A question about the Soddy incircle

Question

Let $a$, $b$, and $c$ be the centers of three circles that are mutually tangent, and let $B_r(s)$ be the Soddy incircle, tangent to all three. My question is whether one can disprove the statement: there is a point $x\neq s$ and radii $d_a$, $d_b$ and $d_c$, such that $B_r(x)$ is tangent to $B_{d_a}(a)$, $B_{d_b}(b)$ and $B_{d_c}(c)$, but none of the latter three circles intersect.

Answer 1

Let $r_a, r_b, r_c$ be the radii of the circles centered at $a, b, c$. If there is a $s$ that satisfy the stated condition, then:

$$d_a + d_b < r_a + r_b,\quad d_b + d_c < r_b + r_c, \quad\text{ and }\quad d_c + d_a < r_c + r_a$$ This implies $d_a + d_b + d_c < r_a + r_b + r_c$ and hence:

$$|s-a| + |s-b| + |s-c| = 3r + d_a+d_b+d_c < 3r + r_a+r_b+r_c\\= |x-a| + |x-b| + |x-c|$$

If you start with a configuration where the center of Soddy incircle coincides with the minimum point of $|x-a| + |x-b| + |x-c|$, (e.g. when $r_a = r_b = r_c$ and $a, b, c$ forms an equilateral triangle ) then you can't find a $s$ the meet your requirement.

UPDATE

It does look like you can never find such a $s$.

The condition $d_a + d_b < r_a + r_b$ is equivalent to $|s−a|+|s−b|<|x−a|+|x−b|$. i.e. $s$ is inside an ellipse with foci $\{a,b\}$ which passes through x. The existence of $s$ is equivalent to the interior of the 3 ellipses with foci $\{a,b\}, \{b,c\},\{c,a\}$ which crosses at $x$ has non-empty intersection.

To attack the problem, let us shift the origin to $x$. Let $\hat{a}, \hat{b}, \hat{c}$ be unit vectors pointing towards the three centers $a, b, c$. It is clear $x = \vec{0}$ is lying inside the interior of the convex hull formed by $a, b, c$. This implies existence of 3 positive numbers $\lambda_a, \lambda_b, \lambda_c$ such that:

$$\lambda_a (r+r_a) \hat{a} + \lambda_b (r+r_b) \hat{b} + \lambda_c (r+r_c) \hat{c} = \vec{0}$$

One consequence of this is the angles among $\hat{a}, \hat{b}, \hat{c}$ are all strictly smaller than $\pi$. Rotate the coordinate axis such that $\hat{c}$ is pointing in the $x$-direction. Flipping with respect to $x$-axis if necessary, let $\alpha \in (0,\pi)$ and $\beta \in (\pi,2\pi)$ such that $\hat{a} = (\cos \alpha, \sin \alpha)$ and $\hat{b} = (\cos \beta, \sin \beta)$. The arrangement is roughly as shown below:

Consider the ellipse with foci $\{a,c\}$ which crosses $x = \vec{0}$. Elementary geometry tell us its inward normal vector at $\vec{0}$ is in the direction of $\hat{a} + \hat{c}$. This in turn implies the interior of the ellipse lies inside the half plane defined by $\vec{s} \cdot ( \hat{a} + \hat{c} ) > 0$. Express everything in polar coordinates $\vec{s} = (\rho \cos \theta, \rho \sin \theta)$. The half plane containing the ellipse corresponds to: $$\theta \in \Theta_{ac} = (\frac{\alpha+3\pi}{2},2\pi) \cup [0,\frac{\alpha+\pi}{2}) $$

Similarly, the interior of the ellipse with focii $\{b,c\}$ lies inside the half plane: $$\theta \in \Theta_{bc} = (\frac{\beta+\pi}{2},2\pi)\cup [0, \frac{\beta-\pi}{2})$$

and the interior of the ellipse with foci $\{a,b\}$ lies inside the half plane: $$ \theta \in \Theta_{ab} = ( \frac{\alpha+\beta-\pi}{2}, \frac{\alpha+\beta+\pi}{2} )$$

The intersection of the admissible angles for the first two cases is:

$$\Theta_{ac} \cap \Theta_{bc} = (\frac{\alpha+3\pi}{2},2\pi)\cup[0,\frac{\beta-\pi}{2})$$

Notice $$\frac{\beta-\pi}{2} < \frac{\alpha + \beta - \pi}{2} \;\text{ and }\;\frac{\alpha+3\pi}{2} > \frac{\alpha+\beta+\pi}{2} \implies \Theta_{ac} \cap \Theta_{bc} \cap \Theta_{ab} = \emptyset$$

We can conclude the intersection of the three ellipses is empty.