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?

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

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

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

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

Solution 4 (Advanced Guesswork)

Looking at the diagram, we see that there must be a circle with a radius smaller than $\frac14$ , and there must be a circle with a radius close to $1$ . Looking at the answers, we can assume that the answer choice is above $1$ , 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 $\frac19$ and $1$ are reasonably close to the radii of the missing circles in the diagram, so we add them up to get $\boxed{\textbf{(C) }\frac{10}{9}}$ .

~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 $r_1, r_2, r_3$ . The curvature of a circle is just the reciprocal of the radius. Then we get, $(r_1 + r_2 + r_3)^2 = 2(r_1^2 + r_2^2 + r_3^2)$ Plugging in the information we already have, we get: $(1 + 4 + r_3)^2 = 2(1^2 + 4^2 + r_3^2)$ $(5 + r_3)^2 = 2(17 + r_3^2)$ $25 + 10r_3 + r_3^2 = 34 + 2r_3^2$ $r_3^2 - 10r_3 - 9 = 0$ $(r_3 - 1)(r_3 - 9) = 0$ So, we get our two curvatures: 1 and 9. The radii equivalent are $1$ and $\frac{1}{9}$ . Adding these up, we get $\boxed{\textbf{(C) }\frac{10}{9}}$ . ~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

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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