A beautiful solution to a math contest problem

I was recently involved in writing a math contest at my university. One of the problems I contributed was the following:

Let $L$ be the line passing through points $(\sqrt{2},0)$ and $(0,\sqrt{2})$ of the XY -plane. Let $C_0$ be the circle of radius 1 centered at (0, 0). Let $C_1, C_2, C_3,\ldots$ be an infinite sequence of circles such that each $C_i$  is centered on the x-axis, each $C_i$ is tangent to $L$ and $C_{i+1}$ is tangent to $C_i$ for all $i\geq 0$. What is the total area of all these circles? 

The solution I had in mind was to find a recursive formula for the radius of $C_i$ and then sum the areas as a geometric series. One student came up with an absolutely stunning method, worthy of Archimedes. Consider the trapezoid $T_0$ formed from the $x$-axis, the line $L$ and the two vertical tangents to $C_0$. The area of the trapezoid is $2\sqrt{2}$ while the area of $C_0$ is $\pi$. So the ratio $\lambda$ of the area of $C_0$ to the area of $T_0$ is $\frac{\pi}{2\sqrt{2}}$. Now observe that the ratio of the area of $C_i$ to the area of the trapezoid $T_i$ formed from the vertical tangents to $C_i$, the line $L$ and the $x$-axis, is still $\lambda$ by similarity. So $$\sum \mathrm{Area}(C_i)=\lambda \sum \mathrm{Area}(T_i).$$
The sum of the areas of $T_i$ is just the area of an isosceles right triangle with side lengths $1+\sqrt{2}$, i.e. $\frac{(1+\sqrt{2})^2}{2}$. So we get $\frac{\pi}{2\sqrt{2}}\cdot \frac{(1+\sqrt{2})^2}{2}$. Magic!