Difference between revisions of "2022 AMC 12B Problems/Problem 21"

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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>?
 
 
 
<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>
 
 
 
==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
 
 
 
==Solution==
 
 
 
We denote by <math>C_1</math> the circle that has the equation <math>x^2 + y^2 = 4</math>.
 
We denote by <math>C_2</math> the circle that has the equation <math>x^2 + y^2 = 64</math>.
 
We denote by <math>C_3</math> the circle that has the equation <math>(x-5)^2 + y^2 = 3</math>.
 
 
 
We denote by <math>C_0</math> a circle that is tangent to <math>C_1</math>, <math>C_2</math> and <math>C_3</math>.
 
We denote by <math>\left( u, v \right)</math> the coordinates of circle <math>C_0</math>, and <math>r</math> the radius of this circle.
 
 
 
From the graphs of circles <math>C_1</math>, <math>C_2</math>, <math>C_3</math>, we observe that if <math>C_0</math> is tangent to all of them, then <math>C_0</math> must be internally tangent to <math>C_2</math>.
 
We have
 
<cmath>
 
\[
 
u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1)
 
\]
 
</cmath>
 
 
 
We do the following casework analysis in terms of the whether <math>C_0</math> is externally tangent to <math>C_1</math> and <math>C_3</math>.
 
 
 
Case 1: <math>C_0</math> is externally tangent to <math>C_1</math> and <math>C_3</math>.
 
 
 
We have
 
<cmath>
 
\[
 
u^2 + v^2 = \left( r + 2 \right)^2  \hspace{1cm} (2)
 
\]
 
</cmath>
 
and
 
<cmath>
 
\[
 
(u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3)
 
\]
 
</cmath>
 
 
 
Taking <math>(2) - (1)</math>, we get <math>r + 2 = 8 - r</math>. Thus, <math>r = 3</math>.
 
We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
 
 
 
Case 2: <math>C_1</math> is internally tangent to <math>C_0</math> and <math>C_3</math> is externally tangent to <math>C_0</math>.
 
 
 
We have
 
<cmath>
 
\[
 
u^2 + v^2 = \left( r - 2 \right)^2  \hspace{1cm} (2)
 
\]
 
</cmath>
 
and
 
<cmath>
 
\[
 
(u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3)
 
\]
 
</cmath>
 
 
 
Taking <math>(2) - (1)</math>, we get <math>r - 2 = 8 - r</math>. Thus, <math>r = 5</math>.
 
We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
 
 
 
Case 3: <math>C_1</math> is externally tangent to <math>C_0</math> and <math>C_3</math> is internally tangent to <math>C_0</math>.
 
 
 
We have
 
<cmath>
 
\[
 
u^2 + v^2 = \left( r + 2 \right)^2  \hspace{1cm} (2)
 
\]
 
</cmath>
 
and
 
<cmath>
 
\[
 
(u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3)
 
\]
 
</cmath>
 
 
 
Taking <math>(2) - (1)</math>, we get <math>r + 2 = 8 - r</math>. Thus, <math>r = 3</math>.
 
We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
 
 
 
Case 4: <math>C_1</math> is internally tangent to <math>C_0</math> and <math>C_3</math> is internally tangent to <math>C_0</math>.
 
 
 
We have
 
<cmath>
 
\[
 
u^2 + v^2 = \left( r - 2 \right)^2  \hspace{1cm} (2)
 
\]
 
</cmath>
 
and
 
<cmath>
 
\[
 
(u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3)
 
\]
 
</cmath>
 
 
 
Taking <math>(2) - (1)</math>, we get <math>r - 2 = 8 - r</math>. Thus, <math>r = 5</math>.
 
We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
 
 
 
 
 
Because the graph is symmetric with the <math>x</math>-axis, and for each case above, the solution of <math>v</math> is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the <math>x</math>-axis.
 
 
 
Therefore, the sum of the areas of
 
all the circles in <math>S</math> is
 
<math></math>
 
\[
 
2 \left( \pi 3^2 + \pi 5^2 + \pi 3^2 + \pi 5^2 \right)
 
= \boxed{\textbf{(E) <math>136 \pi</math>}} .
 
\]
 
<math></math>
 
 
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
 
 
==Video Solution==
 
 
 
https://youtu.be/nqE5QYkzRAw
 
 
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
 
 
{{AMC12 box|year=2022|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2022|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 

Latest revision as of 11:27, 14 December 2022