Suppose I have 3 circles with identical radius,$R$ overlapping each other, with the distance of the center between them to be $d_{12}$, $d_{23}$, $d_{13}$ respectively. What is the largest radius,$r$ of the circle that I can fit in the overlapping areas?
If it is just 2 circles, I think the largest possible circle in the overlapping area is $$r = R-\frac{1}2d$$
given that $d\leq2R$
But I am stumped if it comes to 3 circles. I thought of finding the $r$ if I only look at 2 circles each time, and from there look for the smallest $r$. However, by looking from drawings, I don't think it is correct.
$$r=min(r_{12},r_{23},r_{13})$$
Where $r_{xy}$ is the r relative to 2 circles.
I am not sure if I worded my questions clearly, and I will gladly add any additional .
Trying to find the $r$ of the red circle:
Best Answer
Let $\Delta ABC$ be the triangle with the centres of three circles as its vertices. For the same radii, the radical axes of two circles are the perpendicular bisectors of each sides $a$, $b$ and $c$. Hence, they are concurrent at the circumcentre. Now the circumcentre $O$ is equidistance to the vertices. The required radius is just the difference between $R$ and the circumradius, that is
\begin{align} r &= R-\frac{abc}{4\Delta} \\ &= R-\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}} \\ &= R-\frac{abc}{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}} \\ \end{align}