Seven overlapping circles

Question

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?

Answer 1

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

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, $\angle A =\theta$

Now,

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

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

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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.

EDIT

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.

Answer 2

I believe the smallest radius is..

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

Answer 3

I think it's:

$1/2$

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 $d$.

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 $r = 1/2$ .

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.

Answer 4

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

Answer 5

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).

Answer 6

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

$\sqrt{\frac17}$

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
$\sqrt{\frac17} \le r$