1972 AHSME Problems/Problem 35

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Problem

[asy] draw(unitsquare);draw((0,0)--(.25,sqrt(3)/4)--(.5,0)); label("Z",(0,1),NW);label("Y",(1,1),NE);label("A",(0,0),SW);label("X",(1,0),SE);label("B",(.5,0),S);label("P",(.25,sqrt(3)/4),N); //Credit to Zimbalono for the diagram [/asy]

Equilateral triangle $ABP$ (see figure) with side $AB$ of length $2$ inches is placed inside square $AXYZ$ with side of length $4$ inches so that $B$ is on side $AX$. The triangle is rotated clockwise about $B$, then $P$, and so on along the sides of the square until $P$ returns to its original position. The length of the path in inches traversed by vertex $P$ is equal to

$\textbf{(A) }20\pi/3\qquad \textbf{(B) }32\pi/3\qquad \textbf{(C) }12\pi\qquad \textbf{(D) }40\pi/3\qquad  \textbf{(E) }15\pi$

Solution

$\fbox{D}$