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

(Created page with "==Problem== 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^{...")
 
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-naman12
 
-naman12
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 +
==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}}
 
{{AMC12 box|year=2022|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2022|ab=B|num-b=21|num-a=23}}
 
{{AMC10 box|year=2022|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:03, 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.[/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 $C_1$ 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 (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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