Difference between revisions of "2024 AMC 10B Problems/Problem 21"
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+ | ==Solution 2== | ||
+ | sum of radii = distances from center. Set the center of the big circle be (0,1). The center of the smaller circle is at (x_2, 1/4). (1+1/4)^2 = x_3^2 + (3/4)^2 -> x_2 = 4/4 = 1. | ||
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+ | let (x_2,r_2) be the coordinates of the new circle. Then you have (x_2-0)^2+(r_3-1)^2 = (1+r_3)^2. | ||
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+ | You also have (x_2-1)^2+(r_3-1/4)^2=(1/4 + r_3)^2. | ||
+ | |||
+ | These two you should get a quadratic for r_3, and you get sols 1 and 1/9 >>> 10/9 C | ||
+ | |||
+ | ~mathboy282 | ||
==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)== | ==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)== |
Revision as of 11:58, 14 November 2024
Contents
[hide]Problem
Two straight pipes (circular cylinders), with radii and , lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?
Solution 1
In general, let the left and right outer circles and the center circle have radii . When three circles are tangent as described in the problem, we can deduce by Pythagorean theorem.
Setting we have , and setting we have . Thus our answer is .
~Mintylemon66
Solution 2
sum of radii = distances from center. Set the center of the big circle be (0,1). The center of the smaller circle is at (x_2, 1/4). (1+1/4)^2 = x_3^2 + (3/4)^2 -> x_2 = 4/4 = 1.
let (x_2,r_2) be the coordinates of the new circle. Then you have (x_2-0)^2+(r_3-1)^2 = (1+r_3)^2.
You also have (x_2-1)^2+(r_3-1/4)^2=(1/4 + r_3)^2.
These two you should get a quadratic for r_3, and you get sols 1 and 1/9 >>> 10/9 C
~mathboy282
Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)
https://youtu.be/5fID8UOohr0?feature=shared
~ Pi Academy
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.