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.

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    $\begingroup$ Hint: Can you find an equilateral triangle in your construction? What are its angle measures? $\endgroup$
    – Dzoooks
    Jan 24 '19 at 2:21
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    $\begingroup$ Actually the "natural" compass and straightedge operations can divide a circle into 4,5, or 8 equal parts almost as easily as we divide one into 6 equal parts. But you are correct that we cannot so divide a circle into 7 equal parts. It's related to the construction of regular polygons, and the Gauss-Wantzel theorem tells which numbers of sides are possible with compass and straightedge. $\endgroup$
    – hardmath
    Jan 24 '19 at 2:49
  • $\begingroup$ The title of your question is misleading You can divide a circle into any number of pieces. Haven't you ever had a pizza sliced into 8 slices? :) $\endgroup$
    – Adi
    Jan 24 '19 at 3:27

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.


Construct the first two circles and connect their centers. Notice that the length of this connection is the radius (which we'll define as a unit). Now connect the centers to one of the intersections and note that you have constructed an equilateral triangle (with angles each $\frac \pi 3)$.

Six of these triangles would thus exactly fit in a revolution of $2\pi$.

(Another way of thinking of it is to notice that the length of the arc subtended by the angle at a unit circle center is equal to the angle, i.e. $\frac \pi 3$, and there are six of these lengths in the circumference $2\pi$ of the circle.)


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