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Draw circles of radius $1, \frac12, \frac13, ...$ such that the first two are externally tangent, then starting with the third, each circle is externally tangent to the previous two, with the path of the circle's centres turning in the same direction (clockwise say).

Let $d_n=$ distance between the first circle's centre and the $n$th circle's centre.

Is there a closed form for $L=\lim\limits_{n\to\infty}d_n$ ?

$L$ is the length of the red line segment below.

enter image description here

I superimposed cartesian axes and tried to express the coordinates of the $n^{\text{th}}$ circle's centre in terms of the coordinates of the previous two circles' centres, then take the limit as $n\to\infty$. But the algebra seems to be hopelessly complicated.

By manually drawing circles on desmos, it seems that $L\approx1.116$. Maybe $L=\frac{2}{\ln{6}}$ ?

Dan
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    I believe from observation; as the circles get closer and closer to the centre of the first three circles, they eventually approach the Simson-Steiner point. (I.e. Draw the perpendicular bisectors of the line segments connecting the centers of the circles and their point of tangency. The point where these bisectors intersect is the center of the system of circles, which is referred to as the Simson-Steiner point.). I will the formula in the next comment, I do not have much space here... – Dstarred Jan 21 '23 at 07:08
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    Formula for Simson-Steiner point; $$x_C = \frac{r_1x_1 + r_2x_2 + r_3x_3}{r_1 + r_2 + r_3} \ y_C = \frac{r_1y_1 + r_2y_2 + r_3y_3}{r_1 + r_2 + r_3}$$ Where, $(x_C, y_C)$ is point your looking for. $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ are the coordinates of the centres of first three the circles, and $r_1, r_2, r_3$ are the radii of the first three circles respectively. $$\quad$$ Assume that the first circle centre is $(0, 0)$ and second centre is $(0, \frac{3}{2})$. Calculate the third centre using the cosine laws since you have all three sides of the circum-triangle... – Dstarred Jan 21 '23 at 07:14
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    ... Then compute the Simson-Steiner point as mentioned and find the absolute distance between the first centre and the Simson-Steiner point. This should be the limit you are looking for. – Dstarred Jan 21 '23 at 07:16
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    An interesting question, but I am skeptical about the idea that $L$ admits an elementary closed form. Just for reference, a numerical value of the limit is $$ L = 1.1147919825437510758\ldots, $$ which is accurate to within $10^{-20}$. Inverse symbolic calculators seem unable to find a closed form for this value. – Sangchul Lee Jan 21 '23 at 07:25
  • @Dstarred Using your method, I'm getting $x_C=4\sqrt{11}/99$ and $y_C=11/18$, which gives $L\approx0.626$, but I think that's way off. I think we have to consider more than just the first three circles. – Dan Jan 21 '23 at 08:11
  • @SangchulLee Out of curiosity, were you already skeptical about a closed form before you used the inverse symbolic calculator? – Dan Jan 21 '23 at 08:17
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    @Dan, Yes. My intuition is based on my personal impression that the sequence $(r_n)$ of radii must be fine-tuned in order to yield a closed-form expression for $L$ (which itself is based on the fact the the center of the new circle is a nonlinear function of the previous centers). However, the choice $r_n=1/n$ doesn't seem such one. This is a very algebraic choice, not reflecting any specific properties of planar geometry. – Sangchul Lee Jan 21 '23 at 09:05
  • @SangchulLee. May I ask where the number comes from ? Remember that I cannot see the plot. Thanks and cheers :-) – Claude Leibovici Jan 21 '23 at 09:27
  • Writing $p_n$ for the center of the $n$th circle, I derived a formula for $p_{3m+1} - p_{3m-2}$ in terms of $m$ and then expanded it as a series in $\frac{1}{m}$. Although I was unable to determine the exact formula for the coefficients, it seems that the coefficient is monotone and vanishing as degree grows. So I truncated the series and summed it up for $m = 1, 2, \ldots$ – Sangchul Lee Jan 21 '23 at 12:16

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