Difference between revisions of "2024 AMC 10B Problems/Problem 21"
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\qquad\textbf{(D)}~\frac{11}{9} | \qquad\textbf{(D)}~\frac{11}{9} | ||
\qquad\textbf{(E)}~\frac{19}{9}</math> | \qquad\textbf{(E)}~\frac{19}{9}</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | // By Elephant200 | ||
+ | size(8cm); | ||
+ | draw(circle((0,1),1), linewidth(1.2)); | ||
+ | draw((-1,0)--(3,0), linewidth(1.2)); | ||
+ | draw(circle((1,1/4),1/4), linewidth(1.2)); | ||
+ | draw(circle((2/3,1/9),1/9), red+linewidth(1.2)); | ||
+ | draw(circle((2,1),1), red+linewidth(1.2)); | ||
+ | </asy> | ||
+ | ~Elephant200 | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
+ | 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 have: | ||
+ | <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 | ||
+ | |||
+ | ==Solution 2== | ||
In general, let the left and right outer circles and the center circle have radii <math>r_1,r_2,r_3</math>. When three circles are tangent as described in the problem, we can deduce <math>\sqrt{r_3}=\frac{\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}}</math> by Pythagorean theorem. | In general, let the left and right outer circles and the center circle have radii <math>r_1,r_2,r_3</math>. When three circles are tangent as described in the problem, we can deduce <math>\sqrt{r_3}=\frac{\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}}</math> by Pythagorean theorem. | ||
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~Mintylemon66 | ~Mintylemon66 | ||
+ | ==Solution 3 (Descartes’s Theorem)== | ||
+ | Descartes’s Theorem states that for curvatures <math>k_1,k_2,k_3,k_4</math> we have | ||
− | = | + | <cmath>(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2)</cmath> |
− | + | ||
+ | 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 <math>k_1=1,k_2=4,k_3=0</math>, and we wish to find <math>k_4</math>. Plugging these values into our formula yields: | ||
+ | |||
+ | <cmath>(1+4+0+k_4)^2=2(1^2+4^2+0^2+k_4^2)</cmath> | ||
+ | |||
+ | <cmath>(5+k_4)^2=2(17+k_4^2)</cmath> | ||
+ | |||
+ | <cmath>k_4^2+10k_4+25=2k_4^2+34</cmath> | ||
+ | |||
+ | <cmath>k_4^2-10k_4+9=0</cmath> | ||
+ | |||
+ | <cmath>k_4=1,9</cmath> | ||
+ | |||
+ | The curvature and the radius are reciprocals, so our radii must be <math>1</math> and <math>\frac{1}{9}</math>, and their sum is <math>\boxed{\textbf{(C) }\frac{10}{9}}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
− | + | ==Solution 4 (Advanced Guesswork)== | |
+ | Looking at the diagram, we see that there must be a circle with a radius smaller than <math>\frac14</math>, and there must be a circle with a radius close to <math>1</math>. Looking at the answers, we can assume that the answer choice is above <math>1</math>, eliminating the first two answers. We can also assume that some students will only solve for the smaller circle, while others may only solve for the larger circle. It would be reasonably safe to say that one of the answers must be the radius of the smaller circle, while another must be the radius of the bigger circle. Looking at the answers, we see that <math>\frac19</math> and <math>1</math> are reasonably close to the radii of the missing circles in the diagram, so we add them up to get <math>\boxed{\textbf{(C) }\frac{10}{9}}</math>. | ||
− | + | ~YTH | |
− | + | ==Solution 5 (Essentially Solution 3)== | |
+ | For a problem like this with three externally tangent circles that are all tangent to a line, we can use a specific form of | ||
+ | Descartes's Circle Theorem. Let the curvatures of the three circles be <math>r_1, r_2, r_3</math>. The curvature of a circle is just | ||
+ | the reciprocal of the radius. Then we get, <cmath>(r_1 + r_2 + r_3)^2 = 2(r_1^2 + r_2^2 + r_3^2)</cmath> Plugging in the information we already have, we get: | ||
+ | <cmath>(1 + 4 + r_3)^2 = 2(1^2 + 4^2 + r_3^2)</cmath> | ||
+ | <cmath>(5 + r_3)^2 = 2(17 + r_3^2)</cmath> | ||
+ | <cmath>25 + 10r_3 + r_3^2 = 34 + 2r_3^2</cmath> | ||
+ | <cmath>r_3^2 - 10r_3 - 9 = 0</cmath> | ||
+ | <cmath>(r_3 - 1)(r_3 - 9) = 0</cmath> | ||
+ | So, we get our two curvatures: 1 and 9. The radii equivalent are <math>1</math> and <math>\frac{1}{9}</math>. Adding these up, we get <math>\boxed{\textbf{(C) }\frac{10}{9}}</math>. | ||
− | ~ | + | ~Mr.Lightning |
==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)== | ==Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)== | ||
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~ Pi Academy | ~ Pi Academy | ||
+ | |||
+ | ==Video Solution 2 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=24EZaeAThuE | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2024|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:14, 15 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?
Diagram
~Elephant200
Solution 1
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 have: 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 2
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 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
Solution 4 (Advanced Guesswork)
Looking at the diagram, we see that there must be a circle with a radius smaller than , and there must be a circle with a radius close to . Looking at the answers, we can assume that the answer choice is above , eliminating the first two answers. We can also assume that some students will only solve for the smaller circle, while others may only solve for the larger circle. It would be reasonably safe to say that one of the answers must be the radius of the smaller circle, while another must be the radius of the bigger circle. Looking at the answers, we see that and are reasonably close to the radii of the missing circles in the diagram, so we add them up to get .
~YTH
Solution 5 (Essentially Solution 3)
For a problem like this with three externally tangent circles that are all tangent to a line, we can use a specific form of Descartes's Circle Theorem. Let the curvatures of the three circles be . The curvature of a circle is just the reciprocal of the radius. Then we get, Plugging in the information we already have, we get: So, we get our two curvatures: 1 and 9. The radii equivalent are and . Adding these up, we get .
~Mr.Lightning
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.