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 (Pythagorean Theorem)

Label the center of both circles $O$ . Label the chord in the larger circle as $\overline{ABCD}$ , where $A$ and $D$ are on the larger circle and $B$ and $C$ are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as $M$ . Because a radius that is perpendicular to a chord bisects the chord, $M$ is the midpoint of 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 $\triangle OMB$ , we have $\begin{align*} OM^2 & = OB^2 - BM^2 \\ & = OB^2 - \left( \frac{BC}{2} \right)^2 \\ & = OB^2 - \left( \frac{AD}{4} \right)^2 \\ & = 17^2 - \frac{AD^2}{16} . \end{align*}$

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

~eisthefifthletter and Steven Chen

Solution 2 (Power of a Point)

Draw the diameter perpendicular to the chord. Notice that by symmetry this diameter bisects the chord.

Call the intersection between that diameter and the chord $A$ . In the smaller circle, let the shorter piece of the diameter cut by the chord be $x$ , making the longer piece $34-x.$ In that same circle, let the $y$ be the length of the portion of the chord that is cut by the diameter.

The radius of the larger circle is $2$ more than the radius of the small circle. So, in the larger circle, the shorter piece of the diameter cut by the chord is of length $x+2$ and the longer piece is $36-x.$ As given in the problem, the bisected length of the chord in the larger circle is twice as much, so it is of length $2y$ . By Power of a Point, we can construct a system of equations $x(34-x) = y^2$ $(x+2)(36-x) = (2y)^2.$ Expanding both equations, we get $34x-x^2 = y^2$ $36x-x^2+72-2x = 4y^2,$ 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

Video Solution by TheBeautyofMath

https://youtu.be/ToiOlqWz3LY

~IceMatrix

Video Solution (Logic and Geometry)

https://youtu.be/iG1vVXeTv58

~Education, the Study of Everything

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

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