<s>[INST] Intersection of 3 circles on equilateral triangle given the difference in their radii Imagine we are given three intersecting circles centered on the vertices of an equilateral triangle of side length $1$, with $(0,0)$ arbitrarily placed at the bottom left corner.  The circles have radii $r$, $r+a$, and $r+b$ respectively (going clockwise again arbitrarily), where $r$ is unknown but $a$ and $b$ are given.  For sake of ease we can assume $0<a<b$ and that $a$ and $b$ are  "safe," aka a solution is possible.  What is the intersection $(x,y)$ of the three circles (in terms of $a$ and $b$) inside the triangle?  

I apologize for the crude nature of the diagram, but it was the best I could come up with using MS Paint. Additionally, I'm 99.9% there's a unique solution as long as we restrict our solution set to points inside the triangle, but this may not be the case.  Thank you so much for any help.
Edit: if anybody has any bright ideas in this direction, the solution doesn't need to be exact; a close approximation with nicer algebra would be much appreciated.
 [/INST] Consider a general point $(x,y)$ inside the triangle,
and let the distance of $(x,y)$ from the vertices $(0,0),$
$\left(\frac12,\frac{\sqrt3}2\right),$ and $(1,0)$ be
$r_1$, $r_2,$ and $r_3$ respectively.
(That is, each of $r_1$, $r_2,$ and $r_3$ is a function of $(x,y)$.)
We want to find a point $(x,y)$ such that 
$r_1 = r$, $r_2 = r+a,$ and $r_3 = r+b,$ if such a point exists,
where $r,$ $a,$ and $b$ are given.
Among other things, $(x,y)$ must satisfy the condition $r_3 - r_1 = b.$
This condition defines one branch of
a hyperbola
with foci at $(0,0)$ and $(1,0)$, semi-major axis $\frac b2,$
and eccentricity $\frac 1b.$
In polar coordinates $(\rho,\theta)$, the equation of this hyperbola is
$$
\rho = \frac{\frac b2\left(\left(\frac 1b\right)^2 - 1\right)}
         {1+\frac 1b\cos\theta}
 = \frac{1 - b^2}{2(b + \cos\theta)}
$$
But $(x,y)$ must also satisfy $r_2 - r_1 = a,$ which puts it on a hyperbola
with foci at $(0,0)$ and $\left(\frac12,\frac{\sqrt3}2\right),$
whose equation is
$$
\rho = \frac{1 - a^2}{2\left(a + \cos\left(\theta - \frac\pi3\right)\right)}
$$
Since the desired point $(x,y)$ must satisfy $r_3 - r_1 = b$ and
$r_2 - r_1 = a$ simultaneously, we have
$$
\frac{1 - b^2}{2(b + \cos\theta)}
= \frac{1 - a^2}{2\left(a + \cos\left(\theta - \frac\pi3\right)\right)}.
$$
Cross-multiply, use the fact that 
$\cos\left(\theta-\frac\pi3\right) =
\frac{\sqrt3}2 \sin\theta + \frac12 \cos\theta,$
and collect everything except the $\sin\theta$ terms together,
and we have
$$
\frac{\sqrt3}2(1 - b^2)\sin\theta
= \left(\frac12 - a^2 + b^2\right) \cos\theta + b(1 - a^2) - a(1 - b^2).\tag1
$$
Let $u=\cos\theta$ (so that $\sin^2\theta = 1-u^2$);
let $n=\frac34(1-b^2)^2,$ $m=\frac12 - a^2 + b^2,$
and $k=b(1 - a^2) - a(1 - b^2)$; 
rewrite Equation $1$ in those terms; and square both sides. The result is
$$
n(1-u^2) = (mu +k)^2,
$$
which is equivalent to
$$
(m^2 + n)u^2 + 2mku + k^2 - n = 0.
$$
Solving for $u$ via the quadratic formula,
\begin{align}
u &= \frac{-2mk \pm \sqrt{4m^2k^2 - 4(m^2k^2+k^2n-m^2n-n^2)}}{2(m^2+n)} \\
&= \frac{-mk \pm \sqrt{m^2n+n^2-k^2n}}{m^2+n} \\
\end{align}
The conditions of the problem require $0\leq\theta\leq\frac\pi3$, 
therefore $\cos\theta \geq \frac12.$ Moreover, $m>0$ and $k>0.$
It follows that only the $+$ case of the $\pm$ sign can possibly
lead to a solution to the original problem.
Wolfram Alpha indicates that
$m^2n+n^2-k^2n$ is positive when $0 < a < b < 1$,
so we don't need to check the sign before taking the square root.
So to find $(x,y)$ given $a$ and $b,$ we set $n,$ $m,$ and $k$ as
described above, then compute the following quantities in the sequence shown:
\begin{align}
u &= \frac{-mk + \sqrt{m^2n+n^2-k^2n}}{m^2+n}, \\
\rho &= \frac{1 - b^2}{2(b + u)}, \\
x &= \rho u, \\
y &= \rho \sqrt{1 - u^2}.
\end{align}
Done!
</s>