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)

Video Solution by OmegaLearn

https://youtu.be/jcHeJXs9Sdw

Video Solution by MegaMath

https://www.youtube.com/watch?v=lHyl_JtbSuQ&t=8s

~megahertz13

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=rnuL3sVU5aU

Video Solution by epicbird08

https://youtu.be/mhGblJvYeRs

~EpicBird08

Video Solution

https://youtu.be/BWM8NRQBhIw

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

Video Solution by TheBeautyofMath

https://youtu.be/tD_irI8Yoro

~IceMatrix

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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