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A couple weeks ago during a maths Olympiad I ran across a Geometry problem which i found particularly difficult. I don't have the exact rubric but it goes like this:

Let there be two tangent circles of radius 1 and a line tangent to both, as shown in the picture below. Let $C_1$ be the diameter of the circle tangent to the line and the two circles (the circle between both circles and the line), $C_2$ the diameter of the circle tangent to both radius 1 circles and the previous small circle (but not the line), and so on. Obtain an expression for $C_n$.

a big picture

In the picture, blue and red circles are the radius 1 circles, the x axis represents the line, the green circle is the "1st circle" (so it has radius $\frac{C_1}{2}$, remember $C_n$ is the diameter of the "nth circle", defined from 1 onwards) and the orange one is the 2nd (so its radius is $\frac{C_2}{2}$).

I couldn't solve it in the time that was given, but i suspected the answer to be:

1/(n(n+1))

After googling, i found that using Descartes' kissing circles theorem helps solve it, but it seems too overkill. If anyone can figure out and post an elegant solution, it would be really appreciated!

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Let's define: $$ S_n=\sum_{k=1}^{n-1}C_k, $$ so that $S_1=0$, $S_2=C_1=1/2$, and so on.

The radius $x$ of the $n$-th circle can be computed by applying Pythagoras' theorem to triangle $AOH$ in the diagram below. We have: $$ OA=1+x,\quad OH=1,\quad AH=1-S_n-x, $$ and from $$ (1+x)^2=(1-S_n-x)^2+1 \quad\hbox{we get:}\quad C_n=2x={(1-S_n)^2\over 2-S_n}={1\over 2-S_n}-S_n. $$ enter image description here

But $S_{n+1}=C_n+S_n$, thus we obtain the recursive relation: $$ S_{n+1}={1\over 2-S_n}. $$ This is satisfied by $S_n=(n-1)/n$ and finally: $$ C_n=S_{n+1}-S_n={1\over n(n+1)}. $$

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a set of circles under inversion

Translate everything towards the negative $x$-direction for unit distance. Under this translation,

  • the contact point of the red and blue circle become the new origin.
  • the old $x$-axis becomes the line $y = -1$.

Invert with respect to the unit circle in the new coordinate system. Under this circle inversion,

  • The red circle become the line $x = \frac12$.
  • The blue circle become the line $x = -\frac12$.
  • The contact points $(\pm 1, -1)$ of the red/blue circles and the line $y = -1$ get mapped to $(\pm \frac12, -\frac12)$. Since any straight line will mapped to a circle passing through origin, we find the line $y = -1$ get mapped to a circle of radius $\frac12$ centered at $(0,-\frac12)$.

  • Since the green circle are tangent to the red/blue circles and the line $y = -1$. It get mapped to a circle radius $\frac12$ centered at $(0,-\frac32)$.

  • By a similar argument, one find the orangle circle get mapped to a circle of radius $\frac12$ centered at $(0,-\frac52)$.

Combine these observations, it is not hard to see in general, the $n^{th}$ circles we wish to construct get mapped to a circle of radius $\frac12$ centered at $(0, -(n + \frac12))$. Notice the nearest and farest distance of the inverted $n^{th}$ circle to the origin is $n$ and $n+1$. When one one invert it back, one find the diameter of the original $n^{th}$ circle is simply $$C_n = \left(\frac1n - \frac{1}{n+1}\right) = \frac1{n(n+1)}$$

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Let's consider a small circle with radius $C_{n+1}=r$, touching two circles of radii $1$ and $C_n=R$ and a line. From three right triangles, whose hypotenuses are between circles' centers and legs are parallel and perpendicular to the line, we have: $$\begin{cases} (1-r)^2 + a^2 & = (1+r)^2 \\ (R-r)^2 + b^2 & = (R+r)^2 \\ (1-R)^2 +(a+b)^2 & = (1+R)^2 \end{cases} $$ where $a,b$ are displacement of the center of the $r$-circle from other circles' centers along the line.

That simplifies to $$\begin{cases} a^2 & = 4r \\ b^2 & = 4Rr \\ (a+b)^2 & = 4R \end{cases} $$ hence $$\Big(2\sqrt r + 2\sqrt{Rr}\Big)^2 = 4R$$ and $$C_{n+1} = r = \frac R{(1 + \sqrt R)^2} = \frac {C_n}{\left(1 + \sqrt{C_n}\right)^2}\tag{*}$$

Putting $C_1=1$ we get $$C_2=1/4,\ C_3=1/9,\ C_4=1/16\,\dots$$ so possibly $\boxed{C_n=1/n^2}$ ...?

Indeed, $C_1 = 1 = 1/(1^2)$ and it satisfies the $(*)$ recurrence, too: $$\frac {C_n}{\left(1 + \sqrt{C_n}\right)^2} = \frac {1/n^2}{\left(1 + \sqrt{1/n^2}\right)^2} = \frac {1/n^2}{\left(1 + 1/n\right)^2} = \frac 1{(n+1)^2} = C_{n+1} $$


With the Descartes' theorem we would directly get curvatures $$K_1 = 1 \\ K_{n+1}=1+K_n + 2\sqrt{1\cdot K_n} = \big(1+\sqrt{K_n}\big)^2$$ which resolves to $$K_n = n^2$$ hence the $n$-th radius $$C_n = 1/K_n = 1/n^2$$ as above.

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  • $\begingroup$ That's a nice approach. However, remember the problem asks for the diameter of the nth circle, and i added clarification since i think i was not clear enough and you did not understand it correctly. Refer to i.stack.imgur.com/yqZol.png $\endgroup$ – Juan Cruz de La Torre Sep 28 '17 at 12:24
  • $\begingroup$ That's right, I looked for a next little circle between one of the original circles (radius $1$), the line and the previous small circle. However I'm not able to fix the solution now, Hopefully in next several hours... :) $\endgroup$ – CiaPan Sep 28 '17 at 12:29

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