parmenides51 wrote: The plane has a circle $\omega$ and a point $A$ outside it. For any point $B$, the point $C$ on the circle $\omega$ is defined such that $ABC$ is an equilateral triangle with vertices $A, B$ and $C$ listed clockwise. Prove that if point $B$ moves along the circle $\omega$, then point $C$ also moves along a circle. the point $C$ on the circle $\omega$ if point $B$ moves along the circle $\omega$ Both points $B$ and $C$ on this circle ?

I'm not sure if this is right. But I think there is a typo, and it should be $B$ is on circle $\omega$ and $C$ is the point s.t. $ABC$ is an equilateral triangle. Then we must prove that for all $B$ on circle $\omega$, the locus of points $C$ must trace out a circle.

So like this, but $D$ is point $B$ and $F$ is point $C$. Also I made $A$, $D$, and $F$ counter clockwise accidentally, so forgive my mistake.

Attachments:

A proof of this is noting that $C$ is the rotation of $B$ about $A$ $60^\circ$ clockwise. So the locus of points $C$ is the locus of points $B$ (which is circle $\omega$) rotated about $A$ $60^\circ$ clockwise. Since circles are still circle after rotation, the locus of $C$ must also be a circle. Sorry for triple posting

vanstraelen wrote: parmenides51 wrote: The plane has a circle $\omega$ and a point $A$ outside it. For any point $B$, the point $C$ on the circle $\omega$ is defined such that $ABC$ is an equilateral triangle with vertices $A, B$ and $C$ listed clockwise. Prove that if point $B$ moves along the circle $\omega$, then point $C$ also moves along a circle. the point $C$ on the circle $\omega$ if point $B$ moves along the circle $\omega$ Both points $B$ and $C$ on this circle ? just corrected the wording, indeed it had a typo above, $B$ lies on $\omega$ and not $C$, as natmath suggested

@natmath post #5 If you have used Geogebra, you can rename a point by right clicking.

Given the circle $x^{2}+y^{2}r^{2}$ and the point $A(a,0)$. Take $B_{1}(r,0)$, then $C_{1}(\frac{a+r}{2},\frac{(a-r)\sqrt{3}}{2})$. Take $B_{2}(-r,0)$, then $C_{2}(\frac{a-r}{2},\frac{(a+r)\sqrt{3}}{2})$. Midpoint of our circle, midpoint of $C_{1}C_{2}\ :\ O(\frac{a}{2},\frac{a\sqrt{3}}{2})$. Radius of our circle also $r$.