Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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Finally notice that <math>r_{C_3} + 2 r_{C_4}</math> is really close to <math>\dfrac{\sqrt15}{16}</math>. | Finally notice that <math>r_{C_3} + 2 r_{C_4}</math> is really close to <math>\dfrac{\sqrt15}{16}</math>. | ||
− | <math>r_{C_4} = \dfrac{\sqrt{15} - 12}{16}</math>. This is really close to <math>\dfrac14</math> and the closest answer to that is <math>\boxed{\text{D) }\dfrac{3}{28}</math> | + | <math>r_{C_4} = \dfrac{\sqrt{15} - 12}{16}</math>. This is really close to <math>\dfrac14</math> and the closest answer to that is <math>\boxed{\text{D) }\dfrac{3}{28}}</math> |
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==Video Solution by Little Fermat== | ==Video Solution by Little Fermat== |
Revision as of 11:09, 15 October 2025
- The following problem is from both the 2023 AMC 10A #22 and 2023 AMC 12A #18, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution
- 3 Solution 2: Estimation
- 4 Video Solution by Little Fermat
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution by OmegaLearn
- 7 Video Solution by MegaMath
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 Video Solution by epicbird08
- 10 Video Solution
- 11 Video Solution by TheBeautyofMath
- 12 Video Solution by Problem Solving Channel
- 13 See Also
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
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)
~ap246 (Minor Changes)
Solution 2: Estimation
Draw 2 lines from the middle segment's endpoints to an intersection of and
. Draw the perpendicular bisector of the middle segment until it reaches one of the intersections. We see this becomes a right triangle with a hypotenuse of
and one leg length of
. Using the Pythagorean theorem, the other leg length is
. Also drawing
's radius along the segment of
allows us to find the that radius of the middle circle is
.
Finally notice that is really close to
.
. This is really close to
and the closest answer to that is
Video Solution by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=_zp2L0edaMjl63-P&t=4908 ~little-fermat
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=LdnMT_hCLmgL889h&t=7950
~Math-X
Video Solution by OmegaLearn
Video Solution by MegaMath
https://www.youtube.com/watch?v=lHyl_JtbSuQ&t=8s
~megahertz13
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=rnuL3sVU5aU
Video Solution by epicbird08
~EpicBird08
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Problem Solving Channel
~ProblemSolvingChannel
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.