Problem 25 · 2012 Math Kangaroo
Stretch
Spatial & Visual Reasoning
path-tracing
An equilateral triangle is being rolled around a unit square as shown. How long is the path that the point shown covers, if the point and the triangle are both back at the start for the first time?
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Answer: B — \(\tfrac{28}{3}\pi\)
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Hint 1 of 2
The triangle has the same side length (1) as the square, so each pivot turns it by the exterior angle \(120^\circ\), except at a square corner where an extra \(90^\circ\) is added.
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Hint 2 of 2
The marked point sweeps an arc of radius 1 each flip, unless the point itself is the pivot (then radius 0, no arc); add the arcs over one full return circuit.
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Approach: sum the arcs swept by the marked vertex over one full circuit
- Each tumble pivots the triangle about a vertex; the marked point (another vertex, distance 1 away) sweeps a circular arc of radius 1 — and stays put whenever it is itself the pivot.
- On a flat stretch each flip turns the triangle \(120^\circ = \tfrac{2\pi}{3}\); rounding a square corner adds an extra \(90^\circ\) to that turn.
- Following the rolling until both the point and the triangle first return to their starting position, the swept arcs (each of radius 1) total an angle of \(\tfrac{28}{3}\pi\).
- So the path length is \(\tfrac{28}{3}\pi\), choice B.
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