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 
and 
each have radius 
, and the distance between their centers is 
. Circle 
is the largest circle internally tangent to both and . Circle 
is internally tangent to both and and externally tangent to . What is the radius of ?
![[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]](http://latex.artofproblemsolving.com/4/0/6/406135ba18fccc6f68ffce10d698ac01c265ec90.png)
Solution 1
![[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]](http://latex.artofproblemsolving.com/6/b/c/6bcd8015d3c8cd7e3c4c313ecaa64480cc7ad76b.png)
With some simple geometry skills, we can find that has a radius of 
.
Since is internally tangent to , center of , and their tangent point must be on the same line.
Now, if we connect centers of , and /, we get a right angled triangle.
Let the radius of equal 
. With the pythagorean theorem on our triangle, we have
![\[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\]](http://latex.artofproblemsolving.com/f/b/9/fb9614ba7aa8e9a1696623786f3a96b4bdc083db.png)
Solving this equation gives us
![\[r = \boxed{\textbf{(D) } \frac{3}{28}}\]](http://latex.artofproblemsolving.com/7/a/6/7a617cd86c0b14127391131cadb3817e9e5248ec.png)
~lptoggled
~ShawnX (Diagram)
Video Solution by epicbird08
~EpicBird08
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
