2023 AMC 12A Problems/Problem 18: Difference between revisions

Revision as of 12:11, 11 November 2023

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

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

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

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

2023 AMC 12A (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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 29: / Line 29: @@
 <math>\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}</math>
-==Solution 1==
+==Solution==
 <asy>

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