2024 AMC 10B Problems/Problem 21

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

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

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2024 AMC 10B Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

Video Solution 1 by Pi Academy (In Less Than 5 Mins ⚡🚀)

See also