A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k \ge 1$, the circles in $\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is \[\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?\] [asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.75)); }[/asy]
Let the two circles from $L_0$ be of radius $r_1$ and $r_2$, with $r_1>r_2$. Let the circle of radius $r_1$ be circle $A$ and the circle of radius $r_2$ be circle $B$. Now, let the circle of $L_1$ have radius $r_3$. Let this circle be circle $C$. Draw the radii of the three circles down to the common tangential line and connect the radii. Draw two lines parallel to the common tangential line of the two layers intersecting the center point of circle $B$ and the center point of circle $C$. Now, we have $3$ right triangles with a line of common length (The two parallel lines). Using the pythagorean theorem, we get the formula $\sqrt{(r_2+r_1)^2-(r_2-r_1)^2}=\sqrt{(r_1+r_3)^2-(r_1-r_3)^2}+\sqrt{(r_2+r_3)^2-(r_2-r_3)^2}$ Now we solve for $r_3$. Square both sides, use the identity $(a^2-b^2)=(a+b)(a-b)$ and simplify: $(2r_2)(2r_1) = (2r_1)(2r_3)+2\sqrt{16r_1r_3r_2r_3}+(2r_2)(2r_3)=4(r_1r_3+r_2r_3+2r_3\sqrt{r_1r_2})=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \\ 4r_2r_1=4r_3(r_1+r_2+2\sqrt{r_1r_2}) \implies r_3=\frac{r_2r_1}{r_1+r_2+2\sqrt{r_1r_2}}$ Now, let's change this into a function to clean things up: $f(x,y) = \frac{xy}{x+y+2\sqrt{xy}}=\frac{xy}{(\sqrt{x}+\sqrt{y})^2}$ Let's begin to rewrite the sum we want to find in terms of the radii of the circles, call this $g(x)$: $g(x) = \frac{1}{\sqrt{r}}=\frac{1}{\sqrt{\frac{xy}{(\sqrt{x}+\sqrt{y})^2}}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}} = \frac{\sqrt{y}}{y}+\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}$ Using this, we can find the sum of some layers: $L_0$, $\frac{1}{70}+\frac{1}{73}$, $L_0$ and $L_1$: $\frac{1}{70}+\frac{1}{73}+\frac{1}{70}+\frac{1}{73} = 2(\frac{1}{70}+\frac{1}{73})$ This is interesting, we have that the sum of Layer 0 and Layer 1 is equal to twice of Layer 0. If we continue and find the sum of layers 0, 1 and 2, we see it is equal to $5(L_0)$. This is getting very interesting, there must be some pattern. First of all, we should observe that finding $g(x)$ of a circle is equivalent to adding up those of the 2 larger circles to construct the smaller one. Second, upon further observation, we can draw out the layers. When we're finding the next layer, we can split the current layers across the center, so that each half includes the center circle $L_1$. Now, if we were to find $g(x)$, we notice we are doubling the current sum and including the center circle twice. So, the recursive sum would be $a_n=3a_{n-1}-1$. So, applying this new formula, we get $\sum_{C \in S}\frac{1}{\sqrt{r}} = (3(3(3(3(3(3-1)-1)-1)-1)-1)-1)(\frac{1}{70}+\frac{1}{73})=365\cdot(\frac{1}{70}+\frac{1}{73})=365\cdot\frac{143}{70\cdot73}=\fbox{\frac{143}{14}}$