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2023 AMC 10A Problems/Problem 22: Difference between revisions

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==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>?  


Revision as of 20:44, 9 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 $C_4$?

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

Solution

Connect the centers of $C_1$ and $C_4$, and the centers of $C_3$ and $C_4$. Let the radius of $C_4$ be $r$. Then, from the auxillary lines, we get $(\frac{1}{4})^2 + (\frac{3}{4}+r)^2 = (1-r)^2$. Solving, we get $r = \boxed{\frac{3}{28}}$

-andliu766


Video Solution 1 by OmegaLearn

https://youtu.be/jcHeJXs9Sdw

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

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