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1 sat \ 6 replies \ @south_korea_ln OP 14 Oct 2024 \ parent \ on: [Daily puzzle] Diameter of circle science
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?
For O_2
(1 - \frac{1}{2} - r)^2 + 1 = 1 + 2*r + r^2
For O_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.
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102 sats \ 5 replies \ @Scroogey 14 Oct 2024
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.
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1 sat \ 4 replies \ @south_korea_ln OP 14 Oct 2024
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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1 sat \ 0 replies \ @south_korea_ln OP 14 Oct 2024
You're right, I incorrectly applied Descartes. 1/12 is correct. My bad...
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0 sats \ 2 replies \ @Scroogey 14 Oct 2024
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_2
1 + 1 + 4 + 2*\sqrt{1*1 + 1*4 + 4*1} = 12
Now we can simply repeat with O, O' and O_2 to calculate O_3
1 + 1 + 12 + 2*\sqrt{1*1 + 1*12 + 12*1} = 24
So radius \frac{1}{24} or diameter \frac{1}{12}
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1 sat \ 0 replies \ @south_korea_ln OP 15 Oct 2024
And I guess a direct formula for circle O_n could be
d(n) = \frac{1}{n(n+1)}.
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1 sat \ 0 replies \ @south_korea_ln OP 14 Oct 2024
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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