2022 AMC 12B Problems/Problem 21

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Problem

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$, $x^{2}+y^{2}=64$, and $(x-5)^{2}+y^{2}=3$. What is the sum of the areas of all circles in $S$?

$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$

Solution

[asy]         import geometry;         unitsize(0.5cm);  		void dc(pair x, pen p) {           pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0];           draw(circle(x, abs(x-y)),p);         }          pair O1 = (0,0),O2=(5,0),P1=intersectionpoints(circle(O1,5),circle(O2,3+sqrt(3)))[0],P2=intersectionpoints(circle(O1,3),circle(O2,5+sqrt(3)))[0],P3=intersectionpoints(circle(O1,5),circle(O2,3-sqrt(3)))[0],P4=intersectionpoints(circle(O1,3),circle(O2,5-sqrt(3)))[0];          draw(circle(O1,2));         draw(circle(O1,8));         draw(circle(O2,sqrt(3)));  		dc(P1,blue); 		dc(P2,red); 		dc(P3,darkgreen); 		dc(P4,brown); [/asy] The circles match up as follows: Case 1 is brown, Case 2 is blue, Case 3 is green, and Case 4 is red.[/center] Let x2+y2=64 be circle O, x2+y2=4 be circle P, and (x5)2+y2=3 be circle Q. All the circles in S are internally tangent to circle O. There are four cases with two circle belonging to each:

[*] P and Q are internally tangent to S. [*] P and Q are externally tangent to S. [*] P is externally and Circle Q is internally tangent to S. [*] P is internally and Circle Q is externally tangent to S.

Consider Cases 1 and 4 together. Since circles O and P have the same center, the line connecting the center of S and the center of O will pass through both the tangency point of S and O and the tangency point of S and P. This line will be the diameter of S and have length rP+rO=10. Therefore the radius of S in these cases is 5.

Consider Cases 2 and 3 together. Similarly to Case 1 and 4, the line connecting the center of S to the center of O will pass through the tangency points. This time however, the diameter of S will have length rPrO=6. Therefore, the radius of S in these cases is 3.

The set of circles S consists of 8 circles - 4 of which have radius 5 and 4 of which have radius 3. The total area of all circles in S is 4(52π+32π)=136π(E).

-naman12

2022 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 12 Problems and Solutions
2022 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 10 Problems and Solutions

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