Difference between revisions of "2023 AMC 12A Problems/Problem 18"
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<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> | <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== | ||
| + | |||
| + | With some simple geometry skills, we can find that <math>C_3</math> has a radius of <math>\frac{3}{4}</math>. | ||
| + | |||
| + | Since <math>C_4</math> is internally tangent to <math>C_1</math>, center of <math>C_4</math>, <math>C_1</math> and their tangent point must be on the same line. | ||
| + | |||
| + | Now, if we connect centers of <math>C_4</math>, <math>C_3</math> and <math>C_1</math>or<math>C_2</math>, we get a right angled triangle. | ||
| + | |||
| + | In which we get an equation by pythagorean theorem: | ||
| + | |||
| + | <math>(r+\frac{3}{4})^2+(\frac{1}{4})^2=(1-r)^2</math> | ||
| + | |||
| + | Solving it gives us | ||
| + | |||
| + | <math>r = \boxed{\textbf{(D) } \frac{3}{28}}</math> | ||
| + | |||
| + | ~lptoggled | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2023|ab=A|num-b=17|num-a=19}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 22:27, 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 ?
[someone pls insert diagram]
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 or, we get a right angled triangle.
In which we get an equation by pythagorean theorem:
Solving it gives us
~lptoggled
See Also
| 2023 AMC 10A (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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
