Draw a circle. Draw a new circle with center on the circumference and the same radius. Draw a new circle with center on the intersecting points and the same radius. How many circles can you fit around the original circle? 6. 6 equiliant triangles fit inside a circle, a hexagon inside the circle has the same lengts as the radius, the way the 60 degree angle is constructed. These are all saying the same thing. Why does the circle divide into 6? But why not 4, 5, 7, 8.
Why does the circle naturally divide into $6$
geometry
Related Solutions
You should probably be using Bresenham's Circle Algorithm?
The "distance" between adjacent pixels is either 1 or $\sqrt{2}$, i.e., two adjacent pixels either share a row or column or they are diagonal. This limits how much of the circumference consecutive pixels can consume. So the number of pixels needed, $x$, to complete the circumference of a circles with a radius of $r$ pixels is bounded $\sqrt{2} \pi r \leq x \leq 2 \pi r$, or roughly $4.443 \times r \leq x \leq 6.283\times r$. This gives a first approximation, by which to check the later estimate.
To get more accuracy we see that, by symmetry, only the pixels drawn in the first octant (i.e., $\pi/4$) need to be computed; the remaining can be found by using the same offsets but in different directions. With each consecutive pixel in the first octant progressing upward, there are at least $\lfloor r \sin(\pi/4)\rfloor$ pixels. Multiply this by 8, and you get roughly $5.657\times r$, which is within the earlier bounds.
Let $R$ (=32) be the radius of the large circle and $r$ (=9.5) be the radius of the small disk.
We start with the following simpler problem: Given an annular sector $S$ with inner radius $a$, outer radius $b$ and central angle $\phi$, what is the radius $\rho$ of the smallest circle that covers $S$?
Using the above figure one gets $$\rho^2={a^2+b^2-2 a b\cos\phi\over2(1+\cos\phi)}\qquad(0<\phi<\pi)\ .$$
As suggested by Ross Millikan we now aim at minimizing the number of pieces. To this end I propose the following greedy algorithm:
- Put $r_0:=R$.
- Given $r_k$ for a $k\geq0$ put $b=r_k$, choose a suitable $n$ and put $\phi:={2\pi\over n}$. Then solve the equation $$r^2={a^2+b^2-2 a b\cos\phi\over2(1+\cos\phi)}$$ for $a$ (choose the smaller of the two solutions).
- Repeat step 2. for various values of $n$ until you have found the $n$ maximizing the quantity $$q_n:={b^2-a^2\over n}\ .$$ (Comment: We want to maximize the area gain per piece.)
- If $a<1$ goto 5., else put $r_{k+1}:=a$ and goto 2.
- Cover the center with a last disk.
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Best Answer
I think you are basically asking: Why do exactly $6$ equilateral triangles fit in a circle? and pointing out different instances of it.
In which case: it's because the angles of a triangle sum to $180°$, which is "half a circle".
Since an equilateral triangle has all three angles equal, each angle has to be a third of $180°$—or a "sixth of a circle".
So exactly $6$ of them fit around the circle's centre.