3

Here is the equation I am asked to find. I have researched a lot into the Pappus chain, methods primarily involving circle inversion, but can only find examples of calculating the nth radius directly.

Any hints/tips greatly appreciated.

Thanks

edit to add: Under the inversion, the image maps to two vertical lines, with tangential circles of equal radii vertically poisition between the two. All these circles are of equal radius, so surely the recursion is trivial.. That's assuming the radius is mapped to the radius.... Help :/

Yiorgos S. Smyrlis
  • 83,933
Dennis
  • 91

1 Answers1

1

As you mention: Under inversion in a circle centered at the origin ($O$), the image of the $C_i$ is a vertical stack of congruent circles $C^\prime_i$, say, with centers $P_i$ and common radius $k$.

Let $A_i$ and $B_i$ be the points of intersection of line $\overleftrightarrow{OP_i}$ with $C^\prime_i$. Then $\overline{AB}$ is a diameter of $C^\prime_i$, and the inverse $\overline{A^\prime B^\prime}$ is a diameter of $C_i$. (Note: The inverse of $P$ is not the midpoint of the inverted diameter, so no: radius does not map to radius. That's okay; the diameter gets us what we need.) Therefore, conveniently assuming that we've inverted with respect to a unit circle,

$$2r_i = |\overline{OA^\prime_i}| - |\overline{OB^\prime_i}| = \frac{1}{|\overline{OA_i}|}-\frac{1}{|\overline{OB_i}|} = \frac{1}{|\overline{OP_i}|-k}-\frac{1}{|\overline{OP_i}|+k} = \frac{2k}{|\overline{OP_i}|^2-k^2}$$

so that $$k r^{-1}_i = |\overline{OP_i}|^2-k^2$$ and $$k \left( r^{-1}_{i} - 3r^{-1}_{i+1} + 3r^{-1}_{i+2} - r^{-1}_{i+3}\right) = |\overline{OP_i}|^2 - 3 |\overline{OP_{i+1}}|^2 + 3 |\overline{OP_{i+2}}|^2 - |\overline{OP_{i+3}}|^2 \qquad (\star)$$ We can write $P_i := (\;h,\;(2i-1)k\;)$, for an appropriate $h$. This gives $$|\overline{OP_i}|^2 = h^2 + ( 2 i - 1 )^2 k^2$$ which, as one can verify, makes the right-hand side of $(\star)$ vanish, proving (a). A counterpart to $(\star)$ proves (b).

Blue
  • 75,673