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

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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.
 
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>or<math>C_2</math>, we get a right angled triangle.
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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.
  
 
In which we get an equation by pythagorean theorem:
 
In which we get an equation by pythagorean theorem:

Revision as of 22:28, 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 $C_4$?

[someone pls insert diagram]

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

Solution 1

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

See Also

2023 AMC 10A (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 10 Problems and Solutions

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