Difference between revisions of "2015 AMC 12A Problems/Problem 25"

(Created page with "==Problem== A collection of circles in the upper half-plane, all tangent to the <math>x</math>-axis, is constructed in layers as follows. Layer <math>L_0</math> consists of two ...")
 
(Problem)
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<cmath>\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?</cmath>
 
<cmath>\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?</cmath>
  
DIAGRAM NEEDED
+
<asy>
 +
import olympiad;
 +
size(350);
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defaultpen(linewidth(0.7));
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// define a bunch of arrays and starting points
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pair[] coord = new pair[65];
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int[] trav = {32,16,8,4,2,1};
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coord[0] = (0,73^2); coord[64] = (2*73*70,70^2);
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// draw the big circles and the bottom line
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path arc1 = arc(coord[0],coord[0].y,260,360);
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path arc2 = arc(coord[64],coord[64].y,175,280);
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fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75));
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fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75));
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draw(arc1^^arc2);
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draw((-930,0)--(70^2+73^2+850,0));
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// We now apply the findCenter function 63 times to get
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// the location of the centers of all 63 constructed circles.
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// The complicated array setup ensures that all the circles
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// will be taken in the right order
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for(int i = 0;i<=5;i=i+1)
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{
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int skip = trav[i];
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for(int k=skip;k<=64 - skip; k = k + 2*skip)
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{
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pair cent1 = coord[k-skip], cent2 = coord[k+skip];
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real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);
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real shiftx = cent1.x + sqrt(4*r1*rn);
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coord[k] = (shiftx,rn);
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}
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// Draw the remaining 63 circles
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}
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for(int i=1;i<=63;i=i+1)
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{
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filldraw(circle(coord[i],coord[i].y),gray(0.75));
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}</asy>
  
 
<math> \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}</math>
 
<math> \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}</math>

Revision as of 02:06, 5 February 2015

Problem

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 $k>=1$, 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]

$\textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

Solution

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
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