Can You Squeeze the Bubbles?

Fiddler

Draw a unit circle (i.e., a circle with radius 1). Then draw another unit circle whose center is not inside the first one. Then draw a third unit circle whose center is not inside either of the first two.

Keep doing this until you have drawn a total of seven circles. What is the minimum possible area of the region that’s inside at least one of the circles?

Solution

Circles mimicing a hexafoil, where all centers lie upon the circumference of another circle, seems to create the minimimum possible region.

The following figure was used to determine area of the necessary regions.

• The dark blue sector has an area that is $\dfrac{1}{6}$ of the circle, or $\dfrac{\pi}{6}$.
• The dark blue equilateral triangle has an area that is $\dfrac{\sqrt{3}}{4}$.
• The dark blue or purple half-petals have an area that is $\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4} = \boxed{\frac{2\pi-3\sqrt{3}}{12}}$.
• The teal region is a semicircle minus two half-petals. The area is $\dfrac{\pi}{2} - \dfrac{2\pi-3\sqrt{3}}{6} = \boxed{\dfrac{\pi+3\sqrt{3}}{6}}$.

Answer

The region is a central circle and six teal regions.

$$A = \pi + 6\bigg(\dfrac{\pi+3\sqrt{3}}{6}\bigg) = \boxed{2\pi + 3\sqrt{3}}$$

Extra Credit

Instead of seven unit circles, now suppose you draw N of them. As before, the center of each new circle you draw cannot be inside any of the previous circles.

As N gets very, very large, what is the minimum possible area of the region inside at least one circle in terms of N?

Solution

$\dfrac{3n^2\sqrt{3}}{2}$ is the area of a hexagon.

$f(n) = 3n(n-1) + 1$ represents the number of dots(circles) necessary to draw a hexagon of side length $n$.

Since $N$ is very, very large, $N \approx 3n^2$.

Answer

$$\boxed{\frac{N\sqrt{3}}{2}}$$

Another Solution,

in case the above was too lazy...

• Purple Petals :
  • Each has an area of $\dfrac{2\pi-3\sqrt{3}}{6}$.
  • Each circle has $12$.
  • Each is part of $4$ circles.
  • Relevant area is $N \cdot 12 \cdot \dfrac{1}{4} \cdot \dfrac{2\pi-3\sqrt{3}}{6} = \boxed{N\bigg(\frac{2\pi-3\sqrt{3}}{2}\bigg)}$.
• Yellow Regions :
  • Each is an equilateral triangle minus three half-petals. $\dfrac{\sqrt{3}}{4} - \dfrac{2\pi-3\sqrt{3}}{4} =\sqrt{3}-\dfrac{\pi}{2}$.
  • Each circle has $6$.
  • Each is part of $3$ circles.
  • Relevant area is $N \cdot 6 \cdot \dfrac{1}{3} \cdot \bigg(\sqrt{3}-\dfrac{\pi}{2}\bigg) = \boxed{N\big(2\sqrt{3}-\pi\big)}$.

$$A = N\bigg(\frac{2\pi-3\sqrt{3}}{2}\bigg) + N\big(2\sqrt{3}-\pi\big) = \boxed{\frac{N\sqrt{3}}{2}}$$

Note:

This is $\dfrac{\frac{N\sqrt{3}}{2}}{N\pi} \approx 27.56\%$ of the circles' individual area.

Rohan Lewis

2025.06.09

Code can be found here.