The area of a circle of radius 1 is completely covered by seven smaller circles, all with the same radius as each other. (The circles can overlap - indeed they must!). What is the smallest radius the small circles can have so that it is possible to do this, and why is it not possible for any smaller radius?

enter image description here


Since nobody is providing the proof of minimality, I am going to add one.

A: A circle of radius $r<1/2$ can cover less than $ 1/6 $ of the circumference of the unit circle Circles

The length of the arc $DE$ is maximized when the length of chord $DE$ is maximized, for a circle with radius $r$, this can be at most $2r$. Let, $\angle A =\theta$


$2r \ge DE = \sqrt{AD^2+AE^2-2 AD. AE\ cos \theta} = \sqrt{2-2cos\theta}$

$\implies cos\theta \ge 1-2r^2 \ge 1-2({1 / 2})^2 = 1/2 \implies \theta < 60^{\circ}=1/6 \times 360 ^{\circ}$

Now, It follows that you need all 7 circles to cover the circumference if $r < 1/2$. But a circle with $r < 1/2$ can't cover part of the circumference and the center at the same time. So, None of the 7 circles can cover the center. Hence, $r \ge 1/2$. A construction with $r=1/2$ has been shown before.


Upon further research, I found out that this is actually called the disk covering problem. Formally stated this asks:

Given a unit disk, find the smallest radius $r_n$ required for $n$ disks of equal radius to completely cover the unit disk.

The first few values are given here.


I believe the smallest radius is..

$r = \frac R 2$, i.e. $\frac12$
7 overlapping circles

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    $\begingroup$ Good answer. How do you know it's not possible to cover the big circle if you use a smaller radius? $\endgroup$ – DavidButlerUofA Aug 19 '15 at 21:52
  • $\begingroup$ @DavidButlerUofA Can you at least up-vote my answer? It was the first correct answer, and the only one with a description, and it's still below an incorrect answer. :( $\endgroup$ – Roland Aug 19 '15 at 21:53
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    $\begingroup$ Well, my reasoning was that if you put one circle in the middle, this configuration covers the most area with the six others. If you don't put one circle in the middle, you'd need a configuration similar to @Spacemonkey's answer, and those circles would have to be slightly larger. $\endgroup$ – Sphinxxx Aug 19 '15 at 22:05
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    $\begingroup$ A clue on the formal proof: Why are the boundary circles optimal, what do they maximize/minimize? $\endgroup$ – Rohcana Aug 19 '15 at 22:30
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    $\begingroup$ My clue to the proof: for all points a points inside the circle, at least one of the centers of the little circles must be inside a circle of radius $r$. Consider $r \le \frac{1}{2}$, then contradict the previous by finding a few well chosen points $\endgroup$ – WorldSEnder Aug 20 '15 at 0:17

I think it's:


I'll use this arrangement, with one circle not pictured in the center (reasoning shown below). All blue lines are equal to $r$ in length. The green line needs to be equal to $1$. Call the distance from the origin to the center of an outer circle $d$.

enter image description here

Finding the smallest $r$:

From the image below, we create two triangles.

The first is a right triangle made from the gray line on the left, the green line, and the blue on top. These are of lengths $d$, $1$, and $r$, respectively, and we get: $d^2 + r^2 = 1$

The second is made from the other blue line and the same vertical gray line. To minimize the radii, we let the center circle just reach out to the blue line along the green path. Since all the circles have radius $r$, we are left with an isosceles triangle of sides $r$, $r$, and $d$, and angles of 30, 30, and 120 degrees. From this we acquire: $cos(30^o) = d/2r$

Solving the second for $d$ and substituting, we have $r = 1/2$ .

enter image description here

With no centered circle, and radii of $1/2$, only seven points on the edge of the unit circle are covered. This obviously will not suffice.

Why not smaller circles? This method optimized coverage for a scheme with one centered circle and six surrounding. It features the most symmetry of any seven-circle scheme; six lines of symmetry can be drawn through the origin to 12, 1, 2, 3, 4, and 5 o'clock. This optimizes coverage for circles by the same logic Archimedes used when measuring its area and circumference.

  • $\begingroup$ Actually, you can do it if the radius is smaller than this. $\endgroup$ – DavidButlerUofA Aug 19 '15 at 21:20
  • $\begingroup$ Math error! It's 1/2! $\endgroup$ – Roland Aug 19 '15 at 21:29
  • $\begingroup$ Why is this not accepted? Is it wrong? It explains the answer in greater depth than Sphinx's. $\endgroup$ – warspyking Aug 19 '15 at 23:23
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    $\begingroup$ 1. I am not online all the time. 2. The explanation of why you can't cover with smaller circles is not convincing to me. $\endgroup$ – DavidButlerUofA Aug 19 '15 at 23:51

Backing up the r/2 claim with an alternative visualisation:

enter image description here

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    $\begingroup$ I sorted the spoilertags for you :-) Welcome to Puzzling SE! $\endgroup$ – Rand al'Thor Aug 24 '15 at 14:45

I put this as an answer because I wanted to be able to use the spoiler tags. I believe I know how the circles must be arranged, but do not have the math skills to determine what measurements theyd have.

Because of the nature of the shape (a circle) and the fact you are limited to 7 circles, no pattern or arrangement can efficiently cover the area. You need circles that stretch from the center point of the circle and that cover 1/7th of the circumference of the circle to cover(of radius 1). enter image description here

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    $\begingroup$ Good answer, but it might not be the lowest, for example it might be possible to use Smaller circles such a way that some cover the circumference some cover the middle. $\endgroup$ – Rohcana Aug 19 '15 at 21:32
  • $\begingroup$ Not the smallest circle. Radius will be 0.555. wolframalpha.com/input/…*sin%282pi%2F14%29 - This assumes that I got the math correct this time. $\endgroup$ – Taemyr Aug 20 '15 at 10:11
  • $\begingroup$ I haven't spent huge amounts of time on it since the math was over my head but I guess it's possible to have 5 circles cover the circumference and 2 the middle but then at that point I'm starting to lose sight of how any answer can be hard proven to fit.... (I mean if you go in minute measurements) - Btw I totally think Sphinxx and Roland nailed the arrangement $\endgroup$ – Spacemonkey Aug 20 '15 at 14:03
  • $\begingroup$ @Spacemonkey But you had the right idea. This idea is crucial for the proof. $\endgroup$ – Rohcana Aug 20 '15 at 16:06

I might be misunderstanding this, but based on what I think you're asking, we have the following lower bound:


This is obtained by the following:

If the larger circle has a radius of $1$, then it has an area of $\pi$.

We know the seven smaller circles have enough total area to cover the larger, so we have
$\pi \le 7 \pi r^2$

Simple algebra from there yields
$\sqrt{\frac17} \le r$

  • $\begingroup$ not sure that would work because of the nature of the shapes but I lack the visual/math skills to picture what you just calculated onto a circle of radius 1. $\endgroup$ – Spacemonkey Aug 19 '15 at 21:13
  • $\begingroup$ That's some good reasoning, but it is not actually possible to cover the big circle if your small circles have that area because there would be gaps between the circles. $\endgroup$ – DavidButlerUofA Aug 19 '15 at 21:14
  • $\begingroup$ I see; I might lack the visual skills for this as well. While I rethink, can anyone confirm that the equations came out well? $\endgroup$ – NeedAName Aug 19 '15 at 21:15
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    $\begingroup$ An upper bound seems to be $r \leq \frac{1}{2}$ if you place six circles symmetrically and one circle in the middle. Not sure if this bound can be improved. $\endgroup$ – Carl Löndahl Aug 19 '15 at 21:18
  • $\begingroup$ Equations came out perfectly. :) $\endgroup$ – DavidButlerUofA Aug 19 '15 at 21:21

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