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| − | ==Problem==
| + | #redirect [[2022 AMC 10B Problems/Problem 22]] |
| − | Let <math>S</math> be the set of circles in the coordinate plane that are tangent to each of the three circles with equations <math>x^{2}+y^{2}=4</math>, <math>x^{2}+y^{2}=64</math>, and <math>(x-5)^{2}+y^{2}=3</math>. What is the sum of the areas of all circles in <math>S</math>?
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| − | <math>\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</math>
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| − | ==Solution==
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| − | <asy>
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| − | import geometry;
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| − | unitsize(0.5cm);
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| − | void dc(pair x, pen p) {
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| − | pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0];
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| − | draw(circle(x, abs(x-y)),p);
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| − | }
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| − | 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];
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| − | draw(circle(O1,2));
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| − | draw(circle(O1,8));
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| − | draw(circle(O2,sqrt(3)));
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| − | dc(P1,blue);
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| − | dc(P2,red);
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| − | dc(P3,darkgreen);
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| − | dc(P4,brown);
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| − | </asy>
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| − | The circles match up as follows: Case 1 is brown, Case 2 is blue, Case 3 is green, and Case 4 is red.[/center]
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| − | Let \(x^2 + y^2 = 64\) be circle \(O\), \(x^2 + y^2 = 4\) be circle \(P\), and \((x-5)^2 + y^2 = 3\) be circle \(Q\).
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| − | All the circles in S are internally tangent to circle \(O\).
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| − | There are four cases with two circle belonging to each:
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| − | [*] \(P\) and \(Q\) are internally tangent to S.
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| − | [*] \(P\) and \(Q\) are externally tangent to S.
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| − | [*] \(P\) is externally and Circle \(Q\) is internally tangent to S.
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| − | [*] \(P\) is internally and Circle \(Q\) is externally tangent to S.
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| − | 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 \(r_P + r_O = 10\). Therefore the radius of \(S\) in these cases is 5.
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| − | 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 \(r_P-r_O=6\). Therefore, the radius of \(S\) in these cases is 3.
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| − | The set of circles S consists of 8 circles - 4 of which have radius 5 and 4 of which have radius 3.
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| − | The total area of all circles in S is \(4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}\).
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| − | -naman12
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| − | {{AMC12 box|year=2022|ab=B|num-b=20|num-a=22}}
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| − | {{AMC10 box|year=2022|ab=B|num-b=21|num-a=23}}
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| − | {{MAA Notice}}
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