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1 sat \ 4 replies \ @south_korea_ln OP 14 Oct 2024 \ parent \ on: [Daily puzzle] Diameter of circle science

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.

1 sat \ 0 replies \ @south_korea_ln OP 14 Oct 2024

You're right, I incorrectly applied Descartes. 1/12 is correct. My bad...

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 * 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 * 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 * 1 * 1 + 1 * 12 + 12 * 1 = 24

So radius $\frac{1}{24}$ or diameter $\frac{1}{12}$

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 )} .

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.