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What is the sum of the areas of the grey circles? I have not made any progress so far.

http://s17.postimg.org/iwb126q4b/1408524358160.jpg?noCache=1408526351

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Ooh, that's an interesting problem... –  Shakespeare Aug 20 at 9:28
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Hint: use triangles formed by connecting the radii of different circles to find the values of all the radii. From there, its easy to sum the areas to infinity. ( the sum of an infinite geometric progression will be of help here) –  Assaultous2 Aug 20 at 9:31
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@Assaultous2 What does "connecting the radii" mean here? I see lots of potential triangles to be drawn, but it seems rather daunting to deduce the radius of even the largest circle in a simple way... –  rschwieb Aug 20 at 12:53
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@Assaultous2 It's tempting to guess it might be a geometric series, but not obvious. Could you clearly indicate what you were actually thinking? Or was it just an off-the-cuff guess-hint? –  rschwieb Aug 20 at 16:47
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@KyleGannon I think that formalizing it is part of the problem. The figure has all the information you need. –  becko Aug 20 at 18:14

2 Answers 2

Consider the picture below.

The illustration

On the left hand side we have the original picture which we have put in the complex plane, and on the right hand side is its image under the Möbius transformation $f(z) = \frac{z}{2-z}$. The inverse transformation is $g(z) = \frac{2z}{z+1}$. In the new coordinates the green circles have radius $\frac{1}{2}$ and center at $C_k = \frac{1}{2} + i(k + \frac{1}{2})$, $k=0,1,2,\dots$.

We have $\text{Jac}\, g(z) = |g'(z)|^2 = \frac{4}{|z+1|^4}$. Therefore the area of the disks in the original picture is $$\sum_{k=0}^\infty \int_{B_k} \frac{4}{|(x+iy)+1|^4} \, dx \, dy,$$ where $B_k = \{z : |z - C_k| \le \frac{1}{2}\}$ are the discs in the second picture. Calculating the integral inside the sum turned out to be tedious, however, so I opted for another route. EDIT: achille hui showed how to calculate this integral in an answer to my question at Integral related to a geometry problem. This yields a shorter way to get the answer.

The points where a green circle touches the red line or the blue line are at $i(k + \frac{1}{2})$ and $1 + i(k + \frac{1}{2})$ respectively. Therefore in the original picture they are at $$A = g(i(k+\frac{1}{2})) = \frac{2(k+\frac{1}{2})^2}{1 + (k + \frac{1}{2})^2} + i \frac{2(k+\frac{1}{2})}{1 + (k + \frac{1}{2})^2}$$ and $$B = g(1 + i(k + \frac{1}{2})) = \frac{4 + 2(k+\frac{1}{2})^2}{4 + (k + \frac{1}{2})^2} + i \frac{2(k+\frac{1}{2})}{4 + (k+\frac{1}{2})^2}.$$ Now to find the center of the green circle we calculate the intersection of the lines $1 + t(A-1)$ and $\frac{3}{2} + s(B - \frac{3}{2})$. Real and imaginary part give us two linear equations for $s$ and $t$, and we end up with the solution $$s = \frac{4k^2 + 4k + 17}{4k^2 + 4k + 9}, \quad t = \frac{4k^2 + 4k + 5}{4k^2 + 4k + 9}.$$ Thus the center is at $$1 + t(A-1) = \frac{8k^2 + 8k + 6}{4k^2 + 4k + 9} + i \frac{8k + 4}{4k^2 + 4k + 9}.$$ The radius is then $$|A - 1 - t(A-1)| = |A-1| |t-1| = \frac{4}{4k^2 + 4k + 9}.$$ Hence the answer to the problem is $$\sum_{k=0}^\infty \pi \frac{16}{(4k^2 + 4k + 9)^2},$$ for which Wolfram Alpha gives the closed form $$\frac{1}{16} \pi^2 (\sqrt{2} \tanh(\sqrt{2}\pi) - 2 \pi \text{sech}^2(\sqrt{2}\pi)) \approx 0.8699725\dots$$

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I'm taking the dangerous step of admiring a solution I haven't totally read :) I was hoping there was a way to transform the picture first and then find the solution, and it looks like that's the strategy here. Nice! –  rschwieb Aug 20 at 16:44
    
@rschwieb: That is the strategy indeed. What we have here is basically just a non-linear coordinate transformation after which the circles are of same size and arranged in a vertical strip. By taking the inverse transformation it then becomes possible to calculate the original coordinates of the circles, or alternatively do the area calculation in the transformed plane by using the Jacobian of the inverse transformation. –  J. J. Aug 20 at 16:50
    
I checked numerically using Apollonius problem formula and the answer is approximately .86997 –  Seb Aug 20 at 20:28
    
Where did this Möbius transformation come from? how can you prove it is true? –  nbubis Aug 21 at 0:52
    
@nbubis it's a special transformation of the complex plane that can map circles to lines and circles. That makes it a candidate for "straightening things out" here ;) –  rschwieb Aug 21 at 1:35

This is an alternate approach to derive the area of the circles using Descartes four circle theorem.

WOLOG, assume $R = 1$. Let us call

  • the outer circle (with radius $r_a = 1$) as $C_a$.
  • the inner green circle (with radius $r_b = \frac12$) as $C_b$.
  • the $1^{st}$ gray circle (the largest one at bottom) as $C_0$.
  • the $2^{nd}$ gray circle (the one above $C_b$ and $C_0$) as $C_1$.
  • in general, for any $k > 0$, we will call the gray circle above $C_b$ and $C_{k-1}$ as $C_k$.
  • We can reflect the figure in question vertically. For $k < 0$, we will let $C_k$ be the mirror image of the circle $C_{-(1+k)}$.

The key is for any $k\in\mathbb{Z}$, the four circles $C_a, C_b, C_k$ and $C_{k\pm 1}$ are kissing each other. Let $r_k$ be the radius of $C_k$. We can apply Descartes four circle theorem to obtain:

$$\left(\frac{1}{r_{k\pm 1}} + \frac{1}{r_k} + \frac{1}{r_b} - \frac{1}{r_a} \right)^2 = 2 \left(\frac{1}{r_{k\pm 1}^2} + \frac{1}{r_k^2} + \frac{1}{r_b^2} + \frac{1}{r_a^2}\right)\tag{*1}$$ Since $r_a = 1$ and $r_b = \frac12$, this implies $\displaystyle\;\frac{1}{r_{k \pm 1}}\;$ are the two roots of quadratic equation $$\left(\rho + \frac{1}{r_k} + 2 - 1 \right)^2 = 2 \left(\rho^2 + \frac{1}{r_k^2} + 2^2 + 1^2 \right) $$ and hence $$\frac{1}{r_{k+1}} + \frac{1}{r_{k-1}} = 2\left(\frac{1}{r_k} + 1\right) \quad\iff\quad \frac{1}{r_{k+1}} + \frac{1}{r_{k-1}} - \frac{2}{r_k} = 2$$

The RHS is an inhomogeneous linear recurrence relation on $\displaystyle\;\frac{1}{r_k}\;$ with constant term. Since the characteristic polynomial $(\lambda-1)^2$ has a double root at $1$, we know its solution must have the form $\displaystyle\;\frac{1}{r_k} = k^2 + \lambda k + \mu\;$. By symmetry,

$$r_k = r_{-(1+k)}\quad\implies\quad \lambda = 1 \quad\implies\quad \frac{1}{r_k} = k(k+1) + \mu$$

To fix $\mu$, apply $(*1)$ to the case $C_a, C_b, C_0$ and $C_{-1}$, we get

$$\left(2\mu + 1\right)^2 = 2\left(2\mu^2 + 5\right) \quad\implies\quad \mu = \frac{9}{4} $$ From this, we get $$r_k = \frac{4}{4k^2 + 4k + 9} \quad\implies\quad \text{Area} = \sum_{k=0}^\infty \frac{16\pi}{(4k^2+4k+9)^2}$$

reproducing what's in J.J's answer.

Update

To evaluate the sum, we will use following infinite product expansion of $\cosh(\pi x)$,

$$\cosh\pi x = \prod_{k=0}^\infty \left( 1 + \frac{x^2}{(k+\frac12)^2}\right) $$

Taking logarithm, differentiate w.r.t $x$ and divide by $2x$, we get

$$\sum_{k=0}^\infty \frac{1}{(k+\frac12)^2 + x^2} = \frac{\pi}{2x} \tanh\pi x$$

Differentiate w.r.t $x$ and divide by $-2x$ once more, we get

$$\sum_{k=0}^\infty \frac{1}{((k+\frac12)^2 + x^2)^2} = \frac{\pi}{4x^3}\tanh(\pi x) - \frac{\pi^2}{4x^2\cosh^2(\pi x)} $$ With this, we find $$\text{Area} = \sum_{k=0}^\infty \frac{\pi}{((k+\frac12)^2 + 2)^2} = \frac{\pi^2}{16}\left[\sqrt{2}\tanh(\pi\sqrt{2}) - \frac{2\pi}{\cosh^2(\pi\sqrt{2})}\right] $$ Once again, this agrees with J.J's result.

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(+1) When I saw this problem, I thought of using the Descartes Four Circle Theorem aka Soddy's Theorem. I have a great affinity for this theorem and its extensions. –  robjohn Aug 20 at 19:57

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