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.

First, note that (x,y) is determined by the intersection of the circles with radii r and r+b, Given the "safe" values assumed, so we should be able to express both x and y in terms of r and b only. 
Safe values values are those such that:
  * 2r + a > 1
  * 2r + b > 1
  * r + a < 1
  * r + b < 1  
Since it was specified that a < b, this list can be shorted to:
  * 2r + a > 1 (so that each circle intersects the other 2 circles)
  * r + b < 1  (so even the largest circle has points in the triangle)    
First, we find x in terms of r and b, from the equations for those circles.
Define the coordinates of the vertices as (0,0), $\left(\frac12,\frac{\sqrt3}2\right)$, (1,0) for the circles with radiui r, r+a, and r+b respectively.  
These give the equations:
$$\eqalign{
  x^2+y^2=r^2&\qquad\qquad(1)\cr
  (x-1)^2+y^2=(r+b)^2&\qquad\qquad(2)\cr
  (x-\tfrac12)^2+(y-\tfrac{\sqrt3}2)^2=(r+a)^2&\qquad\qquad(3)\cr}$$  
We will use equations (1) and (2) to find x in terms of r and x.
Expanding (2) we get:
$$x^2 -2x + 1 + y^2 = r^2 + 2rb + b^2$$
Subtracting Equation (1) from this gives:
$$1 - 2x = 2rb +b^2$$ 
Solving for x we obtain:
$$ x = {1-b^2 \over 2} - rb$$
Substituting this for x gives:
$$
\begin{align}
y^2 & = r^2 - x^2  \\
 & =  r^2 - \left({1-b^2 \over 2} - rb \right)^2\\
 & = {(1-b^2)^2 \over 4} - (1 - b^2)rb +r^2b^2 + r^2 \\
 & = {(1-b^2)^2 \over 4} - (1 - b^2)rb +r^2(1 + b^2) \\
 & = {(1-b^2)^2 \over 4} + r(b^3 +b) + r^2(1 +b^2) \\ 
 & = {(1-b^2)^2 + 4r(b^3 +b) + 4r^2(1 +b^2) \over 4} \\ 
 y & = {\sqrt{(1-b^2)^2 + 4r(b^3 +b) + 4r^2(1 +b^2) } \over 2} \\ 
\end{align}
$$
So $(x, y) = \left( {1-b^2 \over 2} - rb , {\sqrt{(1-b^2)^2 + 4r(b^3 +b) + 4r^2(1 +b^2) } \over 2}\right)$.
Done.
Remarks.  
First, we know that since r and b are > 0, the last expression for $y^2$ is positive, so the square root gives a real number for y.
Second, we asserted at the beginning that the point (x,y) is determined only by r and b.  We prove this by showing (x,y) in terms of only r and b, making no use of a.  This also implies that a is determined by r and b.  
By substituting the expressions for x and y in terms of r and b into Equation (3) then solving for a, an expression for a in terms of r and b can be found. 
