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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 1==
+
==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), E = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0);
+
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(B);
 
 
dot(D);
 
dot(D);
 
dot(G);
 
dot(G);
−
draw(B--E, dashed);
+
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));
−
markscalefactor=0.005;
+
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=0.05;
 
</asy>
 
</asy>
  
−
With some simple geometry skills, we can find that <math>C_3</math> has a radius of <math>\frac{3}{4}</math>.
+
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.
  
−
In which we get an equation by pythagorean theorem:
+
Let the radius of <math>C_4</math> equal <math>r</math>. With the pythagorean theorem on our triangle, we have
  
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<cmath>(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2</cmath>
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<cmath>\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2</cmath>
  
−
Solving it gives us
+
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 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==
 +
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==
Line 73: Line 110:
  
 
~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 131:
  
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 17:05, 23 November 2024

The following problem is from both the 2023 AMC 10A #22 and 2023 AMC 12A #18, so both problems redirect to this page.

Problem

Circle $C_1$ and $C_2$ each have radius $1$, and the distance between their centers is $\frac{1}{2}$. Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$. Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$. What is the radius of $C_4$?

[asy] import olympiad; size(10cm); draw(circle((0,0),0.75)); draw(circle((-0.25,0),1)); draw(circle((0.25,0),1)); draw(circle((0,6/7),3/28)); pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), E = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118); dot(B^^C); draw(B--E, dashed); draw(C--F, dashed); draw(B--C); label("$C_4$", D); label("$C_1$", (-1.375, 0)); label("$C_2$", (1.375,0)); label("$\frac{1}{2}$", (0, -.125)); label("$C_3$", (-0.4, -0.4)); label("$1$", (-.85, 0.70)); label("$1$", (.85, -.7)); import olympiad; markscalefactor=0.005; [/asy]

$\textbf{(A) } \frac{1}{14} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{10} \qquad \textbf{(D) } \frac{3}{28} \qquad \textbf{(E) } \frac{1}{9}$

Solution

[asy] import olympiad; size(10cm); draw(circle((0,0),0.75), gray(0.7)); draw(circle((-0.25,0),1), gray(0.7)); draw(circle((0.25,0),1), gray(0.7)); 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), EE = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0), T=(0.75,0); dot(D); dot(G); draw(B--EE, dashed+gray(0.7)); draw(C--F, dashed+gray(0.7)); dot(C, gray(0.9)); draw(B--C, gray(0.7)); draw(B--A); draw(A--D); draw(B--D); draw(B--T); label("$\frac{1}{4}$", (-0.125, -0.125)); label("$r + \frac{3}{4}$", (0.2, 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=0.05; [/asy]

Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$.

Let the point of tangency with the inscribed circle and the right larger circles be $T$.

Then $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.$

Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line.

Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle.

Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have

\[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\]

Solving this equation gives us

\[r = \boxed{\textbf{(D) } \frac{3}{28}}\]

~lptoggled

~ShawnX (Diagram)

~ap246 (Minor Changes)

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

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

https://youtu.be/mhGblJvYeRs

~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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_18&oldid=235366"