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Difference between revisions of "2022 AMC 12B Problems/Problem 21"
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dc(P4,brown);
 
dc(P4,brown);
 
</asy>
 
</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]
+
The circles match up as follows: Case <math>1</math> is brown, Case <math>2</math> is blue, Case <math>3</math> 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 \(Q\).
+
Let <math>x^2 + y^2 = 64</math> be circle <math>O</math>, <math>x^2 + y^2 = 4</math> be circle <math>P</math>, and <math>(x-5)^2 + y^2 = 3</math> be circle <math>Q</math>.
−
All the circles in S are internally tangent to circle \(O\).
+
All the circles in S are internally tangent to circle <math>O</math>.
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There are four cases with two circle belonging to each:
+
There are four cases with two circles belonging to each:
  
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[*] \(P\) and \(Q\) are internally tangent to S.
+
<math>*</math> <math>P</math> and <math>Q</math> are internally tangent to <math>S</math>.
−
[*] \(P\) 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 \(r_P + r_O = 10\). Therefore the radius of \(S\) in these cases is 5.
+
<math>*</math> <math>P</math> and <math>Q</math> are externally tangent to <math>S</math>.
  
−
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.
+
<math>*</math> <math>P</math> is externally and Circle <math>Q</math> is internally tangent to <math>S</math>.
 +
 
 +
<math>*</math> <math>P</math> is internally and Circle <math>Q</math> is externally tangent to <math>S</math>.
 +
 
 +
Consider Cases <math>1</math> and <math>4</math> together. Since circles <math>O</math> and <math>P</math> have the same center, the line connecting the center of <math>S</math> and the center of <math>O</math> will pass through the tangency point of both <math>S</math> and <math>O</math> and the tangency point of <math>S</math> and <math>P</math>. This line will be the diameter of <math>S</math> and have length <math>r_P + r_O = 10</math>. Therefore the radius of <math>S</math> in these cases is <math>5</math>.
 +
 
 +
Consider Cases <math>2</math> and <math>3</math> together. Similarly to Cases <math>1</math> and <math>4</math>, the line connecting the center of <math>S</math> to the center of <math>O</math> will pass through the tangency points. This time, however, the diameter of <math>S</math> will have length <math>r_P-r_O=6</math>. Therefore, the radius of <math>S</math> in these cases is <math>3</math>.
 
      
 
      
−
The set of circles S consists of 8 circles - 4 of which have radius 5 and 4 of which have radius 3.  
+
The set of circles <math>S</math> consists of <math>8</math> circles - <math>4</math> of which have radius <math>5</math> and <math>4</math> of which have radius <math>3</math>.  
−
The total area of all circles in S is \(4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}\).
+
The total area of all circles in <math>S</math> is <math>4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}</math>.
  
 
-naman12
 
-naman12

Revision as of 21:55, 17 November 2022

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. 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

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 $$ (Error compiling LaTeX. Unknown error_msg) \[ 2 \left( \pi 3^2 + \pi 5^2 + \pi 3^2 + \pi 5^2 \right) = \boxed{\textbf{(E) $136 \pi$}} . \] $$ (Error compiling LaTeX. Unknown error_msg)

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

Video Solution

https://youtu.be/nqE5QYkzRAw

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

2022 AMC 12B (Problems • Answer Key • Resources)
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Problem 21 		Followed by
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