Chenying Liu

In the first article we discussed the legend behind the cube root of two through its links to Apollo and Plato. Do you remember the questions raised at the end: can you construct the cube root of two geometrically? In this article we’ll walk through a beautiful solution together – all you need is a single sheet of paper and some clever folding. Let’s begin our adventure into the world of origami!

The first step is as simple as you can expect: turning your piece of paper into a regular square. 

Start with a piece of A4 paper, which measures 210 x 297 milimeters. Hold the top left corner of the paper and fold it towards the lower right side. Then fold along the bottom edge of the triangle. Now tear along the seam of the fold. Finally, unfold the upper part and you’re left with a perfect square of side length 210 mm. Don’t worry if you don’t have a piece of A4 paper at hand, any rectangular sheet of paper will work by following the steps described above.

The next step is almost as easy as the first. We divide the paper into three equal segments.

The key to the problem is to locate trisection points on one side of the square sheet. For a square sheet, it’s quite straightforward to find the midpoint of one side by simply folding it to make two adjacent corners coincide. Once you’ve done this, fold the left bottom corner onto the midpoint. Then, we have our first trisecting point exactly at the intersection of the bottom and right sides (labelled as Q below).

Amazing, right? That’s also what I felt when I first saw this! Let’s see how this works…

Suppose we have a square ABCD with side length of unit 1. We fold the left bottom corner onto the top side at point P, which lies anywhere on AD. That means, P can slide freely between A and D without any restriction, as long as it’s on the line. Following this, you’ll notice the bottom side coincides with the right side at point Q. Suppose that the length of AP is 1/k (representing some proportion of the total length of the side AD), and so PD = 1 – 1/k. Let AR = x. Then PR = BR = 1 – x (since AB has total length 1).

As the triangle APR is right-angled, Pythagoras’ Theorem tells us that we have

PR² = AR² + AP²

or substituting in for the expressions derived above

(1 – x)² = x²  + 1/k²

Rearranging this gives, x = (k² – 1)/2k². Further, the angle RPQ = 90 degrees (as it is a corner of the original square), which implies similarity between the triangles APR and DQP (angles PAR and QDP are right-angled while angles ARP and DPQ are equal as both are complementary angles of angle APR). Thus we can get AR x DQ = AP x DP. That is

(k² – 1)/2k² x DQ = 1/k x (1 – 1/k)

which leads to the result DQ = (k – 1) x 2k²/(k² x (k² – 1) = 2/(k+1).

Now, going back to our situation, the folding process we have used is based on the midpoint P, which means k = 2. Hence we have DQ = 2/3, proving that Q is the lower trisection point of the right-hand side CD.

There are of course numerous methods to divide the square into three equal segments; however, I will leave the choice to you. We are ready to move on to the most exciting and challenging part…

Here comes the final target. Place the well-creased sheet flat on the table. Again, bring the bottom left-hand corner to the top edge of the paper. At the same time, adjust the folding crease and make the left trisection point Q on the bottom meet the right trisection line as shown in the diagram below (this is definitely a good way to exercise your fingers!). And that’s it, we are done! Can you find the cube root of two?

The answer is the ratio of PB/PA. If you don’t believe it, use a ruler to measure the length of each line and you’ll get the ratio to be around 1.26, approximately the value of 2^1/3. Shocked? I know I was the first time i saw this! In fact, when I was a little girl, I would often think to myself ‘how on earth did people come up with this idea’? Was it magic or non-human power? But as we’ve seen here it’s in fact neither – it’s just mathematics. And even better, it’s something that can be learned by anyone. 

Now, before you move on to the next – and final – article, I’ll leave you with another question to ponder. How can you demonstrate by pure geometry that the ratio above is exactly what we are looking for? You may want to start with a similar demonstration procedure to the one we used for the trisection point, in particular using the similarity between two right-angled triangles… I’ll see you in part 3 for the answer!