2021 AMC 12B Spring Problem 8

Attempt Problem 8 of the 2021 AMC 12B Spring below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

Three equally spaced parallel lines intersect a circle, creating three chords of lengths 38,38, 38,38, and 34.34. What is the distance between two adjacent parallel lines?
5125\tfrac{1}{2}
66
6126\tfrac{1}{2}
77
7127\tfrac{1}{2}
Answer: B
Concepts:chordPythagorean Theorem
Small hint:
Equal chords are equidistant from the center, so the two 3838 chords sit at heights ±d2\pm\tfrac{d}{2}
Big hint:
With spacing d,d, use r2−(d2)2=192r^2-\left(\tfrac{d}{2}\right)^2=19^2 and r2−(3d2)2=172r^2-\left(\tfrac{3d}{2}\right)^2=17^2
Solution:
Place the center at height 0.0. Two equal chords lie at equal distances from the center, so the three equally spaced lines are at heights −d2,d2,3d2,-\tfrac{d}{2},\tfrac{d}{2},\tfrac{3d}{2}, with the two 3838-chords at ±d2\pm\tfrac{d}{2} and the 3434-chord at 3d2.\tfrac{3d}{2}.
Half-chord relations give r2−(d2)2=192r^2-\left(\tfrac{d}{2}\right)^2=19^2 and r2−(3d2)2=172.r^2-\left(\tfrac{3d}{2}\right)^2=17^2.
Subtracting, 2d2=192−172=72,2d^2=19^2-17^2=72, so d2=36d^2=36 and d=6.d=6.
Thus, the correct answer is B.
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