Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/tD_irI8Yoro | ||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=21|num-a=23}} | {{AMC10 box|year=2023|ab=A|num-b=21|num-a=23}} |
Revision as of 05:45, 11 November 2023
- The following problem is from both the 2023 AMC 10A #22 and 2023 AMC 12A #18, so both problems redirect to this page.
Contents
[hide]Problem
Circle and each have radius , and the distance between their centers is . Circle is the largest circle internally tangent to both and . Circle is internally tangent to both and and externally tangent to . What is the radius of ?
Solution 1
Let be the center of the midpoint of the line segment connecting both the centers, say and .
Let the point of tangency with the inscribed circle and the right larger circles be .
Then
Since is internally tangent to , center of , and their tangent point must be on the same line.
Now, if we connect centers of , and /, we get a right angled triangle.
Let the radius of equal . With the pythagorean theorem on our triangle, we have
Solving this equation gives us
~lptoggled
~ShawnX (Diagram)
Video Solution by epicbird08
~EpicBird08
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2023 AMC 10A (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 |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.