2021 Fall AMC 12A Problems/Problem 11

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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 (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 statement). 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 $72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{ 8\sqrt{6}}.$

-fidgetboss_4000