2022 AMC 10B Problems/Problem 22

The following problem is from both the 2022 AMC 10B #22 and 2022 AMC 12A #21, so both problems redirect to this page.

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 1

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

Consider Cases $2$ and $3$ together. Similarly to Cases $1$ 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 $3$ .

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 $4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}$ .

-naman12

Solution 2

We denote by $C_1$ the circle that has the equation $x^2 + y^2 = 4$ . We denote by $C_2$ the circle that has the equation $x^2 + y^2 = 64$ . We denote by $C_3$ the circle that has the equation $(x-5)^2 + y^2 = 3$ .

We denote by $C_0$ a circle that is tangent to $C_1$ , $C_2$ and $C_3$ . We denote by $\left( u, v \right)$ the coordinates of circle $C_0$ , and $r$ the radius of this circle.

From the graphs of circles $C_1$ , $C_2$ , $C_3$ , we observe that if $C_0$ is tangent to all of them, then $C_0$ must be internally tangent to $C_2$ . We have $u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1)$

We do the following casework analysis in terms of the whether $C_0$ is externally tangent to $C_1$ and $C_3$ .

Case 1: $C_0$ is externally tangent to $C_1$ and $C_3$ .

We have $u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2)$ and $(u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3)$

Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ . We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$ .

Case 2: $C_1$ is internally tangent to $C_0$ and $C_3$ is externally tangent to $C_0$ .

We have $u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2)$ and $(u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3)$

Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ . We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$ .

Case 3: $C_1$ is externally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$ .

We have $u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2)$ and $(u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3)$

Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ . We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$ .

Case 4: $C_1$ is internally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$ .

We have $u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2)$ and $(u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3)$

Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ . We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$ .

Because the graph is symmetric with the $x$ -axis, and for each case above, the solution of $v$ is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the $x$ -axis.

Therefore, the sum of the areas of all the circles in $S$ is $2\left( 3^2 \pi +5^2 \pi +3^2 \pi +5^2 \pi \right) = \boxed{\textbf{(E) } 136 \pi}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~MrThinker (LaTeX Error)

Video Solution by OmegaLearn using Circular Tangency

https://youtu.be/ZDpmvGmNefQ

~ pi_is_3.14

Video Solution

https://youtu.be/1pkuBlRKt6Q

~ThePuzzlr

https://youtu.be/nqE5QYkzRAw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by TheBeautyofMath

With additional justification reasoning for certain statements made. Also an additional twist on a potential similar alternate problem at the end. https://youtu.be/r-jNrjKIXTU

~IceMatrix

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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 10 Problems and Solutions

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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

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