Difference between revisions of "2018 AMC 10A Problems/Problem 15"
Pi is 3.14 (talk | contribs) (→Video Solution 2) |
(→Video Solution (HOW TO THINK CREATIVELY!)) |
||
(4 intermediate revisions by 2 users not shown) | |||
Line 35: | Line 35: | ||
<cmath>\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.</cmath> | <cmath>\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.</cmath> | ||
Therefore, our answer is <math>65+4= \boxed{\textbf{(D) } 69}</math>. | Therefore, our answer is <math>65+4= \boxed{\textbf{(D) } 69}</math>. | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!)== | ||
+ | https://youtu.be/xFnLbr-qt6I | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution 1== | ==Video Solution 1== |
Latest revision as of 16:10, 15 October 2023
Contents
[hide]Problem
Two circles of radius are externally tangent to each other and are internally tangent to a circle of radius at points and , as shown in the diagram. The distance can be written in the form , where and are relatively prime positive integers. What is ?
Solution
Let the center of the surrounding circle be . The circle that is tangent at point will have point as the center. Similarly, the circle that is tangent at point will have point as the center. Connect , , , and . Now observe that is similar to by SAS.
Writing out the ratios, we get Therefore, our answer is .
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solution 1
- Whiz
https://www.youtube.com/watch?v=llMgyOkjNgU&list=PL-27w0UNlunxDTyowGrnvo_T7z92OCvpv&index=3 - amshah
Video Solution 2 by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=1328
~ pi_is_3.14
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.