2022 AMC 12B Problems/Problem 21
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[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 is brown, Case is blue, Case is green, and Case 4 is red. Let be circle , be circle , and be circle . All the circles in S are internally tangent to circle . There are four cases with two circles belonging to each:
and are internally tangent to .
and are externally tangent to .
is externally and Circle is internally tangent to .
is internally and Circle is externally tangent to .
Consider Cases and together. Since circles and have the same center, the line connecting the center of and the center of will pass through the tangency point of both and and the tangency point of and . This line will be the diameter of and have length . Therefore the radius of in these cases is .
Consider Cases and together. Similarly to Cases and , the line connecting the center of to the center of will pass through the tangency points. This time, however, the diameter of will have length . Therefore, the radius of in these cases is .
The set of circles consists of circles - of which have radius and of which have radius . The total area of all circles in 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
~ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
2022 AMC 12B (Problems • Answer Key • Resources) | |
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2022 AMC 10B (Problems • Answer Key • Resources) | ||
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