1988 AJHSME Problems/Problem 24

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Problem

The square in the first diagram "rolls" clockwise around the fixed regular hexagon until it reaches the bottom. In which position will the solid triangle be in diagram $4$ ?

Solution

Solution 1

The inner angle of the hexagon is $120^\circ$ , and the inner angle of the square is $90^\circ$ . Hence during each rotation the square is rotated by $360^\circ - 120^\circ - 90^\circ = 150^\circ$ clockwise. In the diagram $4$ the square is therefore rotated by $3\cdot 150^\circ = 450^\circ$ clockwise from its original state. Rotation by $450^\circ$ is identical to rotation by $450^\circ - 360^\circ = 90^\circ$ , hence the black triangle in the diagram $4$ will be pointing to the right, and the answer is $\boxed{\text{(A)}}$ .

Solution 2

Alternately, we can simply keep track of the "bottom" side of the square. In the diagrams below, this bottom side is shown in red.

1988 AJHSME (Problems • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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