Geometry Puzzle: Tangent Circles with Integer Radii

Question

Take as a semi-related example a series of circles with radii 10, 9, 8, ..., 2, 1. Place the first (largest) circle in the center and subsequent circles around it, keeping tangency between subsequent circles and the central circle. It would look like this:

The puzzle prompt is this: Find a sequence (any length, not necessarily 10) of strictly decreasing (strictly because otherwise you can pick seven 1s and pack a hex lattice) integer radii such that the final circle perfectly covers the central one. In other words, such that it is also tangent to the second circle (the first non-central). It might look like this:

I ended up posting this over at math SE as well.

Answer 1

I can do it with eight circles:

r0 = 1346015551088116451588241696826457667928305
r1 = 1285983548348276483652942082564013466760320
r2 = 1238768816482055499967626824541529444769070
r3 = 1156239292571620556013720831252688362612000
r4 = 990358148968489476126926154954692437512561
r5 = 961539350943357825433592453776542736631070
r6 = 908374343417739169873995450234402764477000
r7 = 780821295607168322239085153945384792689735

Unfortunately I'm not well-versed in the technology to render beautiful pictures as the OP did, so I'll stick to the numbers for now.

Explanation:

Tom Sirgedas' answer on MSE has set up some key variables for the problem.

Here we use $r_0$ to mean the radius of the center circle, and $r_1>\ldots>r_n$ to mean the surrounding ones, in decreasing order although the ordering is not important.

By Tom's calculation, the central circle's central angle between directions towards two neighboring circles $r_i,r_{i+1}$ is given by $$\theta_i=\arccos\left(1-2s_is_{i+1}\right)=2\arcsin\left(\sqrt{s_i}\sqrt{s_{i+1}}\right),$$ where $s_i=\frac{r_i}{r_0+r_i}$.

Since $r_i=\frac{s_i}{1-s_i}r_0$, we know the radii $r_0,\ldots,r_n$ can all be integers as long as $s_1,\ldots,s_n$ take rational values. We name some new variables $t_1,\ldots,t_n$ whose values shall be determined, such that they satisfy the property $$\sqrt{s_i}\sqrt{s_{i+1}}=\frac{2t_i}{1+t_i^2}$$ and moreover $t_1,\ldots,t_n$ are rational.

Then we derive $$\theta_i=2\arcsin\left(\frac{2t_i}{1+t_i^2}\right)=4\arctan\left(t_i\right).$$ In other words, the rational numbers $t_1,\ldots,t_n$ will need to satisfy the properties $$\left\{\begin{aligned}&0<t_i<\tan\frac{\pi}{12}\\&\sum_{i=1}^n\arctan t_i=\frac{\pi}{2}\end{aligned}\right.$$ to ensure that the corresponding values $\theta_i$ are less than $60^\circ$ each and at the same time sum up to $2\pi$.

For example, a brute-force search might easily help one come up with the combination $$\begin{gathered}\arctan\frac{1}{5}+\arctan\frac{4}{19}+\arctan\frac{2}{9}+\arctan\frac{3}{13}+{}\\\arctan\frac{7}{29}+\arctan\frac{9}{37}+\arctan\frac{1}{4}=\frac{\pi}{2},\end{gathered}$$ but these values do not yet guarantee that the radii $r_1,\ldots,r_n$ will be in decreasing order. So I used Python to generate about a hundred groups of values of $t_i$, and found the solution $$(t_1\ldots t_7)=\left(\frac{8}{31},\frac{1}{4},\frac{7}{30},\frac{13}{59},\frac{3}{14},\frac{1}{5},\frac{2}{9}\right)$$ which happens to work out.

Then we can plug in the values of $t_i$ back to solve for $s_i$, thus for example $$s_1=\frac{2t_1}{1+t_1^2}\cdot\frac{1+t_2^2}{2t_2}\cdot\frac{2t_3}{1+t_3^2}\cdots\frac{2t_7}{1+t_7^2}=\frac{15748992}{32233175},$$ and after that use the formula $r_1=\frac{s_1}{1-s_1}r_0$ from the beginning to get $r_1=\frac{15748992}{16484183}r_0$, and then finally clear the common denominators of $\frac{r_1}{r_0},\ldots,\frac{r_7}{r_0}$ to get $r_0$.

Edit from OP:

It looks like this:

Answer 2

This puzzle has already been solved using 30-digit numbers, but I'm interested in using smaller numbers.

I wrote some code to start with a central unit circle and the loop over values for the next-largest circle. Then I do a depth-first search, picking decreasing radii, and keeping track of the remaining angle and the least-common-multiple of denominators. The search is surprisingly narrow when the denominator is constrained and the tangent of all half-angles is required to be rational (update: or a rational multiplied by the root of some fixed integer). The search allows for an unlimited number of circles. The last radius in a valid configuration can be calculated instead of searched.

I found dozens of 6-digit solutions (see edit history for a few). But this 4-digit solution with fifteen circles is the best by far, and only barely avoids an overlap:

Zoomed view:

Other solutions

Here's 5-digit solution with central radius $42120$:

Answer 3

edwardh's answer above gives eight circles with radii circa $10^{42}$:

rs = [
  1346015551088116451588241696826457667928305,
  1285983548348276483652942082564013466760320,
  1238768816482055499967626824541529444769070,
  1156239292571620556013720831252688362612000,
  990358148968489476126926154954692437512561,
  961539350943357825433592453776542736631070,
  908374343417739169873995450234402764477000,
  780821295607168322239085153945384792689735,
]

I coded up his approach in Python and threw a lot of laptop time at it. Today the computer finally found two smaller solutions, involving radii circa $10^{30}$:

rs = [
  3351747005653424648383741138532,
  3205723049598255148555654977125,
  2775693993058987197622141299968,
  2750071286250985696293256343360,
  2576265167388890063586881775735,
  2355770023351368908545664347968,
  2298393837394654407377433698465,
  2168159590729803674819757284343
]

# The corresponding t values are:
ts = [1/4, 16/67, 7/30, 2/9, 3/14, 39/187, 3/13]

and slightly smaller:

rs = [
  2941564115506288009572255050208,
  2641663701806844064724293810176,
  2588185711710249680183593951791,
  2354857694119310401212561211842,
  2236188368634576996067318957950,
  2089644243395382075169644129792,
  1996023217448211952755320125300,
  1967470772556389886966582369792
]
ts = [1/4, 7/29, 3/13, 2/9, 3/14, 29/138, 11/48]

They look like this:

r0 = 3351747005653424648383741138532	r0 = 2941564115506288009572255050208

I made the above images as SVG using this script. Sadly, StackExchange doesn't support rendering the SVG directly, so the images you see above are PNG screenshots.

I've blogged about this in "Tangent circles of integer radius" (2024-06-10), and updated the post to include Tom's much much smaller answers too (although I still don't know precisely how he's finding them).

Answer 4

The closest I've got so far with some very brute force searching is this sequence of radii: 20, 19, 18, 17, 16, 14, 8, 5, 4, 3, 2. The final circle overlaps with the rightmost one by about 5.39x10e-6. I think I'll close this and post a similar question over in math.

Answer 5

In Figure 12, one possible solution is depicted, with radii $1/146$, $1/27$, $1/23$, and $1/18$, respectively. This can easily be scaled by $90666$ times (the LCM) to become integers, namely $621$, $3358$, $3942$, $5037$. Judging from the magnitude of the numbers, its no wonder brute force computer search wasn't fruitful.