1972 AMC 12 Problem 35

Attempt Problem 35 of the 1972 AMC 12 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 1972 AMC 12 solutions, or check the answer key.

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

35.

Equilateral triangle ABPABP with side ABAB of length 22 inches is placed inside square AXYZAXYZ with side of length 44 inches so that BB is on side AX.AX. The triangle is rotated clockwise about B,B, then P,P, and so on along the sides of the square until P,P, A,A, and BB all return to their original positions. The length of the path in inches traversed by vertex PP is equal to:
20π3\dfrac{20\pi}{3}
32π3\dfrac{32\pi}{3}
12π12\pi
40π3\dfrac{40\pi}{3}
15π15\pi
Answer: D
Concepts:equilateral triangletransformationarcprocess simulation
Small hint:
Track the orientation after one eight-pivot circuit of the square
Big hint:
Three circuits are needed; classify the pivots according to whether PP is fixed or moves through a 120∘120^\circ or 30∘30^\circ arc
Solution:
One circuit of the square uses 88 pivots and changes the triangle’s orientation by 23\frac{2}{3} of a full turn. Therefore 33 circuits, or 2424 pivots, are required to restore the entire triangle. In 88 of those pivots the rotation is about P,P, so PP does not move. In the other 16,16, eight arcs subtend 120∘120^\circ and eight subtend 30∘,30^\circ, all with radius 2.2. Hence the total path length is 8(13⋅4π)+8(112⋅4π)=32π3+8π3=40π3. \begin{aligned} &8\left(\frac13\cdot4\pi\right) +8\left(\frac1{12}\cdot4\pi\right)\\ &\qquad=\frac{32\pi}{3}+\frac{8\pi}{3}\\ &\qquad=\frac{40\pi}{3}. \end{aligned}
Therefore, the correct answer is D.
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