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Difference between revisions of "2022 AMC 10B Problems/Problem 22"

(Redirected page to 2022 AMC 12B Problems/Problem 21)
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(→‎Video Solution by The Power of Logic(#20-#21))
 
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#redirect [[2022 AMC 12B Problems/Problem 21]]
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{{duplicate|[[2022 AMC 10B Problems/Problem 22|2022 AMC 10B #22]] and [[2022 AMC 12B Problems/Problem 21|2022 AMC 12B #21]]}}
 +
 
 +
==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^{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 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+linewidth(2));
 +
        }
 +
 
 +
        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,mediumgreen);
 +
dc(P4,brown);
 +
</asy>
 +
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 <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 <math>O</math>.
 +
There are four cases with two circles belonging to each:
 +
 
 +
<math>*</math> <math>P</math> and <math>Q</math> are internally tangent to <math>S</math>.
 +
 
 +
<math>*</math> <math>P</math> and <math>Q</math> are externally tangent to <math>S</math>.
 +
 
 +
<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 <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 <math>S</math> is <math>4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}</math>.
 +
 
 +
-naman12
 +
 
 +
==Solution 2==
 +
 
 +
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>2\left( 3^2 \pi +5^2 \pi +3^2 \pi +5^2 \pi \right) = \boxed{\textbf{(E) } 136 \pi}</math>.
 +
 
 +
~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
 +
 
 +
==Video Solution by The Power of Logic(#20-#21)==
 +
https://youtu.be/7FiTsDNMmgg
 +
==Video Solution by Interstigation==
 +
https://youtu.be/GWEfdjTiXSQ
 +
 
 +
~Interstigation
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2022|ab=B|num-b=21|num-a=23}}
 +
{{AMC12 box|year=2022|ab=B|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 00:15, 10 June 2023

The following problem is from both the 2022 AMC 10B #22 and 2022 AMC 12B #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+linewidth(2)); } 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,mediumgreen); 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

Video Solution by The Power of Logic(#20-#21)

https://youtu.be/7FiTsDNMmgg

Video Solution by Interstigation

https://youtu.be/GWEfdjTiXSQ

~Interstigation

See Also

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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