Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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| + | ==Video Solution by epicbird08== | ||
| + | https://youtu.be/mhGblJvYeRs | ||
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| + | ~EpicBird08 | ||
==See Also== | ==See Also== | ||
Revision as of 23:07, 9 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:
Solving it gives us
~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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
