Intersection of 3 circles on equilateral triangle given the difference in their radii

Question

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

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.

Answer 1

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.

Answer 2

This can be done, but unless someone has a bright idea, the algebra is pretty revolting. Start with $$\eqalign{ x^2+y^2=r^2&\qquad\qquad(1)\cr (x-1)^2+y^2=(r+b)^2&\qquad\qquad(2)\cr (x-\tfrac12)^2+(y-\tfrac{\sqrt3}2)^2=(r+a)^2&\qquad\qquad(3)\cr}$$ If you now take $(a-b)$ times $(1)$, minus $a$ times $(2)$, plus $b$ times $(3)$, then assuming my algebra is correct you get $$(2a-b)x-b\sqrt3\,y=(a-b)(ab+1)\ .$$ Now solve for $y$ in terms of $x$; use $(1)$ to get $r$ in terms of $x$; then use these and $(2)$ to get an equation in $x,a,b$ only. If, again, my algebra is correct (highly unlikely), you end up with the quadratic $$(12-12b^2-16a^2+16ab-4b^2)x^2+(-12+12b^2+8(2a-b)(a-b)(ab+1))x+(3+3b^4-6b^2-4(a-b)^2(ab+1)^2)=0\ ,$$ which hopefully will turn out to have real solutions.

Sorry for the cop-out but I am definitely leaving the rest to you :)

Answer 3

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!

Answer 4

Let the coordinates of the triangle be $(0,0),(1,0),\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. Then, we can write three equations of the form

$$(x-x_0)^2+(y-y_0)^2=r_0^2$$

for various $x_0,y_0,r_0$. What happens if you assume a solution to each of these concurrently and then solve for $r$?