Difference between revisions of "2022 AMC 12B Problems/Problem 21"
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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
Contents
[hide]Problem
Let be the set of circles in the coordinate plane that are tangent to each of the three circles with equations , , and . What is the sum of the areas of all circles in ?
Solution
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
[*]
Consider Cases 1 and 4 together. Since circles
Consider Cases 2 and 3 together. Similarly to Case 1 and 4, the line connecting the center of
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
-naman12
Solution
We denote by the circle that has the equation . We denote by the circle that has the equation . We denote by the circle that has the equation .
We denote by a circle that is tangent to , and . We denote by the coordinates of circle , and the radius of this circle.
From the graphs of circles , , , we observe that if is tangent to all of them, then must be internally tangent to . We have
We do the following casework analysis in terms of the whether is externally tangent to and .
Case 1: is externally tangent to and .
We have and
Taking , we get . Thus, . We can further compute (omitted here) that there exist feasible with this given .
Case 2: is internally tangent to and is externally tangent to .
We have and
Taking , we get . Thus, . We can further compute (omitted here) that there exist feasible with this given .
Case 3: is externally tangent to and is internally tangent to .
We have and
Taking , we get . Thus, . We can further compute (omitted here) that there exist feasible with this given .
Case 4: is internally tangent to and is internally tangent to .
We have and
Taking , we get . Thus, . We can further compute (omitted here) that there exist feasible with this given .
Because the graph is symmetric with the -axis, and for each case above, the solution of is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the -axis.
Therefore, the sum of the areas of all the circles in is $$ (Error compiling LaTeX. Unknown error_msg) \[ 2 \left( \pi 3^2 + \pi 5^2 + \pi 3^2 + \pi 5^2 \right) = \boxed{\textbf{(E) }} . \] $$ (Error compiling LaTeX. Unknown error_msg)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
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 |
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