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0 sats \ 12 replies \ @Scroogey 14 Oct 2024 \ on: [Daily puzzle] Diameter of circle science

For $O_{1}$ looking at the right triangle

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 $O_{n}$

1 sat \ 11 replies \ @south_korea_ln OP 14 Oct 2024

Looks good. I'm not sure about your value for $O_{3}$ though.

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?

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

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.

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 $O_{3}$ .

102 sats \ 5 replies \ @Scroogey 14 Oct 2024

For $O_{3}$

circles2

(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 $O_{3}$ .

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.

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.

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

I'll check my notes tomorrow.