How To Really Solve The Kissing Circles Puzzle

I first came across the kissing circles puzzle when I was studying the topic of hyperspheres. The puzzle as such is quite simple, but the insights it leads to are arguably profound and insightful.

In this essay, I will start by giving an introduction to the topic of hyperspheres. Following this, we will look at what is special about the notion of kissing circles in the context of hyperspheres. Then, we will jump straight into the kissing circles puzzle. Finally, we will extend our insights to solve a bonus puzzle (I don’t want to spoil it for you just yet). Let us begin.

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What are Hyperspheres?

Hyperspheres are fascinatingly difficult abstractions to visualize or imagine. Before we try and understand them, let us start with the definition of a circle. Given a fixed point on a plane, a circle is the locus of all points that lie at an equal distance from the said fixed point. Now here is an interesting question:

Can we extend this concept of the locus of all equidistant points from a fixed point to higher dimensions?

The answer is: of course, yes! However, before we tackle higher dimensions, let us try to scale one dimension down from a circle (which is 2-dimensional). In one dimension, we just have a line to work with. It turns out that a 1-dimensional circle would be just two points located at equal distances on either side of the central fixed point.

In the mathematical world, such 1-dimensional circles are known as 1-D spheres. Similarly, the circle is known as a 2-D sphere. Now, we all know about and can visualize the 3-D sphere, which is typically what we refer to using the word “sphere”.

Hyperspheres are geometrical abstractions that are constructed by merely extending this concept to higher dimensional spaces (like 4D, 5D, etc).

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Interesting Properties of Hyperspheres

I still remember trying to visualize or imagine hyperspheres to the point that my head started hurting. The harsh truth is that since we live in a 3-dimensional world, higher dimensional spheres are very, very difficult (if not, impossible) to visualize.

However, we could begin by imagining some of their properties by merely extending the properties of lower order spheres. A circle is capable of rotating around its centre (a point). A sphere is capable of rotating around its axis (a line). So, it could be generalized that an n-dimensional sphere would be capable of rotating around an (n-2)-dimensional entity (axis).

So, by extension, a 4D-sphere would rotate around a plane. A 4D-sphere’s rotation goes beyond just extended properties though. It turns out that a 4D-sphere is capable of double rotation around two mutually perpendicular planes. To save you the trouble of a headache, I would suggest not to try and imagine this.

Moving on to other interesting properties, projection pops up! If you project a circle onto a line, you get a line segment. Each point of this line segment, except the end-points, represents two points on the circle. Similarly, if you project a sphere onto a surface, you get an elliptical disc. Here again, every point on the disc, except the boundary points, represents two points on the sphere.

By extension, we could say that if we project a 4D-sphere onto our 3D-space, we would get a solid sphere, with every internal point representing two points on the 4D-sphere’s hypersurface. As an example, I imagined what would happen if such a 4D-sphere is composed of our usual 3-dimensional space and a fourth dimension of time. Then, each projected point (except the boundary points) would represent two different states of time.

We could go on and on with exploring such interesting properties, but it is one particular property that interests us in this essay.

How are Kissing Circles Related to Hyperspheres?

You see, the notion of kissing circles represents one of the most smoothly scalable properties of hyperspheres. In more technical terms, it is rather easy to generalize the formula for radii of the maximum number of mutually touching nD-spheres (where n is any arbitrary dimension).

Although this property was known to René Descartes specifically for circles, it was British chemist Frederick Soddy who popularized it and generalized it to spheres. Let us start with circles first.

The Mathematics of Kissing Circles

The inverse of the radius of a circle (1/r) is known as its curvature (this is also valid for a sphere). If we exclude the degenerate cases (like a collapsed circle that features infinite curvature), Soddy noted that for kissing circles, the sum of the squares of all kissing curvatures is half the square of their sum.

Furthermore, he noted that the maximum number of kissing circles is 4. There are two possible configurations as you can see below: one where three bigger circles enclose a smaller circle, and one where one bigger circle encloses three smaller circles.

Consequently, if a, b, c, and d are the curvatures of each of the circles involved, then, Soddy’s formula is as follows:

2(a² + b² + c² + d²) = (a + b + c + d)²

Soddy went on to establish that for spheres, the maximum number of kissing spheres is 5. So, the corresponding relationship of their curvatures (considering a, b, c, d, and e as the curvatures respectively) is as follows:

3(a² + b² + c² + d² + e²) = (a + b + c + d + e)²

This relationship was then generalized to n-dimensional spheres by amateur mathematician Thorold Gosset. However, what we have seen so far is sufficient for us to jump directly into the kissing circles puzzle.

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The Kissing Circles Puzzle — Revealed

You can see the two cases of kissing circles below. One thing to note here is that the bigger circle that encloses three smaller circles is said to have a concave curvature, which is mathematically denoted by a negative sign. On the other hand, the smaller circle that is enclosed by three bigger circles is said to have convex curvature, which is mathematically denoted by a positive sign.

Given this setting, if the circle-1 features a 1 unit radius, circle-2 features a 2 unit radius, and circle-3 features a 3 unit radius, calculate the radius of the fourth circle (circle-x) in both the presented cases.

Spoiler Alert:

If you wish to solve this puzzle on your own, I suggest that you tune off of this essay for now. After you are done with your attempt, you may return and continue reading. Beyond this section, I will be explicitly discussing solutions to the puzzle.

How to Really Solve the Kissing Circles Puzzle?

All we need to do is to plug in the radii numbers into Soddy’s formula (I’ve assigned x units for the unknown radius/radii), and we would end up with the following expression:

When we resolve this expression, we get the following result:

You might be a little perplexed by one of the results being negative. But when you remember that concave curvature features a negative sign, it becomes clear that we have obtained the result for both the cases we were trying to solve in one go:

At the beginning of the essay, I promised that you’d get to solve a bonus puzzle. That is precisely what you are about to get next.

The Bonus Puzzle — The Kissing Spheres Puzzle

We are now about to extend the concepts that we have seen so far from 2-dimensions to 3-dimensions. If all goes well, you should be able to solve this one relatively easily.

Imagine that three mutually kissing watermelons of a radius of 1 unit each are held on a flat plate. Now, a spherical orange is placed on the table such that it kisses the three watermelons. The puzzle you will need to solve is:

What is the radius of the orange?

Spoiler Alert:

Like before, beyond this section, I will be explicitly discussing solutions. So, if you wish to solve this puzzle on your own, now is the time.

The Solution to the Bonus Puzzle of Kissing Spheres

The first key insight you will need is that the flat plate on which the spherical fruits are placed could be considered the fifth sphere, albeit with an infinite radius. So, when you apply Soddy’s formula, the component of the fifth sphere can be neglected, and we end up with the following expression:

When you solve this expression as we did before, you would end up with the following solution:

So, we see that the orange features convex curvature and has a radius that is one-third of that of the watermelons.

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Reference and credit: Martin Gardner.

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Further reading that might interest you: How To Really Solve The Monkey And The Coconuts Puzzle? and The Helix Puzzle — A Simple Geometric Challenge.

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