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If two equal circles intersect so that each circle's centre lies on the other circle's circumference, what is the ratio of the part of a circle's circumference that overlaps, to its whole circumference?

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Hint: if you draw segments from each center to the intersection points of the circles and from one center to the other, you make a pair of triangles. Is there anything interesting about them?

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intersecting circles

Segments $BC$, $BA$, and $BD$ are all radii of circle $c_1$, and segments $AC$, $AB = BA$, and $AD$ are all radii of circle $c_2$. Since the circles are equal in size, their $radii$ are equal, and so all five segments (blue dashed lines) are equal in length.

In particular, $\triangle ABC$ and $\triangle ABD$ are equilateral triangles, and so the measure of all the angles of each triangle are equal: 3 equal angles totaling $180^\circ \rightarrow\;$ each angle of each triangle must be $60^\circ$. If we focus on circle $c_1$, we can see that $$m(\angle CBD) = m(\angle CBA) + m(\angle ABD) = 60 + 60 = 120^\circ.$$

Now, a full revolution of a circle corresponds to $360^\circ$. So the angle in $c_1$ by point $B$ and the points at which the two circles intersect is $\frac {1}{3}$ of a complete revolution. The ratio of length of the arc $CD$ to the circumference of the circle must therefore be $1:3$.

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I was leaving that to OP as it looked like homework. You are spot on. – Ross Millikan May 18 '11 at 3:08

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