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Looks good. I'm not sure about your value for
O_3 though.reply
For
O_2(1 - \frac{1}{2} - r)^2 + 1 = 1 + 2*r + r^2For
O_3(1 - \frac{1}{2} - \frac{1}{6} - r)^2 + 1 = 1 + 2*r + r^2(\frac{6}{6} - \frac{3}{6} - \frac{1}{6} - r)^2 + 1 = 1 + 2*r + r^2(\frac{1}{3} - r)^2 + 1 = 1 + 2*r + r^2\frac{1}{9} - \frac{2}{3}*r + r^2 + 1 = 1 + 2*r + r^2\frac{1}{9} - \frac{2}{3}*r = 2*r\frac{1}{9} = \frac{8}{3}*r\frac{3}{9*8} = r\frac{1}{24} = rDiameter is twice that, so
\frac{1}{12}No?
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That would give diameters
\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30}, ...For the sequence of divisors there are multiple candidates
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This is a great website. Didn't know about it. Made me check out the infamous https://oeis.org/A000127 sequence, for which there seem to be many more cases showing this unexpected change from 32 to 31.
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Aren't you using the same symbol
r to denote the different radii of O_1 (as in the drawing) and O_2 (as in the right-hand side of the formula below)? And same comment for the subsequent O_3 equation?ForO_2(1 - \frac{1}{2} - r)^2 + 1 = 1 + 2*r + r^2ForO_3(1 - \frac{1}{2} - \frac{1}{6} - r)^2 + 1 = 1 + 2*r + r^2
I used a different approach to solve this problem (I'll post it as a solution with the next iteration), but that one gives me
1/18 for O_3.reply
For
O_3
(the top-left corner of the triangle is supposed to be the center of
O)The hypotenuse is 1 +
r, where r denotes the radius of O_3.The horizontal leg is simply 1 (this is true for all
O_n).The vertical leg is
1 minus the sum of all diameters O_1 and O_2 minus r of O_3.So, Pythagoras gives
r of O_3.reply
Ok got it. This seems correct. I'll have to figure out why my approach is giving a different result. I likely must have applied it incorrectly. Please don't spend more time on this, I'm assuming the error must be on my side at this point.
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You're right, I incorrectly applied Descartes. 1/12 is correct. My bad...
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Oh, neat, I didn't know about Descartes' theorem!
So you're calculating
k_4 = k_1 + k_2 + k_3 + 2*\sqrt{k_1*k_2 + k_2*k_3 + k_3*k_1}where curvature
k_n is \frac{1}{r_n} (the inverse of the radius).If I pick
O, O' and O_1 to calculate O_21 + 1 + 4 + 2*\sqrt{1*1 + 1*4 + 4*1} = 12Now we can simply repeat with
O, O' and O_2 to calculate O_31 + 1 + 12 + 2*\sqrt{1*1 + 1*12 + 12*1} = 24So radius
\frac{1}{24} or diameter \frac{1}{12}reply
And I guess a direct formula for circle
O_n could bed(n) = \frac{1}{n(n+1)}.reply
Yes, exactly.
Related to Apollonian gaskets
a fractal generated by starting with a triple of circles, each tangent to the other two, and successively filling in more circles, each tangent to another three.
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I'll check my notes tomorrow.
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O_1looking at the right triangle(1-r)^2 + 1^2 = (1+r)^2O_1is\frac{1}{2}.O_2giving\frac{1}{6}andO_3giving\frac{1}{12}etc.O_n