2021 Fall AMC 12A Problems/Problem 11

Problem

Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?

$\textbf{(A)}\ 12\sqrt{2} \qquad\textbf{(B)}\ 10\sqrt{3} \qquad\textbf{(C)}\ \sqrt{17 \cdot 19} \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 8\sqrt{6}$

Solution 1 (Power of a Point)

Draw the diameter perpendicular to the chord. Call the intersection between that diameter and the chord $A$ . In the circle of radius $17$ , let the shorter piece of the diameter cut by the chord would be of length $x$ , making the longer piece $34-x.$ In that same circle, let the $y$ be the length of the portion of the chord in the smaller circle that is cut by the diameter we drew. Thus, in the circle of radius $7$ , the shorter piece of the diameter cut by the chord would be of length $x+2$ , making the longer piece $36-x,$ and length of the piece of the chord cut by the diameter would be $2y$ (as given in the problem). By Power of a Point, we can construct the system of equations $x(34-x) = y^2$ $(x+2)(36-x) = (y+2)^2$ Expanding both equations, we get $34x-x^2 = y^2$ and $36x-x^2+72-2x = 4y^2,$ in which the $34x$ and $-x^2$ terms magically cancel when we subtract the first equation from the second equation. Thus, now we have $72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{\textbf{(E)} \: 8\sqrt{6}}$ .

-fidgetboss_4000

Solution 2 (Pythagorean Theorem)

Label the center of both circles $O$ . Label the chord in the larger circle $\overline{AD}$ ; let this chord intersect the smaller circle at $B$ and $C$ , where $B$ is between $A$ and $C$ . Construct the radius perpendicular to the chord, and label its intersection with the chord as $M$ . $M$ is the midpoint of $AB$ because a radius that is perpendicular to a chord bisects the chord.

Construct segments $\overline{AO}$ and $\overline{BO}$ . These are radii with lengths 17 and 19 respectively.

Then, use the Pythagorean Theorem. In $\triangle OMA$ , we have $\begin{align*} OM^2 & = OA^2 - AM^2 \\ & = OA^2 - \left( \frac{AD}{2} \right)^2 \\ & = 19^2 - \frac{AD^2}{4} . \end{align*}$

In $\triangle OMC$ , we have $\begin{align*} OM^2 & = OC^2 - CM^2 \\ & = OC^2 - \left( \frac{BC}{2} \right)^2 \\ & = OC^2 - \left( \frac{AD}{4} \right)^2 \\ & = 17^2 - \frac{AD^2}{16} . \end{align*}$

Equating these two expressions, we get $19^2 - \frac{AB^2}{4} = 17^2 - \frac{AB^2}{16}$ and $AD=\boxed{\textbf{(E) }8 \sqrt{6}}$ .

~eisthefifthletter and Steven Chen

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 12 Problems and Solutions

Page

Toolbox

Search

2021 Fall AMC 12A Problems/Problem 11

Contents

Problem

Solution 1 (Power of a Point)

Solution 2 (Pythagorean Theorem)

See Also