof, but before that, let’s recall how to do the paper fold.

Paper Folding Instructions

1. Get a rectangular piece of paper and fold it in the middle, horizontally and vertically, and let the creases (see green segments in the applet) represent the coordinate axes.
2. Let denote (0,1) and let R denote (-r,0).
3. Make a single fold that places M on y = -1 and R on x=r.
4. The x-intercept of the fold is $\sqrt[3]{r}$.
The GeoGebra applet below visualizes the fold. Drag points P and Q to satisfy the conditions above. Note that you can also move point R

Theorem: Prove that if M on $y = -1$ and $R$ is on $x = r$, the intersection of the fold and the x-axis is at $\sqrt[3]{r}$.

Proof:

One possible fold satisfying the conditions above is shown in the figure below. Checking the Hint check box reveals three triangles: MNO, POR, and PON.

It is clear that the three triangles are similar (Can you see why?).This implies that

$\frac{ON}{OM} = \frac{OP}{ON} = \frac{OR}{OP}.$

Equating the first and second ratios, we have  $\frac{ON}{1} = \frac{OP}{ON}$ which means that $OP = ON^2$.  Also, equating ratios 1 and 3, we have $\frac{ON}{1} = \frac{OR}{OP}$ which results to $ON = \frac{OR}{ON^2}$. This means that $OR = ON^3$ giving us $ON = \sqrt[3]{OR}$. But the length of $OR = r$, therefore $ON = \sqrt[3]{r}$ and we are done with our proof.