0
$\begingroup$

So I stumbled across this problem a while ago, and have been unable to make any progress with it. Exhaustive searching has found no answers elsewhere.

With four solid balls $A$, $B$, $C$, and $D_1$, all of which are tangent to each other, you construct ball $D_2$ to be tangent to all four of them. Assume that $D_1$ has radius 1. After constructing $D_2$, you create ball $D_3$ tangent to $A$, $B$, $C$, and $D_2$. What is the radius of $D_{2017}$?

Approach

I started by just examining the two dimensional case. In this instance, we would ignore sphere $C$, because you can only have up to 4 mutually tangent spheres in a plane. Remembering that curvature is 1/radius, we So, we let curvatures of $D_1$ equal 1, and call the other two radii a and b respectively. Then, by Descartes's theorem, the curvature _$k_2$ of $D_2$ is defined by $$r_2 = 1+a+b\pm2\sqrt{a+b+ab} $$ So then, the next step would be to iterate this process 2016 times to get $k_2017$. This is where I'm stuck. I can't seem to manipulate the equation into a form that doesn't immediately turn ugly.

After this, the approach for spheres would probably be similar, using the generalization $$\left(\sum _{{i=1}}^{{5}}k_{i}\right)^{2}=3\,\sum _{{i=1}}^{{5}}k_{i}^{2}$$ of Descartes's theorem as applied to spheres. Again, though, I don't see how to manipulate this into a useful form.

$\endgroup$
  • $\begingroup$ What are the radii of $A,B,C?$ are they 1 as well? $\endgroup$ – Doug M Jun 28 '17 at 20:24
2
$\begingroup$

Here is an approach for the 2D case that should extrapolate to the 3D case.

We create a "circle of inversion."

Project every point on circles $A,B,C$ based on the inverse of the distance from a point on $A$ (marked with the fuzzy dot) i.e. if some point on $B$ is $d$ units from the fuzzy dot, project along the same ray to the point $\frac {1}{d}$ units away.

Circles that intersect the fuzzy dot will project to lines. Circles that don't intersect the fuzzy dot will project to circles. Points of tangency will project to points of tangency.

It is then easy to find the sequence of purple circles. Then project them back.

enter image description here

In the 3D case it will be a sphere of inversion. Nonetheless find a sequence of spheres all tangent to one plane sounds easier than finding a sequence of sphere all tangent to another sphere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.