2023 AMC 12A Problems/Problem 18

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

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2023 AMC 12A Problems/Problem 18

Problem

Solution 1

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