Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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+ | {{duplicate|[[2023 AMC 10A Problems/Problem 22|2023 AMC 10A #22]] and [[2023 AMC 12A Problems/Problem 18|2023 AMC 12A #18]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
Circle <math>C_1</math> and <math>C_2</math> each have radius <math>1</math>, and the distance between their centers is <math>\frac{1}{2}</math>. Circle <math>C_3</math> is the largest circle internally tangent to both <math>C_1</math> and <math>C_2</math>. Circle <math>C_4</math> is internally tangent to both <math>C_1</math> and <math>C_2</math> and externally tangent to <math>C_3</math>. What is the radius of <math>C_4</math>? | Circle <math>C_1</math> and <math>C_2</math> each have radius <math>1</math>, and the distance between their centers is <math>\frac{1}{2}</math>. Circle <math>C_3</math> is the largest circle internally tangent to both <math>C_1</math> and <math>C_2</math>. Circle <math>C_4</math> is internally tangent to both <math>C_1</math> and <math>C_2</math> and externally tangent to <math>C_3</math>. What is the radius of <math>C_4</math>? | ||
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<math>\textbf{(A) } \frac{1}{14} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{10} \qquad \textbf{(D) } \frac{3}{28} \qquad \textbf{(E) } \frac{1}{9}</math> | <math>\textbf{(A) } \frac{1}{14} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{10} \qquad \textbf{(D) } \frac{3}{28} \qquad \textbf{(E) } \frac{1}{9}</math> | ||
− | ==Solution | + | ==Solution== |
<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
size(10cm); | size(10cm); | ||
− | draw(circle((0,0),0.75)); | + | draw(circle((0,0),0.75), gray(0.7)); |
− | draw(circle((-0.25,0),1)); | + | draw(circle((-0.25,0),1), gray(0.7)); |
− | draw(circle((0.25,0),1)); | + | draw(circle((0.25,0),1), gray(0.7)); |
− | draw(circle((0,6/7),3/28)); | + | draw(circle((0,6/7),3/28), gray(0.7)); |
− | pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), | + | pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), EE = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0), T=(0.75,0); |
− | |||
dot(D); | dot(D); | ||
dot(G); | dot(G); | ||
− | draw(B-- | + | draw(B--EE, dashed+gray(0.7)); |
− | draw(C--F, dashed); | + | draw(C--F, dashed+gray(0.7)); |
− | draw(B--C); | + | dot(C, gray(0.9)); |
+ | draw(B--C, gray(0.7)); | ||
+ | draw(B--A); | ||
draw(A--D); | draw(A--D); | ||
draw(B--D); | draw(B--D); | ||
+ | draw(B--T); | ||
label("$\frac{1}{4}$", (-0.125, -0.125)); | label("$\frac{1}{4}$", (-0.125, -0.125)); | ||
label("$r + \frac{3}{4}$", (0.2, 3/7)); | label("$r + \frac{3}{4}$", (0.2, 3/7)); | ||
label("$1 - r$", (-0.29, 3/7)); | label("$1 - r$", (-0.29, 3/7)); | ||
+ | label("$O$",A,S); | ||
+ | label("$A$",B,S); | ||
+ | dot("$B$",C,S); | ||
+ | dot("$T$",T,E); | ||
+ | label("$1$", (-.85, 0.70)); | ||
+ | label("$1$", (.85, -.7)); | ||
markscalefactor=0.005; | markscalefactor=0.005; | ||
</asy> | </asy> | ||
− | + | Let <math>O</math> be the center of the midpoint of the line segment connecting both the centers, say <math>A</math> and <math>B</math>. | |
+ | |||
+ | Let the point of tangency with the inscribed circle and the right larger circles be <math>T</math>. | ||
+ | |||
+ | Then <math>OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.</math> | ||
Since <math>C_4</math> is internally tangent to <math>C_1</math>, center of <math>C_4</math>, <math>C_1</math> and their tangent point must be on the same line. | Since <math>C_4</math> is internally tangent to <math>C_1</math>, center of <math>C_4</math>, <math>C_1</math> and their tangent point must be on the same line. | ||
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Now, if we connect centers of <math>C_4</math>, <math>C_3</math> and <math>C_1</math>/<math>C_2</math>, we get a right angled triangle. | Now, if we connect centers of <math>C_4</math>, <math>C_3</math> and <math>C_1</math>/<math>C_2</math>, we get a right angled triangle. | ||
− | + | Let the radius of <math>C_4</math> equal <math>r</math>. With the pythagorean theorem on our triangle, we have | |
− | <cmath>(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2</cmath> | + | <cmath>\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2</cmath> |
− | Solving | + | Solving this equation gives us |
<cmath>r = \boxed{\textbf{(D) } \frac{3}{28}}</cmath> | <cmath>r = \boxed{\textbf{(D) } \frac{3}{28}}</cmath> | ||
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~ShawnX (Diagram) | ~ShawnX (Diagram) | ||
+ | |||
+ | ~ap246 (Minor Changes) | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/jcHeJXs9Sdw | ||
+ | |||
+ | ==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== | ==Video Solution by epicbird08== | ||
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~EpicBird08 | ~EpicBird08 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/BWM8NRQBhIw | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/tD_irI8Yoro | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by Problem Solving Channel== | ||
+ | https://youtu.be/7Wg-_79LepU | ||
+ | |||
+ | ~ProblemSolvingChannel | ||
==See Also== | ==See Also== | ||
Line 79: | Line 123: | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 23:54, 20 April 2024
- The following problem is from both the 2023 AMC 10A #22 and 2023 AMC 12A #18, so both problems redirect to this page.
Contents
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)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.