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Difference between revisions of "2023 AMC 12A Problems/Problem 18"

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==Solution 1==
 
==Solution 1==
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With some simple geometry skills, we can find that <math>C_3</math> has a radius of <math>\frac{3}{4}</math>.
 
−
 
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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.
 
−
 
−
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:
 
−
 
−
<cmath>(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2</cmath>
 
−
 
−
Solving it gives us
 
−
 
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<cmath>r = \boxed{\textbf{(D) } \frac{3}{28}}</cmath>
 
  
 
<asy>
 
<asy>
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markscalefactor=0.005;
 
markscalefactor=0.005;
 
</asy>
 
</asy>
 +
 +
With some simple geometry skills, we can find that <math>C_3</math> has a radius of <math>\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.
 +
 +
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:
 +
 +
<cmath>(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2</cmath>
 +
 +
Solving it gives us
 +
 +
<cmath>r = \boxed{\textbf{(D) } \frac{3}{28}}</cmath>
  
 
~lptoggled
 
~lptoggled
  
−
~ShawnX (Graph)
+
~ShawnX (Diagram)
  
 
==Video Solution by epicbird08==
 
==Video Solution by epicbird08==

Revision as of 01:54, 10 November 2023

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 1

[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), G = (0,0); dot(B); dot(D); dot(G); draw(B--E, dashed); draw(C--F, dashed); draw(B--C); draw(A--D); draw(B--D); label("$\frac{1}{4}$", (-0.125, -0.125)); label("$r + \frac{3}{4}$", (0.2, 3/7)); label("$1 - r$", (-0.29, 3/7)); markscalefactor=0.005; [/asy]

With some simple geometry skills, we can find that $C_3$ has a radius of $\frac{3}{4}$.

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.

In which we get an equation by pythagorean theorem:

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

Solving it gives us

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

~lptoggled

~ShawnX (Diagram)

Video Solution by epicbird08

https://youtu.be/mhGblJvYeRs

~EpicBird08

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

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