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:


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!


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: $$ C_n=S_{n+1}-S_n={1\over n(n+1)}. $$


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 $$C_n = \left(\frac1n - \frac{1}{n+1}\right) = \frac1{n(n+1)}$$


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.