A necklace of circles

Relevant wiki: Triangles - Exterior Angles

The sum of the angles for the different parts between the blue and yellow areas should always equal to 2 times the sum of exterior angle of the polygon.

Both red and green part are the exterior angle of the 10-sided polygon.

The sum of exterior angle of the polygon always be 360 degree(angle on yellow side will count as negative values).

Sum of red and green part exterior angle will be 2 times 360.

So the positive difference between the blue and yellow areas should be 2 times the area of one circle.

As the radius is 2, the positive difference between the blue and yellow areas will be $2\times\pi r^2=2\times\pi2^2=\boxed{8}\pi$

Relevant wiki: General Polygons - Angles

The total degrees in 10 circles: $3600^\circ$

The interior angle of a decagon (the yellow area): $1440^\circ = \color{#CEBB00}\dfrac{2}{5} \color{#333333}3600$

The full external angle of a decagon (the blue area): $3600^\circ - 1440^\circ = 2160^\circ = \color{#3D99F6}\dfrac{3}{5} \color{#333333}3600$

There are 10 identical circles of radius $2$ , creating a total area: $10 \pi \times 2^2 = 40\pi$

The blue area: $\color{#3D99F6}B\color{#333333} = \color{#3D99F6}\dfrac{3}{5} \color{#333333}40\pi = 24\pi$

The yellow area: $\color{#CEBB00}Y\color{#333333} = \color{#CEBB00}\dfrac{2}{5} \color{#333333}40\pi = 16\pi$

---

$\color{#3D99F6}B\color{#333333} - \color{#CEBB00}Y\color{#333333} = k\pi$

$24\pi - 16\pi = \boxed{8}\pi$

Relevant wiki: General Polygons - Area

The sum of interior angles equals 1440°

Sum of the exterior angles equals 2160°

This gives a ratio of blue:red = 2160:1440 = 6:4

Total area of circles = 10 × pi × 2^2 or 40 × pi.

So blue area = 24 pi, and yellow area = 16 pi.

24 - 16 = 8

Relevant wiki: Polygon triangulation / Grids

The sum of interior and exterior degrees do not depend on the shape (you can see this by triangulating). So put the circles in two rows of 5 so that the resulting polygon is a rectangle. 6 circles are then cut in half (so they do not contribute to the difference). The remaining 4 circles are 1/4 in and 3/4 out: a difference of 1/2 each. So the difference is in total $4 \cdot \tfrac{1}{2} =2$ circles. The Area of two circles is $2 \cdot \pi \cdot 2^2 = 8 \pi$ .

The arrangement is not quantitatively specified, implying the answer is invariant.

Arrange them in a 4x3 rectangle with two center circles removed; connect their centers. Six of the circles are bisected, contributing zero to the area difference. The other four are quartered, contributing 1/2 of their area to the difference. Four half-areas totals 8 pi.

I imagined moving the balls into a 2-across by 5 deep pattern (still all touching at their radii) such that the center 3 pairs were cut in half, and the outer 4 (2 on top and 2 on the bottom) were sectioned into 1/4 yellow and 3/4 blue. Then, the center 3 pairs cancel out having equal area blue and yellow, and the outer 4 having total area equal to 4 pi, pi of which is yellow, and 3 pi of which is blue. (3 pi - pi) * 4 = 8 pi, thus k = 8

The number of circles doesn't matter. You can follow a path through the circles, and every time that path changes direction, the difference between the yellow and blue areas is a wedge of twice the angle of the change in direction. The total of all the changes in radians is τ. So the wedges add up to 2τ (or 2 complete circles.) With a radius of 2, that's 4τ, or 8π.

Imagine that each circle begins evenly split between blue and yellow.

Each circle must "give up" degrees to contribute to the total 360° that will be the sum of the yellow polygon's exterior angles.

Because there are 10 circles, each circle must give up an average of 36°.

For each circle, contributing these 36° also means converting 10% of its yellow area to blue area.

So, on average, circles must be 40% yellow and 60% blue by area.

That is, the average difference between blue and yellow area in each circle will be 20% of the circle's area.

20% * 4π * 10 = 8π

Each circle has an area of 4PI. Rearrange the circles in a 2x5configuration. Each of the 6 circles NOT at a corner has equal blue and yellow areas, so the difference is 0. The remaining 4 circles at a corner have 3PI blue an 1PI yellow for a difference of 2PI each. 4x2PI=8PI.

• // Java solution for any number of circles:
• int circles = 10;
• int radius = 2;
• // PI not included so devision later is not required.
• double circleSurface = Math.pow(radius, 2);
• double removedSurface = (circleSurface / 2) - circleSurface / circles;
• double remainingSurface = circleSurface - removedSurface;
• double blue = remainingSurface * circles;
• double yellow = removedSurface * circles;
• double k = blue - yellow;

Doing a displacement of the cirles so that they are aranged ind two lines of 4 circles, we can pair the four edges in pairs of two of Pi each. Then we see that the rest of the circles have yellow area of Pi each. Those are 6 circles not cornered, and 2 pairs of corners. Which sums up to 8 Pi. You can expand this to a general solution of k circles being (k-2)*Pi