*Chenying Liu*

In the last article, together we mastered origami techniques to create the cube root of two. The discussion ended with a question: how can we show mathematically that the ratio we have constructed is indeed the cube root of two? Stumped? Check out the answer below.

To demonstrate the ratio PBPA (as shown in the diagram below) is the cube root of 2, we first assume PA to be of unit length 1 and call the length PB = x. Our task is then to show x = 2^{1/3}.

Let AR = k, then PR = CR = 1 + x – k (since the total length of one side is 1 + x). In triangle APR, Pythagoras’ Theorem tells us PR^{2} = AR^{2} + AP^{2}. From here we can solve for k as (x^{2} + 2x)/(2x + 2) and thus PR = 1 + x – k = 1 + x – (x^{2} + 2x)/(2x + 2).

Using the similarity between triangles APR and triangle EQP (if you don’t remember why these are similar triangles, check the last article – corresponding angles and sides are labelled in the diagram above), we have

PR/PQ = AR/EP

Now remember point Q is a trisection point on line CD (1/3 of the way along) and point E is a trisection point on AB (2/3 of the way along). Therefore, we know PQ = (1+x)/3, EP = AE – AP = (2x+2)/3 – 1. Substituting this into the expression above gives:

If we multiply up by the denominators on both sides and simplify the expressions to make the equation look neater, we have x^{3} + 3x^{2} + 2x = 2x^{3} + 3x^{2} + 2x – 2. Cancelling terms this gives x^{3} = 2, which proves the result.

Now, are you convinced origami is not only the art of paper folding, but also a useful tool for mathematical problems? Indeed, there are lots of fantastic examples of how origami is used by mathematicians, including to solve two of the three classical Greek geometric problems (the other being angle trisection). Nonetheless, we’ll leave the origami discussion here as it’s time to **draw** the cube root of two on a stone surface which refuses to fold over…

Imagine you were Plato and wanted to extend the altar’s side to 2^{1/3} times its current length. Can it be done with only a compass and ruler? You might be confused, as I’ve previously mentioned that it’s impossible with an unmarked straight-edge, but here we are going to use a marked ruler instead, and it’s this slight difference of adding a measure which makes all the difference. Here are the exact steps…

Grab a pen and mark the ruler with the given altar length. To simplify, we assume this to be the unit length of 1. For example, if you have a 15cm ruler, we can set 5cm (or any other number smaller than 15) as unit 1. Fortunately, it doesn’t need to be as long as the actual altar’s length, only a projection of the real world (which is good, as the actual altar length is 10 feet and it’s quite unlikely you have a ruler that long!).

Next, draw an equilateral triangle ABC with each side of the given unit length. This can be done as follows. First, draw a line BC of the given length. Place your compass point on B and measure the distance to point C. Swing an arc of this size above (or below) the line. Without changing the span on the compass, place the compass point at C and swing the same arc, intersecting with the first one. The intersection of the two arcs is point A which is the third vertex of an equilateral triangle with three angles of 60-degrees.

Next, extend AB by the unit length to point D. Also extend BC and DC as much as you can to the points E and F shown below.

All done? The last challenge is coming! Place the ruler so it goes through point A and adjust its direction to make one end of the marked length fall on line CF and another end on line CE. In other words, the distance along the ruler between the lines CF and CE should be the ‘unit length’ (if you’re still unsure see the diagram below). Thus, the intersection points form a line GH with the given unit length. Then, the length of line AG (shown in the diagram below) is exactly what we want: 2^{1/3} times the given length.

So, the big question is why is this true? To answer this, we need Menelaus’ theorem. This says that when a triangle is trisected by a line, the ratio of the lengths AD and DB, multiplied by the ratio of lengths BE and EC, and the ratio of lengths of CF and FA (where the points are as shown in the diagram), will always give 1.

For example, given AB = 5, AC = 4, AD= 2, and AF = 3, we’ll know the length of BC as long as the length of CE is given. So, if CE = 2, then we can have BC = 7. The triangle and trisecting line is illustrated below.

If you are still curious about how this works, see the demonstration below (skip if you are not interested as this won’t influence the demonstration of the cube root of two).

We start by drawing three perpendiculars from points A, B, and C to the transversal (yellow line) at points H, G, and I respectively. This creates three pairs of similar triangles: BGD and AHD, AHF and CIF, CIE and BGE. They are similar because each pair has a right angle and the same vertex or shared angle. In particular:

- Triangle BGD is similar to triangle ADH, so we have AD/DB = AH/BG;
- Triangle AHF is similar to triangle CIF, so we have CF/FA = CI/AH;
- Triangle CIE is similar to triangle BGE, so we have BE/EC = BG/CI.

Multiplying the left-hand sides and right-hand sides of the three equations respectively, gives us Menelaus’ theorem.

Now, let’s apply this to the diagram we’ve constructed:

You can hopefully see that both triangles ACD and ACG are right-angled. Because the triangle ABC is equilateral, we know each of its inner angles are 60 degrees. Points A, B, and D are collinear and thus we have angle DBC = 120 degrees. Since triangle BCD is isosceles, then angle BCD must be 30 degrees. Now it’s clear that line AC is perpendicular to line DG and then we have two right angles. Thus, by Pythagoras’ Theorem, CD^{2} = AD^{2} – AC^{2} = 4 – 1 = 3. If we now set AG = x, then again by Pythagoras’ Theorem we have CG^{2} = x^{2}-1.

Nearly there! Taking stock of all of the lengths that we have calculated so far, and spotting that the line BCH goes through triangle ADG, we can apply Menelaus’ theorem and solve the resulting equations to reach the final solution. To check your answer, scroll down below the avatar of Plato…

According to Menelaus’ theorem, we have AB/BD * DC/CG * GH/HA = 1 for triangle ADG. That is: 1/1 * (3/(x^{2}-1))^{1/2} * 1/(x+1) = 1.

Squaring both sides of the equation and rearranging we have x^{4} + 2x^{3} – 2x -4 = 0 which can be factorised as (x + 2)(x^{3} – 2) = 0.

The only positive real root is therefore 2^{1/3 } – exactly as we wanted!

So there we have it, we’ve succeeded where even the great Plato failed by drawing the cube root of 2. I hope you had a pleasant journey through this series of articles – thanks for reading and hopefully see you soon! 👋

[…] point, in particular using the similarity between two right-angled triangles… I’ll see you in part 3 for the […]

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