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Difference between revisions of "2024 AMC 10B Problems/Problem 21"

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==Solution 1==
 
==Solution 1==
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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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Setting <math>r_1=1, r_2=\frac14</math> we have <math>r_3=\frac19</math>, and setting <math>r_1=1,r_3=\frac14</math> we have <math>r_2=1</math>. Thus our answer is <math>\boxed{\textbf{(C)}}</math>.
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~Mintylemon66
  
 
==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 08:39, 14 November 2024

Problem

Two straight pipes (circular cylinders), with radii $1$ and $\frac{1}{4}$, 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?

[asy] size(6cm); draw(circle((0,1),1), linewidth(1.2)); draw((-1,0)--(1.25,0), linewidth(1.2)); draw(circle((1,1/4),1/4), linewidth(1.2)); [/asy]

$\textbf{(A)}~\frac{1}{9} \qquad\textbf{(B)}~1 \qquad\textbf{(C)}~\frac{10}{9} \qquad\textbf{(D)}~\frac{11}{9} \qquad\textbf{(E)}~\frac{19}{9}$

Solution 1

In general, let the left and right outer circles and the center circle have radii $r_1,r_2,r_3$. When three circles are tangent as described in the problem, we can deduce $\sqrt{r_3}=\frac{\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}}$ by Pythagorean theorem.

Setting $r_1=1, r_2=\frac14$ we have $r_3=\frac19$, and setting $r_1=1,r_3=\frac14$ we have $r_2=1$. Thus our answer is $\boxed{\textbf{(C)}}$.

~Mintylemon66

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

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