Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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In which we get an equation by pythagorean theorem: | In which we get an equation by pythagorean theorem: | ||
| − | < | + | <cmath>(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2</cmath> |
Solving it gives us | Solving it gives us | ||
| − | < | + | <cmath>r = \boxed{\textbf{(D) } \frac{3}{28}}</cmath> |
~lptoggled | ~lptoggled | ||
Revision as of 00:10, 10 November 2023
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
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.
In which we get an equation by pythagorean theorem:
![\[(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2\]](http://latex.artofproblemsolving.com/8/3/7/837a01ec5843d0f5fbd97e80ea14d54c04373ced.png)
Solving it gives us
![\[r = \boxed{\textbf{(D) } \frac{3}{28}}\]](http://latex.artofproblemsolving.com/7/a/6/7a617cd86c0b14127391131cadb3817e9e5248ec.png)
~lptoggled
Video Solution by epicbird08
~EpicBird08
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. 
