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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>?
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| − | | |
| − | <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>
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| − | | |
| − | ==Solution 1==
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| − | <asy>
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| − | import geometry;
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| − | unitsize(0.5cm);
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| − | | |
| − | void dc(pair x, pen p) {
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| − | pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0];
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| − | draw(circle(x, abs(x-y)),p);
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| − | }
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| − | | |
| − | 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];
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| − | | |
| − | draw(circle(O1,2));
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| − | draw(circle(O1,8));
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| − | draw(circle(O2,sqrt(3)));
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| − | | |
| − | dc(P1,blue);
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| − | dc(P2,red);
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| − | dc(P3,darkgreen);
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| − | dc(P4,brown);
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| − | </asy>
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| − | 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.
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| − | 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>.
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| − | All the circles in S are internally tangent to circle <math>O</math>.
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| − | There are four cases with two circles belonging to each:
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| − | | |
| − | <math>*</math> <math>P</math> and <math>Q</math> are internally tangent to <math>S</math>.
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| − | | |
| − | <math>*</math> <math>P</math> and <math>Q</math> are externally tangent to <math>S</math>.
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| − | | |
| − | <math>*</math> <math>P</math> is externally and Circle <math>Q</math> is internally tangent to <math>S</math>.
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| − | | |
| − | <math>*</math> <math>P</math> is internally and Circle <math>Q</math> is externally tangent to <math>S</math>.
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| − | | |
| − | 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>.
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| − | | |
| − | 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>.
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| − |
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| − | 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>.
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| − | 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>.
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| − | | |
| − | -naman12
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| − | | |
| − | ==Solution 2==
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| − | | |
| − | We denote by <math>C_1</math> the circle that has the equation <math>x^2 + y^2 = 4</math>.
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| − | We denote by <math>C_2</math> the circle that has the equation <math>x^2 + y^2 = 64</math>.
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| − | We denote by <math>C_3</math> the circle that has the equation <math>(x-5)^2 + y^2 = 3</math>.
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| − | | |
| − | 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>.
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| − | 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.
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| − | | |
| − | 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>.
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| − | We have
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| − | <cmath>
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| − | \[
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| − | u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1)
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| − | \]
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| − | </cmath>
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| − | | |
| − | 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>.
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| − | | |
| − | Case 1: <math>C_0</math> is externally tangent to <math>C_1</math> and <math>C_3</math>.
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| − | | |
| − | We have
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| − | <cmath>
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| − | \[
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| − | u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2)
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| − | \]
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| − | </cmath>
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| − | and
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| − | <cmath>
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| − | \[
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| − | (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3)
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| − | \]
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| − | </cmath>
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| − | | |
| − | Taking <math>(2) - (1)</math>, we get <math>r + 2 = 8 - r</math>. Thus, <math>r = 3</math>.
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| − | We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
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| − | | |
| − | 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>.
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| − | | |
| − | We have
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| − | <cmath>
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| − | \[
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| − | u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2)
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| − | \]
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| − | </cmath>
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| − | and
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| − | <cmath>
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| − | \[
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| − | (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3)
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| − | \]
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| − | </cmath>
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| − | | |
| − | Taking <math>(2) - (1)</math>, we get <math>r - 2 = 8 - r</math>. Thus, <math>r = 5</math>.
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| − | We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
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| − | | |
| − | 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>.
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| − | | |
| − | We have
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| − | <cmath>
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| − | \[
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| − | u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2)
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| − | \]
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| − | </cmath>
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| − | and
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| − | <cmath>
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| − | \[
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| − | (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3)
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| − | \]
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| − | </cmath>
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| − | | |
| − | Taking <math>(2) - (1)</math>, we get <math>r + 2 = 8 - r</math>. Thus, <math>r = 3</math>.
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| − | We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
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| − | | |
| − | 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>.
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| − | | |
| − | We have
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| − | <cmath>
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| − | \[
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| − | u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2)
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| − | \]
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| − | </cmath>
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| − | and
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| − | <cmath>
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| − | \[
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| − | (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3)
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| − | \]
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| − | </cmath>
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| − | | |
| − | Taking <math>(2) - (1)</math>, we get <math>r - 2 = 8 - r</math>. Thus, <math>r = 5</math>.
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| − | We can further compute (omitted here) that there exist feasible <math>(u,v)</math> with this given <math>r</math>.
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| − | | |
| − | | |
| − | 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.
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| − | | |
| − | 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>.
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| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | | |
| − | ~MrThinker (LaTeX Error)
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| − | == Video Solution by OmegaLearn using Circular Tangency ==
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| − | https://youtu.be/ZDpmvGmNefQ
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| − | | |
| − | ~ pi_is_3.14
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| − | | |
| − | ==Video Solution==
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| − | | |
| − | https://youtu.be/1pkuBlRKt6Q
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| − | | |
| − | ~ThePuzzlr
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| − | | |
| − | https://youtu.be/nqE5QYkzRAw
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| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | | |
| − | {{AMC12 box|year=2022|ab=B|num-b=20|num-a=22}}
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| − | {{AMC10 box|year=2022|ab=B|num-b=21|num-a=23}}
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| − | {{MAA Notice}}
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