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2021 AMC 12A Fall Problem 11

Below is the professionally curated solution for Problem 11 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

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Concepts:chordPythagorean Theorem

Difficulty rating: 1590

11.

Consider two concentric circles of radius 1717 and 19.19. 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?

12212\sqrt{2}

10310\sqrt{3}

17⋅19\sqrt{17 \cdot 19}

1818

868\sqrt{6}

Solution:

Let the chord lie at distance dd from the common center. Its total length is 2361−d2,2\sqrt{361 - d^2}, and the portion inside the smaller circle has length 2289−d2.2\sqrt{289 - d^2}.

Since half the chord lies inside, 2289−d2=12⋅2361−d2.2\sqrt{289 - d^2} = \tfrac{1}{2}\cdot 2\sqrt{361 - d^2}. Squaring gives 4(289−d2)=361−d2,4(289 - d^2) = 361 - d^2, so 3d2=7953d^2 = 795 and d2=265.d^2 = 265.

The chord length is 2361−265=296=86.2\sqrt{361 - 265} = 2\sqrt{96} = 8\sqrt{6}.

Thus, the correct answer is E.

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