Difference between revisions of "2024 AMC 10B Problems/Problem 21"
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==Solution 2== | ==Solution 2== | ||
− | sum of radii | + | Notice that the sum of radii of two circles tangent to each other will equal to the distance from center to center. Set the center of the big circle be at <math>(0,1).</math> Since both circles are tangent to a line (in this case, <math>y=0</math>), the y-coordinates of the centers are just its radius. |
− | + | Hence, the center of the smaller circle is at <math>\left(x_2, \frac14\right)</math>. From the the sum of radii and distance formula, we have: | |
+ | <cmath>1+\frac14 = \sqrt{x_2^2 + \left(\frac34\right)^2} \Rightarrow x_2 = 1.</cmath> | ||
− | + | So, the coordinates of the center of the smaller circle are <math>(1, \frac14).</math> Now, let <math>(x_3,r_3)</math> be the coordinates of the new circle. Then, from the fact that sum of radii of this circle and the circle with radius <math>1</math> is equal to the distance from the two centers, you have: | |
− | + | <cmath>\sqrt{(x_3-0)^2+(r_3-1)^2} = 1+r_3.</cmath> | |
− | + | Similarly, from the fact that the sum of radii of this circle and the circle with radius <math>\frac14</math>, you ahve: | |
+ | <cmath>\sqrt{(x_3-1)^2+\left(r_3-\frac14\right)^2}= \frac14 + r_3.</cmath> | ||
+ | Squaring the first equation, you have: | ||
+ | <cmath>x_3^2+r_3^2-2r_3+1=1+2r_3+r_3^2 \Rightarrow 4r_3=x_3^2 \Rightarrow x_3=2\sqrt{r_3}.</cmath> | ||
+ | Squaring the second equation, you have: | ||
+ | <cmath>x_3^2-2x_3+1+r_3^2-\frac{r_3}{2}+\frac{1}{16}=\frac{1}{16}+\frac{r_3}{2}+r_3^2 \Rightarrow x_3^2-2x_3+1=r_3</cmath> | ||
+ | Plugging in from the first equation we have | ||
+ | <cmath>r_3-1=x_3^2-2x_3=4r_3-2\sqrt{r_3} \Rightarrow 3r_3-2\sqrt{r_3}+1=0 \Rightarrow (3\sqrt{r_3}-1)(\sqrt{r_3}-1)=0 \Rightarrow r_3=1, \frac19.</cmath> | ||
+ | Summing these two yields <math>\boxed{\frac{10}{9}}.</math> | ||
~mathboy282 | ~mathboy282 |
Revision as of 19:19, 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
Notice that the sum of radii of two circles tangent to each other will equal to the distance from center to center. Set the center of the big circle be at Since both circles are tangent to a line (in this case, ), the y-coordinates of the centers are just its radius.
Hence, the center of the smaller circle is at . From the the sum of radii and distance formula, we have:
So, the coordinates of the center of the smaller circle are Now, let be the coordinates of the new circle. Then, from the fact that sum of radii of this circle and the circle with radius is equal to the distance from the two centers, you have: Similarly, from the fact that the sum of radii of this circle and the circle with radius , you ahve: Squaring the first equation, you have: Squaring the second equation, you have: Plugging in from the first equation we have Summing these two yields
~mathboy282
Solution 3 (Descartes’s Theorem)
Descartes’s Theorem states that for curvatures we have
with a curvature being the reciprocal of the radius of a circle, being positive if it is externally tangent to other circles, negative if it is internally tangent to other circles, and zero if we consider a line as a degenerate circle. Here our curvatures are , and we wish to find . Plugging these values into our formula yields:
The curvature and the radius are reciprocals, so our radii must be and , and their sum is .
~eevee9406
Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)
https://youtu.be/5fID8UOohr0?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
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.