Mathnasium Journal

Place three unit circles ( $r = 1$ ) at alternating vertices ( $A, C, E$ ) of a regular hexagon.
We choose the hexagon size so the circles are just tangent (touching, not overlapping).

If the hexagon side length is $n$ , then the distance between alternating vertices is:

3 n

Tangency of unit circles means center distance $= 2$ , so:

3 n = 2 \Rightarrow n = \frac{2}{3}

Now compute area inside the hexagon but outside those circles.

Hexagon area: $\frac{3\sqrt3}{2}\cdot1.15^2=3.464$

Center distance: $d=\sqrt3\,n\approx2.000$

Target area: $A(1.15)=\frac{3\sqrt3}{2}n^2-\pi\approx0.323$

Adjust hexagon radius n

$n_{\text{tan}}=\frac{2}{\sqrt3}\approx1.155$ . This is the tangent setup: each pair of unit circles touches at one point. Removed sector total remains $\pi\approx3.142$ .

1) Sum of interior angles (general formula)

Sides: 6

Triangles from one vertex: $n-2=4$

Interior-angle sum: $(n-2)\cdot180^\circ=720^\circ$

Number of sides n

For any $n$ -gon:

Sum of interior angles = (n - 2) \cdot 18 0^{\circ}

Example: for a pentagon ( $n = 5$ ), the sum is $(5 - 2) \cdot 18 0^{\circ} = 54 0^{\circ}$ .

2) Single interior angle for a regular polygon

Total interior sum: $(n-2)\cdot180^\circ=720^\circ$

Single interior angle: $\theta_n=\frac{720^\circ}{6}=120.0^\circ$

Number of sides n

θ_{n} = \frac{( n - 2 ) \cdot 18 0 ^{\circ}}{n}

For a regular hexagon ( $n = 6$ ):

θ_{6} = \frac{( 6 - 2 ) \cdot 18 0 ^{\circ}}{6} = 12 0^{\circ}

3) Sector fraction of a unit circle

One sector area: $\frac{120^\circ}{360^\circ}\pi\cdot1^2=1.047$

Total with 3 sectors: $3\cdot1.047=3.142$

Theta (degrees) Number of equal sectors

Since $12 0^{\circ}$ is one third of $36 0^{\circ}$ , one interior sector at a chosen vertex is:

\frac{θ}{36 0 ^{\circ}} π r^{2} \Rightarrow \frac{12 0 ^{\circ}}{36 0 ^{\circ}} π (1)^{2} = \frac{π}{3}

Three such sectors give total circle contribution:

3 \cdot \frac{π}{3} = π

4) Hexagon area from six equilateral triangles

Height: $h=\sqrt{1.15^2-\left(\frac{1.15}{2}\right)^2}=1.000$

One triangle area: $\frac{1}{2}\cdot1.15\cdot1.000=0.577$

Hexagon area: $6\cdot0.577=3.464$

Hexagon radius and side length n

Default is the tangent setup: $n=\frac{2}{\sqrt3}\approx1.155$

A regular hexagon splits into 6 equilateral triangles, each with side length $n$ .

Using a right-triangle split:

h = n^{2} - (\frac{n}{2})^{2} = \frac{3}{2} n

Triangle area:

\frac{1}{2} \cdot n \cdot \frac{3}{2} n = \frac{3}{4} n^{2}

Hexagon area:

6 \cdot \frac{3}{4} n^{2} = \frac{3 3}{2} n^{2}

For the tangent-unit-circle configuration $n = \frac{2}{3}$ :

H_{tan} = \frac{3 3}{2} (\frac{2}{3})^{2} = 23

5) Subtract circle-sector total

Hexagon area: $\frac{3\sqrt3}{2}\cdot1.15^2=3.464$

Center distance: $d=\sqrt3\,n\approx2.000$

Target area: $A(1.15)=\frac{3\sqrt3}{2}n^2-\pi\approx0.323$

Adjust hexagon radius n

$n_{\text{tan}}=\frac{2}{\sqrt3}\approx1.155$ . This is the tangent setup: each pair of unit circles touches at one point. Removed sector total remains $\pi\approx3.142$ .

Using the sector total from step 3:

A (n) = \frac{3 3}{2} n^{2} - π

At tangency $n = \frac{2}{3}$ :

A_{tan} = 23 - π

This is exact at tangency (the circles only meet at points, so there is no overlap area to correct).