Math Puzzle for Quarantining Humans: 6 Inscribed Circles

7 min readJul 20, 2020

This post is part of a series of math puzzles. Here is the list of puzzles for reference:

3333
6 Inscribed Circles (this post)
Slicing the Parabola
4444
[…coming soon…]
[…coming soon…]

Now, let’s continue with 6 Inscribed Circles:

6 Inscribed Circles: PROBLEM

The premise of this problem is very simple. Essentially, the outer circle has radius 1, and it has 6 circles with equal radius inscribed inside it in such a way that they both touch the outer circle and touch each other. Now, the question is, if we were to put a final, 7th circle in the center of the outer circle in such a way that this 7th circle touches all the other 6 circles, what would be radius of this 7th circle?

Note that you could, if you wanted, try to guess the answer. However, for this question you must prove beyond doubt that this circle has the radius that you give. If you get stuck, scroll down to see a hint (don’t scroll too fast or else you’ll get to the solution).

6 Inscribed Circles: HINT

Here’s a hint for you. The 7th circle has the same radius as the 6 inscribed circles. Remember, you still have to prove whatever radius you found. The only thing you know are that outer circle has radius 1, and all 6 inner circles touch each other and touch the outer circles (but they do not overlap).

6 Inscribed Circles: SOLUTION

6 Inscribed Circles with the center circle drawn

Tools

Before we get to the solution, we must make ourselves aware of a few formulas and rules. These are going to be our tools to solve this problem. We’ll use the following triangle to illustrate the tools below.

Triangle with angles A, B, C and a segments a, b, c.

All of the following laws/rules/theorems can be proved with geometry, trigonometry, algebra, or other techniques, and such proofs can be easily found through a quick wikipedia search. We’re going to assume that they’re true.

Law of Cosines (LoC) and Law of Sines

The Law of Cosines says that c^2 = a^2 + b^2 — 2 * a * b * cos(C)

The Law of Sines says that sin(A)/a = sin(B)/b = sin(C)/c

Angle Sum of a Polygon (ASP)

This formula essentially says that the sum of angles of a polygon with n sides is (n-2) * 180 degrees. For a triangle, this number is (3-2) * 180 = 180 degrees.

Isosceles Triangles Property (ITP)

If our triangle is an isosceles triangle—i.e., a = b and c may have a different length(without loss of generalization), then angles A and B are equal as well.

Two Nested Square Roots (TNS)

Here’s a screenshot from a latex document I wrote (because it’s much cleaner seeing this in latex):

In the above, Squares is essentially the set {q^2 | q \in Q} . I.e., the set of squares of rational numbers. For example, 4/9 is in that set because 4/9 = (2/3)^2 , but 3/9 is not in that set because it can’t be expressed as the square of another rational number.

The Solution

Now that we’ve defined our tools, finding the solution should be a piece of cake—if we approach it the right way. One thing that makes this problem particularly hard is the fact that how you approach it is very open ended, so it makes it easy to go about it the wrong way. I myself struggled a bit before solving it again. However, with the following approach things should make sense.

First, let’s imagine that we draw two rays from the center of the outer circle to the center of one of the tangent smaller circles and in between two of the other circles (Trust me, it’s totally possible to do this. See [1]).

With that, we can connect the intersections between the two rays and the outer circle. Then, we can make note of two things. First, since the circles are equidistant apart and we know that the total span of a circle is 360º, we can find the angle the two rays make by dividing that with however many pieces there are. In this case, there are 12 pieces, so the angle between the two rays is going to be 30º. Second, since both rays up to the intersection go from outer circle center to the outer circles, they both have length 1. Using the isosceles triangles property (ITP), both remaining angles must be equal. Then, using the angle sum of polygon (ASP), we find that the two angles must be 180º minus 30º divided by two, which gives us 75º.

Given that, we’re able to deduce one other thing: the length of the segment between the two intersection. We can do it using the Law of Cosines (LoC). Let's call this segment y . y^2 = 1^2 + 1^2 — 2 * 1 * 1 * cos(30) = 2 — sqrt(3) . Then, y = sqrt(2 — sqrt(3)) . Then, using the Two Nested Square Roots (TNS) formula, we can find that y = (sqrt(3) — 1) / sqrt(2) . You can check this using Desmos if you want.

On the diagram, I will keep this variable as y . Then, with the few steps (which I will explain), we get to the following:

What happened now is that we drew a segment from the center of this inner circle down to the point where this inner circle intersect with the other inner circle. We can use a similar trick to [1] to find the exact place where this intersection is. One thing that we note is that the angle between the bottom ray and this segment has to be 90º because the two circles are tangent to each other and the bottom ray is tangent to both circles. Since this intersection touches the outer edge of our inner circle and it goes through its center, it has to be perpendicular to whatever line is tangent to that circle at that point. If it’s perpendicular, it has to be a 90º angle. This allows us to deduce the angle above to be 60º from ASP, which means the angle next to it is 120º (180º minus 60º)as shown in the diagram above.

Let’s denote the radius of the inner circle to be R . In a similar fashion, let’s denote the radius of the inside circle to be Q , which we draw out in the next diagram. At this point we cannot assume that R and Q are the same (even though we know this from the hint. Regardless, we’ll prove that here).

What we’re going to do next is to draw a segment from the bottom, middle point and the top right point in our diagram. After we do that, we’re able, in very similar fashion to what we’ve done so far, deduce the angles of neighboring corners (I will leave it as an exercise to deduce those angles):

With that done, I have given that newly added segment the length x . If you look carefully, our problem is almost solved! Essentially, we can find x using LoC again in terms of R . That will give us x^2 = R^2 + R^2 — 2 R^2 cos(120) , which gives:

Now, using the law of sines (LoS) finally, we can solve for x with sin(75)/x = sin(60)/y . Using that, and knowing that

(there’s a trick to get this, you can look it up :D)

we get the equation

Then, if you replace x with sqrt(3R^2) and solve for R , you’ll get the answer that R = 1/3!

Now, we’re technically not done yet. What we wanted was to solve for Q . We can get a formula for Q to be Q = 1 — 2R (since the radius Q is equal to the the radius of outer circle minus the diameter of inner circle, and this diameter is twice the radius of inner circle, which is R), and if you solve for this, you’ll get that Q = 1/3 as well!

So, this is how we solve this problem. You can now see that Q and R are equal 😃.

End Notes

[1] it is possible to find the center of either circles with some geometry using a compass. To find the line in between the two circles, we can create a line between the centers of two tangent, smaller circles and find the point in between using the same trick. Then, connect the center of the outer circle with this newly found in-between point.