Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2015 AMC 12A Problems/Problem 25

Revision as of 13:51, 15 June 2021 by Ericpatrick2005 (talk | contribs) (→‎Solution 1)

Problem

A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as followsLayer $L_0$ 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]

$\textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}$

Solution 1

Let us start with the two circles in $L_0$ and the circle in $L_1$. Let the larger circle in $L_0$ be named circle $X$ with radius $x$ and the smaller be named circle $Y$ with radius $y$. Also let the single circle in $L_1$ be named circle $Z$ with radius $z$. Draw radii $x$, $y$, and $z$ perpendicular to the x-axis. Drop altitudes $a$ and $b$ from the center of $Z$ to these radii $x$ and $y$, respectively, and drop altitude $c$ from the center of $Y$ to radius $x$ perpendicular to the x-axis. Connect the centers of circles $x$, $y$, and $z$ with their radii, and utilize the Pythagorean Theorem. We attain the following equations. \[(x - z)^2 + a^2 = (x + z)^2 \implies a^2 = 4xz\] \[(y - z)^2 + b^2 = (y + z)^2 \implies b^2 = 4yz\] \[(x - y)^2 + c^2 = (x + y)^2 \implies c^2 = 4xy\]

We see that $a = 2\sqrt{xz}$, $b = 2\sqrt{yz}$, and $c = 2\sqrt{xy}$. Since $a + b = c$, we have that $2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}$. Divide this equation by $2\sqrt{xyz}$, and this equation becomes the well-known relation of Descartes's Circle Theorem $\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.$ We can apply this relationship recursively with the circles in layers $L_2, L_3, \cdots, L_6$.

Here, let $S(n)$ denote the sum of the reciprocals of the square roots of all circles in layer $n$. The notation in the problem asks us to find the sum of the reciprocals of the square roots of the radii in each circle in this collection, which is $\textstyle\sum_{n=0}^{6}S(n)$. We already have that $S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}$. Then, $S(2) = 2S(1) + S(0) = 3S(0)$. Additionally, $S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)$, and $S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)$. Now, we notice that $S(n + 1) = 3S(n)$ because $S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)$, which is a power of $3.$ Hence, our desired sum is $(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)$. This simplifies to $365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{(D)}$.

Note that the circles in this question are known as Ford circles.

Solution 2 (Pythagorean Theorem)

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 $\sum_{C \in S}\frac{1}{\sqrt{r}} = (3(3(3(3(3(3-1)-1)-1)-1)-1)-1)(\frac{1}{70}+\frac{1}{73})=365\cdot(\frac{1}{70}+\frac{1}{73})=365\cdot\frac{143}{70\cdot73}=\boxed{\frac{143}{14}}$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc12a/402

~ dolphin7

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_25&oldid=156019"