Can you turn a beach ball into a rubber ring?

Alba Sendón Blanco

13 November 2026

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Imagine that you are sunbathing at the beach: the smell of the salty water, the sound of the waves… The peace is broken by a child crying: they cannot swim because they’ve forgotten their inflatable ring. “I have some scissors and glue in my bag,” you tell your friend. “Do you think we can turn our beach ball into a ring?”

“No, that is mathematically impossible!”, your friend answers. It turns out that this challenge is related to a famous problem posed by David Hilbert. The resolution involves the Euler characteristic, so we have a lot to cover before we can go back to relaxing.

Practising with cardboard

After seeing your astonished face, your friend hides the beach ball, takes a small cardboard sheet and says “Well, let’s first do this. Can you cut this sheet into polygonal pieces and glue them back edgewise to form any other polygon? For example, could you form an octagon as tall as a person?”

There is not enough cardboard, right? You can only possibly form polygons that have the same area as your initial sheet.

This seemingly harmless observation yields a very deep mathematical principle. In many fields in mathematics, we want to know when two objects are so similar that we can call them equivalent. For example, at school we learned that two triangles are equivalent if we can scale, rotate, translate and mirror one to make the other. Notice that we do not allow cutting and pasting operations here. However, in the cardboard example, we say that two polygons are equivalent when we can cut and paste one into the other. In general, it is a very difficult problem to know when two given objects are equivalent.

One thing one can try to do is to find properties that equivalent objects share: these are called invariants. For example, equivalent triangles have the same angles, and the area is a cut-and-paste invariant of polygons. This means that if two triangles have different angles they are not equivalent, and that it is not possible to cut and paste a polygon into another polygon with a different area. In general, if two objects do not share an invariant they cannot be equivalent. Mathematicians use invariants for distinguishing apart objects and hence contributing to their classification.

In the early 19th century, William Wallace, Farkas Bolyai and Paul Gerwien proved that given any two polygons with the same area, there is an algorithm to cut and paste one into the other. Their algorithm, illustrated below, turns any polygon $P$ into a rectangle with sides of length $1$ and $\operatorname{area}(P)$:

1. Cut $P$ into triangles.
2. Cut (orange dotted lines) and paste (blue lines) each triangle into a rectangle.
3. Cut and paste each rectangle into a rectangle with one side of length $1$.
4. Glue all rectangles together.

A polygon $P$ can always be cut and pasted into a rectangle with sides of length 1 and $\operatorname{area}(P)$ using this algorithm

Observe that all these steps are always possible: there are algorithms to execute them. Notice that all these movements can be reversed, so we can cut and paste any rectangle into any other polygon with the same area. This means that the area is the only cut-and-paste invariant for polygons. So with your cardboard sheet, scissors and glue, you can form any other polygon of the same area.

Going up one dimension

Now, your friend hides the cardboard sheet and shows a wooden block. “What about now? Can you cut this into three-dimensional pieces and glue them back, forming any shape you like?”

The answer is clear for you now: “No! They must have the same volume”. Very good, volume is a three-dimensional cut-and-paste invariant! But is it the only one? Given any two three-dimensional pieces of equal volume, is it always possible to cut and paste one into the other? This question is one of the 23 problems that the German mathematician David Hilbert posed in 1900.

One year later, Hilbert’s student Max Dehn answered the question negatively: he found a new three-dimensional cut-and-paste invariant, now known as the Dehn invariant. The cube and the regular tetrahedron have a different Dehn invariant, so we cannot cut and paste one into the other even when they have the same volume. In 1965, Sydler proved that the volume and the Dehn invariant are the only cut-and-paste invariants for three-dimensional solids.

Nothing stops mathematicians from asking the same questions for higher dimensions, but the difficulty of the answer increases with the dimension. In the 70s, Jessen extended Sydler’s results to four dimensions, and nowadays it is an open problem whether the volume and the Dehn invariant are the only cut-and-paste invariants for higher dimensions. See? Our initial beach game led to very deep open problems in mathematics!

Triangulating

Now, we are ready to go back to the beach ball. Since it is made of rubber, you can stretch it and contract it and hence the area is no longer preserved. Notice that when we were cutting and pasting polygons, our pieces were also polygons. The ball and the rubber ring can be seen as mathematical objects called surfaces.

Mathematicians call the ball a sphere and the rubber ring a torus. Another example of a surface is a disc; this has a boundary, a curve that separates it from the outside. We allow our surfaces to have a boundary, but we do not allow corners: a square would not be a surface for us. While cutting and pasting, we want our pieces to still be surfaces and hence we need to be careful and not allow for corners nor T-shapes. Throwing pieces away is not permitted, and we must glue back everything that we have cut. These rules are depicted above.

Creating corners, gluing into a T-shape and overlapping pieces are all illegal at this beach

Surfaces, like the ball and the ring, are mathematical objects present in various disciplines, such as dynamical systems, general relativity, geometry or quantum field theory. Mathematicians have developed a whole theory to understand surfaces, which can be challenging. One thing we can do is approximate surfaces by simpler structures, like triangles. This method is called triangulating the surface. Here you can see two different ways of triangulating one hemisphere of the beach ball (notice that the triangles can be curved):

And here you can see a triangulation of the upper half of the rubber ring:

There are infinitely many ways of triangulating the beach ball and the ring: given any triangulation, you can find a new one by dividing one of the triangles further into several triangles.

Nowadays, triangulations are not only used in abstract mathematics. For example, triangulations are extensively used in video games, and researchers might use the method of triangulation to make computer models of objects varying from vases to human organs.

The Euler characteristic

With this approximation of the surfaces by triangles, we can extract information about them. For instance, we can count how many vertices, edges and triangles our triangulation has. This, of course, depends on the triangulation we are considering: for example, if we triangulate both hemispheres of the beach ball using way 1 from the previous page, we get a triangulation of the beach ball with 17 vertices, 45 edges and 30 triangles (be careful not to count the vertices and edges in the equator twice!). If we do the same but with way 2, we get a triangulation of the beach ball with 5 vertices, 9 edges and 6 triangles. However, in both cases, if we add the number of vertices and of triangles and we subtract the number of edges, we get the same number: $17+30-45=5+6-9=2$.

This not only happens in this example, this quantity is really intrinsic to the surface: no matter which triangulation you choose, you always get the same number. We call this number the Euler characteristic $\chi$ of the surface:

We have just seen that the Euler characteristic of the beach ball is $2$: $\chi(\text{ball})=2$. On the other hand, our triangulation of the rubber ring has 12 vertices, 36 edges and 24 triangles, and hence the Euler characteristic of the ring is $\chi(\text{ring})=12-36+24=0$.

The Euler characteristic owes its name to the Swiss mathematician Leonhard Euler, who proved that the Euler characteristic of all Platonic solids (tetrahedron, cube, octahedron, dodecahedron and icosahedron) is 2. Notice that all Platonic solids are polyhedral approximations of the beach ball.

Now let us go back to our starting problem: suppose that you have a surface and that you cut it and paste it back into another surface. Then, you can choose a triangulation of your initial object in such a way that you only cut and paste through vertices and edges.

For example, if you want to cut a disc out of one hemisphere of your beach ball, our first triangulation of the beach ball, way 1, does the trick:

Now, you have a new intermediate triangulated surface. This triangulation will have the same number of triangles as before, but now the vertices and edges that you have touched with the scissors get duplicated after cutting, because you count them in both pieces that get formed. In our example, three vertices and three edges have been touched by the scissors and then duplicated (in orange). We keep track of this using the Euler characteristic. Here’s how it changes:

For example, the Euler characteristic of an annulus and a disc will be the same as the Euler characteristic of a disc, plus three vertices that get duplicated, minus three edges that get duplicated: $$\chi(\text{annulus and disc})=\chi(\text{disc})+3-3.$$ Let us rewrite the previous equation as:

The same happens if you cut two discs out of the beach ball, one in each hemisphere:

In this case, 6 vertices and 6 edges get touched by scissors, so then: $$\chi\left(\text{ball}\right)=\chi\left(\text{annulus and two discs}\right)-6+6.$$ If you glue the cut edges to make our final surface (a ring plus a ball), we have the same number of triangles as before, but now the vertices and edges that you have touched with the glue merge together. We halve the total number to take this gluing into account.

Notice that in our cut beach ball example, the final surface is disconnected: it consists of a ball and a ring, but this is perfectly allowed and we still can compute its Euler characteristic, which will be the sum of the Euler characteristic of the ball and the Euler characteristic of the ring. In this case, twelve vertices and twelve edges have been touched by the glue and then merged (in blue). Then, the Euler characteristic of the final surface will be:

Now, remember that we need to re-glue everything we have cut. This means that the number of vertices touched by glue is twice the number of vertices touched by scissors, because these vertices were duplicated by the cutting and we need to glue all of them back together. The same happens to the number of edges touched by glue. In other words,

As a consequence of this, we get:

With this example, we have learned something: the Euler characteristic does not change after cutting and pasting: the Euler characteristic is a cut-and-paste invariant for surfaces. This means that if we are able to cut and paste a surface into another surface, then their Euler characteristics must be the same. Since, as we have seen, this is not the case for the beach ball and the rubber ring, we cannot succeed in our initial mission. However, there is something better that we can do: we have seen that we can cut a beach ball and glue it into a beach ball and a ring, so we are able to help the kid that could not swim while still keeping our beach ball.

A triangulation of the double ring

Are there any other cut-and-paste invariants?

Your friend is still not happy and points at an inflatable double ring. “What about now? Can you build a single ring with the beach ball and this double ring?” Now you know what to do: take a triangulation of the double ring and compute its Euler characteristic. For example, we can consider the triangulation shown below for each half of the double ring.

In total, we have 22 vertices, 72 edges and 48 triangles. Doing the computations, we find that $\chi(\text{double~ring})=22-72+48=-2$. Hence, the Euler characteristic of the double ring and the beach ball together is equal to 0, the Euler characteristic of the ring. This means that the Euler characteristic invariant does not distinguish these surfaces, but is there possibly another invariant that distinguishes them?

In the 1970s, Ulrich Karras, Matthias Kreck, Walter Neumann and Erich Ossa answered this question in the negative: the Euler characteristic is the only cut-and-paste invariant for surfaces. Unlike in the case of polygons, it is still the only cut-and-paste invariant for the generalisation of surfaces to higher dimensions.

So you actually can cut the ball and the double ring and glue them back into a single ring. How? Well, that is for you to figure out so that your friend lets you go back to sunbathing!

Alba Sendón Blanco

Alba is a fourth-year mathematics PhD student at the Vrije Universiteit Amsterdam. Her advisers, Renee Hoekzema and Thomas Rot, introduced her to the world of cutting and pasting and of science communication, respectively. While attending a workshop mentored by Nicos Starreveld, she decided to write this article, inspired by her own work together with Mona Merling, Ming Ng, Julia Semikina and Lucas Williams on scissors congruence K-theory for equivariant manifolds. Whenever Alba is not doing maths, she might be knitting or playing the bagpipes.

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