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

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 $(0,1).$ Since both circles are tangent to a line (in this case, $y=0$ ), the y-coordinates of the centers are just its radius.

Hence, the center of the smaller circle is at $\left(x_2, \frac14\right)$ . From the the sum of radii and distance formula, we have: $1+\frac14 = \sqrt{x_2^2 + \left(\frac34\right)^2} \Rightarrow x_2 = 1.$

So, the coordinates of the center of the smaller circle are $(1, \frac14).$ Now, let $(x_3,r_3)$ be the coordinates of the new circle. Then, from the fact that sum of radii of this circle and the circle with radius $1$ is equal to the distance from the two centers, you have: $\sqrt{(x_3-0)^2+(r_3-1)^2} = 1+r_3.$ Similarly, from the fact that the sum of radii of this circle and the circle with radius $\frac14$ , you have: $\sqrt{(x_3-1)^2+\left(r_3-\frac14\right)^2}= \frac14 + r_3.$ Squaring the first equation, you have: $x_3^2+r_3^2-2r_3+1=1+2r_3+r_3^2 \Rightarrow 4r_3=x_3^2 \Rightarrow x_3=2\sqrt{r_3}.$ Squaring the second equation, you have: $x_3^2-2x_3+1+r_3^2-\frac{r_3}{2}+\frac{1}{16}=\frac{1}{16}+\frac{r_3}{2}+r_3^2 \Rightarrow x_3^2-2x_3+1=r_3$ Plugging in from the first equation we have $r_3-1=x_3^2-2x_3=4r_3-2\sqrt{r_3} \Rightarrow 3r_3-2\sqrt{r_3}+1=0 \Rightarrow (3\sqrt{r_3}-1)(\sqrt{r_3}-1)=0 \Rightarrow r_3=1, \frac19.$ Summing these two yields $\boxed{\frac{10}{9}}.$

~mathboy282

Solution 2

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 3 (Descartes’s Theorem)

Descartes’s Theorem states that for curvatures $k_1,k_2,k_3,k_4$ we have

$(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2)$

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 $k_1=1,k_2=4,k_3=0$ , and we wish to find $k_4$ . Plugging these values into our formula yields:

$(1+4+0+k_4)^2=2(1^2+4^2+0^2+k_4^2)$

$(5+k_4)^2=2(17+k_4^2)$

$k_4^2+10k_4+25=2k_4^2+34$

$k_4^2-10k_4+9=0$

$k_4=1,9$

The curvature and the radius are reciprocals, so our radii must be $1$ and $\frac{1}{9}$ , and their sum is $\boxed{\textbf{(C) }\frac{10}{9}}$ .

~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

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
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