How to fold a square of paper into perfect thirds, and how to prove it works

How to fold a square of paper into perfect thirds, and how to prove it works

I never liked maths lessons when I was younger, but I do like origami. I've been making origami for fun for years, but I've only recently started learning that as well as being a fun creative pursuit, it can also function as a tool for solving problems in maths, just like the geometry you were taught at school.

The kind of geometry you usually learn at school is called "plane geometry". It's the geometry of flat surfaces, and it comes with its own set of rules, or axioms, which define how the system works, and what you can and can't do in order to solve problems. At school you usually learn about plane geometry through challenges where you're asked to solve some problem just using a drawing compass and a special unmarked ruler called a straight edge. Classic puzzles include how to split an angle or a line exactly in two, or how to draw a perfect equilateral triangle.

But guess what: straight edge and compass construction is just one of many ways to study geometry. We can explore all these problems and more using the solving power of a simple piece of paper ... no additional equipment needed!

Some problems that take a bit of thought using the straight edge and construct approach are trivial to solve using origami. For example, bisecting a line can be as easy as folding a sheet of paper in half! Origami geometry still deals with flat surfaces, but it actually offers an extended problem solving tool kit, simplifying many challenges.

In this lesson I'll show you how to use origami to solve the problem of dividing a square of paper into thirds, and then I'll show you how to construct an algebraic proof to demonstrate that this technique really works.

Along the way we'll learn or refresh useful maths skills, like working with quadratic equations and solving problems with Pythagoras’ Theorem.

You can work through this lesson just by looking at the diagrams on screen, but you may find it more interesting if you grab a square of paper and follow along in real life. You can usually buy specially made squares of origami paper from craft stores, or you can make your own square using this simple technique. You might also want to keep a pen and paper handy to help with some of the algebra.

1. The solution

We're going to start at the end of this problem by looking at the solution first. Follow these instructions to divide your sheet into thirds:

By folding the paper in this way we are creating an intersection where the paper overlaps itself at the bottom right of the sheet. This intersection occurs e $\frac{2}{3}$ of the way down the sheet at the point where the bottom edge of the sheet crosses the right edge. We can now mark this point with a second fold before unfolding everything.

Okay, so you think you're finished? Wrong! We've completed the task, but how do we know this technique really works? It certainly looks like we've split the sheet into equal thirds. If you check with a ruler you'll get values that seem to confirm this. But we need to show that this solution is foolproof, and not just a very close estimate of the correct answer.

2. Line segments

Let's start by assuming our square has sides of length 1. This could mean 1 centimetre, 1 inch, 1 metre, or even 1 mile. This proof is meant to work for a square of any size, so it actually doesn't matter what value we use, but the maths will be easier if we just pick "1" for now. We'll also add some letters to our diagram to make it easier to define some of the line segments.

In this diagram, point $E$ is crucial, because this is the intersection which is supposed to mark the $\frac{2}{3}$ point along the edge of our sheet. To prove that our method works, we need to demonstrate that the line segment $D E$ is exactly equal to $\frac{2}{3}$ . For this proof to be valid, we can’t make any assumptions about the length of $D E$ yet. Instead we are going to figure out the lengths of the other line segments and use this information to prove that $D E$ must be equal to $\frac{2}{3}$ .

Let’s see how many line segments we can evaluate right now. We already know that $A D = 1$ , but what else can we figure out?

Here’s an easy one to start with: what is the value of length $A C$ ?

Not quite. We made point $C$ by folding the sheet in half, so $C$ must be at the exact halfway point of the line $A D$ .

Not quite. We made point $C$ by folding the sheet in half, so C must be at the exact halfway point of the line $A D$ .

Correct! By folding the sheet in half right at the start we found the halfway point of $A D$ . We also know that $C D$ = 1/2 as well.

Here's a trickier one: how long is the line legment $BC$ ?

Perfect. We dont know the value of $BC$ outright, but we do know something about it. The line segments $A B$ and BC fall along the same edge of our sheet. Since the sheet is a square with sides of length 1, $A B + BC$ must add up to 1. If we unfold the sheet, we can see that these two segments make up a whole edge. In algebra lingo, we're expressing $BC$ in terms of $A B$ . This is helpful because later on all we will need to do is find the value of $A B$ to solve $BC$ as well.

Sorry! This is cheeky, because we don't know the value of $BC$ outright. However, we do know something about it. The line segments $A B$ and $BC$ fall along the same edge of our sheet. Since the sheet is a square with sides of length 1, $A B + BC$ must add up to 1. If we unfold the sheet, we can see that these two segments make up the a whole edge. In algebra lingo, we're expressing $BC$ in terms of $A B$ . This is helpful because all we need to do is find the value of $A B$ to solve $BC$ as well.

Take another look at the sheet. You'll notice that we've actually formed two triangles; one in each of the top corners. We'll call these $X$ and $Y$ . Since this is a square, we know that these must be right angled triangles. This is important, because we can use a special tool to help understand some more about them.

3. Side lengths of triangles

Forget the sheet for a moment and look at the triangle below. What's the length of side $r$ ?

That's right. I bet you solved this using Pythagoras’ Theorem. This is a helpful tool for calculating the length of a side on a right angled triangle if you know the length of the other two sides already. $a^{2} + b^{2} = c^{2}$ , where $c$ is the hypotoneuse, or longest side, of the triangle (hint: the longest side is always the side opposite the right angle). In this case:

$3^{2} + 4^{2} = r^{2}$

$9 + 16 = r^{2}$

$r^{2} = 25$

$r = 25$

$r = 5$

Pythagoras’ Theorem uses the fundamental properties of all right angled triangles to let us calculate the length of a side on a right angled triangle if you know the length of the other two sides already. $a^{2} + b^{2} = c^{2}$ , where $c$ is the hypotoneuse, or longest side, of the triangle (hint: the longest side is always the side opposite the right angle). In this case:

$3^{2} + 4^{2} = r^{2}$

$9 + 16 = r^{2}$

$r^{2} = 25$

$r = 25$

$r = 5$

Pythagoras’ Theorem uses the fundamental properties of all right angled triangles to let us calculate the length of a side on a right angled triangle if you know the length of the other two sides already. $a^{2} + b^{2} = c^{2}$ , where $c$ is the hypotoneuse, or longest side, of the triangle (hint: the longest side is always the side opposite the right angle). In this case:

$3^{2} + 4^{2} = r^{2}$

$9 + 16 = r^{2}$

$r^{2} = 25$

$r = 25$

$r = 5$

Alright, let’s try another one of those. What’s the length of side $s$ in the triangle below?

Good work. Let's check this over:

$3^{2} + s^{2} = 8^{2}$

$9 + s^{2} = 64$

$s^{2} = 64 - 9$

$s^{2} = 55$

$s = 55$

$\approx 7.42$

One we've rearranged the equation to put $s$ on the left hand side, we can figure out the answer quite easily. We can use a calculator to figure out that it's roughly 7.42, or we can just leave $55$ for now, since this is more accurate.

Good work. Let's check this over:

$3^{2} + s^{2} = 8^{2}$

$9 + s^{2} = 64$

$s^{2} = 64 - 9$

$s^{2} = 55$

$s = 55$

$\approx 7.42$

One we've rearranged the equation to put $s$ on the left hand side, we can figure out the answer quite easily. We can use a calculator to figure out that it's roughly 7.42, or we can just leave $55$ for now, since this is more accurate.

Try again. This time we're not calculating the longest side. try writing out the formula and subsituting the numbers in:

$3^{2} + s^{2} = 8^{2}$

$9 + s^{2} = 64$

$s^{2} = 64 - 9$

$s^{2} = 55$

$s = 55$

$\approx 7.42$

One we've rearranged the equation to put $s$ on the left hand side, we can figure out the answer quite easily. We can use a calculator to figure out that it's roughly 7.42, or we can just leave $55$ for now, since this is more accurate.

Now that we've practiced Pythagoras’ Theorem, lets return to the original problem.

Take a good look at triangle X. What's the length of side $A B$ ? You need to express this as a either a whole number or a fraction.

Don’t worry, this step isn’t easy. We need to start by substituting our values into Pythagoras’ Theorem like we did before. This will give us:

$A B^{2} + (\frac{1}{2})^{2} = (1 - A B)^{2}$

We can expand this equation by multiplying out the brackets on the right hand side. If you need a bit of help here check out this guide.

$A B^{2} + \frac{1}{4} = (1 - A B) (1 - A B)$

$A B^{2} + \frac{1}{4} = 1 - 2 A B + A B^{2}$

Now we can balance up the equation. Notice that we can cancel out the $A B^{2}$ on each side to make this much easier to solve.

$2 A B + \frac{1}{4} = 1$

$A B = \frac{3}{8}$

See how important it was to rewrite $BC$ as $1 - A B$ ? This is what allows us to solve the equation.

Don’t worry, this step isn’t easy. We need to start by substituting our values into Pythagoras’ Theorem like we did before. This will give us:

$A B^{2} + (\frac{1}{2})^{2} = (1 - A B)^{2}$

We can expand this equation by multiplying out the brackets on the right hand side. If you need a bit of help here check out this guide.

$A B^{2} + \frac{1}{4} = (1 - A B) (1 - A B)$

$A B^{2} + \frac{1}{4} = 1 - 2 A B + A B^{2}$

Now we can balance up the equation. Notice that we can cancel out the $A B^{2}$ on each side to make this much easier to solve.

$2 A B + \frac{1}{4} = 1$

$A B = \frac{3}{8}$

See how important it was to rewrite $BC$ as $1 - A B$ ? This is what allows us to solve the equation.

That’s great work. If we use Pythagoras’ Theorem here we get:

$A B^{2} + (\frac{1}{2})^{2} = (1 - A B)^{2}$

We can expand this equation by multiplying out the brackets on the right hand side. If you need a bit of help here check out this guide.

$A B^{2} + \frac{1}{4} = (1 - A B) (1 - A B)$

$A B^{2} + \frac{1}{4} = 1 - 2 A B + A B^{2}$

Now we can balance up the equation. Notice that we can cancel out the $A B^{2}$ on each side to make this much easier to solve.

$2 A B + \frac{1}{4} = 1$

$A B = \frac{3}{8}$

See how important it was to rewrite $BC$ as $1 - A B$ ? This is what allows us to solve the equation.

Okay, so now we’ve calculated the length of the line $A B$ , let’s look back at the original diagram and figure out the length of side $BC .$

Just like with $A B$ , we need to express $BC$ as a whole number or a fraction.

There’s no need to overthink this! We know that $BC = 1 - A B$ and $A B = \frac{3}{8}$ , so $BC = \frac{5}{8}$ .

Perfect; no tricky equations to solve here. We know that $BC = 1 - A B$ and $A B = \frac{3}{8}$ , so $BC = \frac{5}{8}$ .

There’s no need to overthink this! We know that $BC = 1 - A B$ and $A B = \frac{3}{8}$ , so $BC = \frac{5}{8}$ .

Now let’s update our diagram to include all the side lengths we’ve calculated:

4. Similar triangles

We’ve done a great job with triangle $X$ , but triangle $Y$ is still causing us some problems. Remember, eventually we must find out the length of line of segment $D E$ to complete our proof.

It looks like we’re a bit stuck here. For Pythagoras’ Theorem to work we need to know the lengths of two sides of a triangle, but we only know the length of one side in triangle $Y$ . It’s time to open the toolkit and bring out another concept to get us through this next step: similar triangles.

Similar triangles are triangles which have the same shape, although their sizes may vary. When triangles are similar, their angles will always be the same, and their corresponding side lengths will always be in proportion to one another.

Here’s a quick test: an equilateral triangle is a triangle whose sides are all the same length. Are all equilateral triangles similar?

Incorrect. All equilateral triangles have internal angles of 60, 60 and 60 degrees. If these angles were any different, the side lengths wouldn’t be the same. You can vary the size of equilaterial triangles, but they will always be similar.

Spot on. All equilateral triangles have internal angles of 60, 60 and 60 degrees. If these angles were any different, the side lengths wouldn’t be the same. You can vary the size of equilaterial triangles, but they will always be similar.

Have a look at the triangles below. We can immediately see that they are similar, because their internal angles are the same. Can you figure out the missing values for sides $j$ and $k$ ?

Not quite right. In similar triangles, the lengths of corresponding sides are always proportional. We can find the missing values $k$ and $j$ by dividing the values in the larger triangle by 2. If we were starting with the values from the smaller triangle we would multiply them by 2 instead.

We can say that the reduction scale factor between these two triangles is 0.5

Likewise, we can say that the enlargement scale factor between them is 2.

Not quite right. In similar triangles, the lengths of corresponding sides are always proportional. We can find the missing values $k$ and $j$ by dividing the values in the larger triangle by 2. If we were starting with the values from the smaller triangle we would multiply them by 2 instead.

We can say that the reduction scale factor between these two triangles is 0.5

Likewise, we can say that the enlargement scale factor between them is 2.

That’s correct. In similar triangles, the lengths of corresponding sides are always proportional. We can find the missing values $k$ and $j$ by dividing the values in the larger triangle by 2. If we were starting with the values from the smaller triangle we would multiply them by 2 instead.

We can say that the reduction scale factor between these two triangles is 0.5

Likewise, we can say that the enlargement scale factor between them is 2.

Okay, let’s make sure we’ve got that. Here’s another question:

Triangles $A$ and $B$ are similar. Triangle $A$ has side lengths of 7, 13, and 15. The enlargement scale factor between these two triangles is 1.6. What are the side lengths for triangle $B$ ?

Take another look at the scale factor. We need multiply the original side lengths by 1.6 to get the new values.

Let’s go back to our original sheet of paper and take another look at triangles $X$ and $Y$ .

We know that triangles are similar if they share the same internal angles, so are $X$ and $Y$ similar? To help you with this, you’ll want to bear in mind two other facts:

The angles inside any triangle always add up to 180 degrees.

The angle of a straight line is always 180 degrees.

Not quite; we need to solve this with some lateral thinking. We can figure out the answer by adding in values like in the diagram below. The angles across the top of the sheet form a straight line (180 degrees). We don’t know the exact values of the angles on either side of the right angle, but we can call them $A$ and $90 - A$ for now, since $90 + A + 90 - A = 180$ .

Triangles $X$ and $Y$ both contain right angles. We can figure out from this that the remaining angles must correspond with each other!

$90 - A$ is the missing value that allows the internal angles of triangle $X$ to add up to 180 degrees. Meanwhile, $A$ is the missing value that does the same for triangle $Y$ .

Notice that we don’t need to know the exact values of these angles; we just need to know that they are the same in both triangles.

That’s correct; we need to solve this with some lateral thinking. We can figure out the answer by adding in values like in the diagram below. The angles across the top of the sheet form a straight line (180 degrees). We don’t know the exact values of the angles on either side of the right angle, but we can call them $A$ and $90 - A$ for now, since $90 + A + 90 - A = 180$ .

Triangles $X$ and $Y$ both contain right angles. We can figure out from this that the remaining angles must correspond with each other!

$90 - A$ is the missing value that allows the internal angles of triangle $X$ to add up to 180 degrees. Meanwhile, $A$ is the missing value that does the same for triangle $Y$ .

Notice that we don’t need to know the exact values of these angles; we just need to know that they are the same in both triangles.

Now that we’ve confirmed triangles $X$ and $Y$ are similar, we have a method for figuring out the value of length $D E$ in the diagram below. We just need to figure out the right scale factor.

What is the enlargement scale factor between triangles $X$ and $Y$ ?

Sorry! The side of length $\frac{3}{8}$ in triangle $X$ corresponds to the side of length $\frac{1}{2}$ in triangle $Y .$ The englargement scale factor is the amount we need to multiply $\frac{3}{8}$ by to get to $\frac{1}{2}$ .

Let scale factor = $s$

$\frac{3}{8} s = \frac{1}{2}$

$s = \frac{1}{2} \div \frac{3}{8}$

$s = \frac{8}{6} = \frac{4}{3}$

Sorry! The side of length $\frac{3}{8}$ in triangle $X$ corresponds to the side of length $\frac{1}{2}$ in triangle $Y .$ The englargement scale factor is the amount we need to multiply $\frac{3}{8}$ by to get to $\frac{1}{2}$ .

Let scale factor = $s$

$\frac{3}{8} s = \frac{1}{2}$

$s = \frac{1}{2} \div \frac{3}{8}$

$s = \frac{8}{6} = \frac{4}{3}$

Perfect. The side of length $\frac{3}{8}$ in triangle $X$ corresponds to the side of length $\frac{1}{2}$ in triangle $Y .$ The englargement scale factor is the amount we need to multiply $\frac{3}{8}$ by to get to $\frac{1}{2}$ .

Let scale factor = $s$

$\frac{3}{8} s = \frac{1}{2}$

$s = \frac{1}{2} \div \frac{3}{8}$

$s = \frac{8}{6} = \frac{4}{3}$

So, what’s the value of $D E$ ?

Try again. To figure out $D E$ we need to multiply the corresponding side in triangle $X$ by our scale factor $s$ .

$D E = \frac{1}{2} \times s$

$D E = \frac{1}{2} \times \frac{4}{3}$

$D E = \frac{2}{3}$

Try again. To figure out $D E$ we need to multiply the corresponding side in triangle $X$ by our scale factor $s$ .

$D E = \frac{1}{2} \times s$

$D E = \frac{1}{2} \times \frac{4}{3}$

$D E = \frac{2}{3}$

Correct! To figure out $D E$ we need to multiply the corresponding side in triangle $X$ by our scale factor $s$ .

$D E = \frac{1}{2} \times s$

$D E = \frac{1}{2} \times \frac{4}{3}$

$D E = \frac{2}{3}$

Check out the diagram below to see how everything fits together:

Congratulations! Let’s remind ourselves of what we’ve achieved here: we’ve found a way to describe lines, shapes and folded bits of paper using algebra; we’ve then modelled a geometrical problem using equations; we’ve then solved these equations using simple arithmetic to create a proof.

Along the way we’ve learned about the properties of triangles, practiced a couple of techniques to help us evaluate triangles, and got lots of practice working with equations. But that’s just the start! Check out future lessons to see how origami can help you to sharpen your maths skills.