# 2023 AMC 12A Problems/Problem 18

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

$[asy] import olympiad; size(10cm); draw(circle((0,0),0.75), gray(0.7)); draw(circle((-0.25,0),1), gray(0.7)); draw(circle((0.25,0),1), gray(0.7)); draw(circle((0,6/7),3/28), gray(0.7)); pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), EE = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0), T=(0.75,0); dot(D); dot(G); draw(B--EE, dashed+gray(0.7)); draw(C--F, dashed+gray(0.7)); dot(C, gray(0.9)); draw(B--C, gray(0.7)); draw(B--A); draw(A--D); draw(B--D); draw(B--T); label("\frac{1}{4}", (-0.125, -0.125)); label("r + \frac{3}{4}", (0.2, 3/7)); label("1 - r", (-0.29, 3/7)); label("O",A,S); label("A",B,S); dot("B",C,S); dot("T",T,E); label("1", (-.85, 0.70)); label("1", (.85, -.7)); markscalefactor=0.005; [/asy]$

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)

~megahertz13

~EpicBird08

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~IceMatrix

## Video Solution by Problem Solving Channel

~ProblemSolvingChannel

 2023 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 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 byProblem 17 Followed byProblem 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