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[Daily puzzle] Diameter of circle
2154 sats \ 13 comments \ @south_korea_ln 14 Oct 2024 science
The two circles O and O^\prime with radii of 1 are tangent with each other and with the horizontal line at the bottom. Circle O_1 is inscribed with a maximum possible radius inside O, O^\prime, and the horizontal line. O_2, O_3,... are successively inscribed following the same rule maximizing their radius.
Find the diameter of a circle O_n.
Previous iteration: #720410 (quite a few different valid solutions were found for this one, congrats~~)
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0 sats \ 12 replies \ @Scroogey 14 Oct 2024
For O_1 looking at the right triangle
Pythagoras says
(1-r)^2 + 1^2 = (1+r)^2
which means the diameter of O_1 is \frac{1}{2}.
This can be repeated of O_2 giving \frac{1}{6} and O_3 giving \frac{1}{12} etc.
I don't yet have the generic formula for O_n
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1 sat \ 11 replies \ @south_korea_ln OP 14 Oct 2024
Looks good. I'm not sure about your value for O_3 though.
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0 sats \ 10 replies \ @Scroogey 14 Oct 2024
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
(\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} = r
Diameter is twice that, so \frac{1}{12}
No?
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106 sats \ 1 reply \ @Scroogey 14 Oct 2024
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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1 sat \ 0 replies \ @south_korea_ln OP 14 Oct 2024
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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1 sat \ 6 replies \ @south_korea_ln OP 14 Oct 2024
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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1 sat \ 0 replies \ @south_korea_ln OP 14 Oct 2024
I'll check my notes tomorrow.
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