Compass and straightedge contruction of an equilateral triangle inscribed in a given circle with unknown center I haven't found a proper solution for this problem, found in Hartshorne's "Geometry: Euclid and Beyond": 

(4.3) Given a circle, but not given its center, construct an inscribed equilateral triangle in as few steps as possible. 

I managed to construct it in $9$ steps (use of compass or straightedge) but I can't get any lower. Finding circle center takes $5$ of those $9$ uses, and then I need $1$ more to get vertices and $3$ for constructing the triangle.
 A: Two circles + five lines = 7 steps:



*

*Select an arbitrary point $P_0$ on a given circle $C_0$ 
and draw a circle $C_1$
with an arbitrary radius $r$, which is small enough to intersect $C_0$.

*Draw a circle $C_2$ centered at the intersection point $P_1$ with the same radius $r$
to get the intersection points $P_2$ and $P_3$.

*Draw a line through points $P_2,P_1$, intersecting $C_0$ at point $P_4$.

*Draw a line through points $P_1,P_3$, intersecting $C_0$ at point $P_5$.

*Draw a side $P_5 P_4$ of the inscribed equilateral triangle.

*Draw a side $P_4 P_0$ of the inscribed equilateral triangle.

*Draw a side $P_0 P_5$ of the inscribed equilateral triangle.
A: This problem can be found in the online/android/ios puzzle game Euclidea. I came across this thread while researching alternative solutions.
As constructing points doesn't count as a step, the most efficient solution I have found is six steps.
Given circle $c1$ with unknown center:


*

*Mark arbitrary points $P1$ and $P2$ on $c1$

*Draw circle $c2$ with center $P1$ and radius $\overline{P1P2}$

*Mark point $P3$ at the intersection of $c1$ and $c2$

*Draw circle $c3$ with center $P2$ and radius $\overline{P2P3}$

*Mark point $P4$ at the intersection of $c2$ and $c3$

*Draw circle $c4$ with center $P4$ and radius $\overline{P4P2}$

*Mark points $P5$ and $P6$ at the intersections of $c3$ and $c4$

*Draw line through $P5$ and $P2$, intersecting $c1$ at $P7$

*Draw line through $P6$ and $P2$, intersecting $c1$ at $P8$

*Draw line segment $\overline{P7P8}$ to complete $\triangle P2P7P8$



A: If the par score is 7, I'm afraid the best I've managed so far is this bogey 8!
Choose an arbitrary point $A$ on the given circle $\Gamma_0$, and an arbitrary radius, strictly less than the diameter of $\Gamma_0$. All the following circles $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$, $\Gamma_5$ are constructed with this radius.


*

*Draw circle $\Gamma_1$, centre $A$, cutting $\Gamma_0$ at points $B$, $B'$.

*Draw circle $\Gamma_2$, centre $B$, cutting $\Gamma_1$ at point $C$, on same side (of diameter through $A$) as $B'$.

*Draw circle $\Gamma_3$, centre $B'$, cutting $\Gamma_1$ at point $C'$, on same side (of diameter through $A$) as $B$.

*Draw circle $\Gamma_4$, centre $C$, cutting $\Gamma_3$ at point $D \ne A$.

*Draw circle $\Gamma_5$, centre $C'$, cutting $\Gamma_2$ at point $D' \ne A$.

*Draw $AD$, cutting $\Gamma_0$ at $E$.

*Draw $AD'$, cutting $\Gamma_0$ at $E'$.

*Draw $EE'$.
The triangle $AEE'$ is equilateral, and inscribed in $\Gamma_0$.

My justification of this construction (in rough, with a blunt pencil, on a very old sheet of graph paper, covered with previous failed attempts) is as follows:
Draw the equilateral triangle $ABC$, and its reflection on the other side of $AB$, whose apex is the other intersection (call it $F$) of $\Gamma_1$ and $\Gamma_2$; and similarly, the equilateral triangle $AB'C'$, and its reflection on the other side of $AB'$, whose apex is the other intersection (call it $F'$) of $\Gamma_1$ and $\Gamma_3$.
With the tangent to $\Gamma_0$ at $A$, the segments $AF$, $AC'$,  $AB$, $AB'$, $AC$, $AF'$ make a series of angles:
$$
\alpha + \left(\frac{\pi}{3} - 2\alpha\right) + 2\alpha + \left(\frac{\pi}{3} - 2\alpha\right) + 2\alpha + \left(\frac{\pi}{3} - 2\alpha\right) + \alpha = \pi.
$$
By bisecting the two angles $2\alpha$, we construct two line segments making angles of $\pi/3$ with each other and with the tangent to $\Gamma_0$ at $A$. These suffice to construct the inscribed equilateral triangle.
I hope this sketch of a proof will be enough; it doesn't seem worth labouring, as it didn't make par.
A: 0: Given a circle [c, the black one] (with center NOT given) but radius = r.

1: Start from any point C on c, draw another equal circle (d) cutting c at E and F.
2: Draw 2 other equal circles (e and f, centered at E and F respectively) cutting c at A and B respectively.
3: Join AB, BC, CA.
It takes 3 circles + 3 line = 6 steps.
A: This depends on what "given a circle $C$" means. Suppose you must start with just 2 points $P,Q$ in the plane, neither of them on $C,$ and suppose $C$ is "invisible". And suppose that if any line  that is constructible from  $P$ and $Q$ intersects $C$ at a point or points, then you can "see" the point(s) of intersection. That is, the point(s) of the intersection are deemed to have been constructed. 
Draw lines through $P,$ evenly radially spaced at a radial separation of $\pi2^{-n}$ for $n=1,2,3,...$ until $n$ is large enough that two of these lines intersect $C$  at points $E$ and $F,$ or until one of these lines intersects $C$ at 2 pointes $E$ and $F$. Now that  you have 2 points $E,F$ on $C,$ the right bisector of $EF$ will intersect $C$ at  points $G,H$. The mid-point of $GH$ is of course the center of $C.$ 
