Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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import olympiad; | import olympiad; | ||
size(10cm); | size(10cm); | ||
− | draw(circle((0,0),0.75)); | + | draw(circle((0,0),0.75), gray(0.7)); |
− | draw(circle((-0.25,0),1)); | + | draw(circle((-0.25,0),1), gray(0.7)); |
− | draw(circle((0.25,0),1)); | + | draw(circle((0.25,0),1), gray(0.7)); |
− | draw(circle((0,6/7),3/28)); | + | 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); | 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(B); | ||
dot(D); | dot(D); | ||
dot(G); | dot(G); | ||
− | draw(B--E, dashed); | + | draw(B--E, dashed+gray(0.7)); |
− | draw(C--F, dashed); | + | draw(C--F, dashed+gray(0.7)); |
− | draw(B--C); | + | dot(C, gray(0.9)); |
+ | draw(B--C, gray(0.7)); | ||
+ | draw(B--A); | ||
draw(A--D); | draw(A--D); | ||
draw(B--D); | draw(B--D); |
Revision as of 20:14, 10 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 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 ?
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
~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.