Find a geometric solution for a equilateral tringle within 3 circles For the task of constructing an equilateral triangle with a vertex on each of three concentric circles (the subject of a previous question), I found a solution on: http://mathafou.free.fr/pbg_en/sol113.html, which is hard to understand.

Red triangle is the solution. Quoting from the reference:

With point P as center, draw circles (A), (B) and (C) with radii a,b,c. We can fix A anywhere on circle (A).

So in the graphic is A fixed on the bottom of the outter circle.
A circle is drawn with the radius AP and cuts at M.

The locus of C is then (C). B is deduced from C by rotating with center A and angle π/3.

There is no B in the graph. Just confusing.

The locus of B is then circle (C') image of (C) using this rotation. But it is also (B) from definition. Point B is then the intersection point of (B) and (C').

Why is the circle (C') with middle point drawn here?

Point C is then deduced by inverse rotation.

Ok. We have B1 and now we rotate back 60 degrees to have C.

There are two intersection points B1 and B2 giving the two triangles obtained from analytic solution.
Only one of them contains point P.

Yes, B1, but why? "We must have angle ABP < π/3"
Please, I need some help.
 A: Comment: Experimental approach
$r_1=40, r_2=55, r_3=70$, in this case the difference between radii is equal.
The locus of center of triangle ABC , Q is a circle center on O and radius OQ and we have:
$OQ\approx \frac{\sqrt{40+55+70}+\sqrt{40}+\sqrt{55}+\sqrt{70}}2\approx 17.48$
This value is close to $17.88$ shown in figure.

In this figure the difference of radii is not equal. $r_1=47, r_2=60, r_3=80$ and we have:
$OQ\approx\frac{\sqrt{47+60+80}+\sqrt{47}+\sqrt{60}+\sqrt{80}}2\approx 18.6$
This value is close to $20.8$ shown in figure. So for a rough drawing:
1- calculate OQ and draw a circle $r=OQ$ center on O.
2- take an arbitrary point on this circle (Q). this is the center of triangle. In a particular case one of altitudes of triangle crosses the center O. So draw a line connecting O and Q, this line is perpendicular bisector of BC.
3- The angles both QB and QC make with OQ is $60^o$.By drawing the rays of these angles you get vertices B and C on circles with radii $r_2$ and $r_1$ respectively. Vertex A is the intersection of OQ with third circle with radius $r_3$.
May be by analytic geometry we can find a more accurate formula for $OQ$ in terms of $r_1, r_2$ and $r_3$.
