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Suppose we take 3 blocks shaped like equilateral triangles that measure 18 units on a side,and place them centrally upon axles that are themselves the vertices of an equilateral triangle measuring 30 units on a side. Suppose we wrap a durable rubber-band about the three blocks(as depicted by the bold black triangle) and paint a dot on the portion of the band that intersects the top vetex of the "top block". Then the three blocks are rotated at equal speeds- the rubber band rotates with them without slipping or expanding. The question: how far has the dot painted on the band travelled by the time it returns to its original position at the top vertex?

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2 Answers

It appears to be the perimeter of the outer triangle. A point traveling on a loop path of unchanging length and returning to the same position must surely travel the distance of the path on which it travels. In this case the path length is the perimeter of the triangle. That would be 3 * (30 + (2 * 9)) where 9 is half the side length of the smaller triangles. Which means it's 144 units.

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If we consider the bottom left vertex on the top triangle and the bottom left vertex on the lower left triangle. As the triangles rotate together, do the points get closer or further apart? I thought neither, which is why I took the view that the rubber band didn't stretch and so the total distance was simply the perimeter of the larger triangle. –  Alan Gee Sep 30 '12 at 21:11
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It doesn’t stretch, but each point moves through a distance greater than the length of the band. –  Brian M. Scott Sep 30 '12 at 21:43
    
At first glance that sounds like a paradox. I'm not saying your wrong though. –  Alan Gee Sep 30 '12 at 22:36
    
And while I can't argue with your calculations, it does lead to the rather weird implication that a triangular wheeled tank will move further in one revolution of its tread than the length of the tread, even though the tread doesn't stretch. Or am I missing something? –  Alan Gee Oct 1 '12 at 16:54
    
@user43196: Your argument would apply if the band as a set would stay fixed in the process, but this is not the case. –  Christian Blatter Oct 2 '12 at 9:40
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Consider what happens to the dot as the triangles rotate one-third of a turn: it must move through one-third of a circle of radius equal to the distance from the axle to the vertex. That radius is $6\sqrt3$, so the dot moves a circular distance of $4\pi\sqrt3$ units. It moves only $18$ linear units around the large triangle, however.

When the triangles rotate another third of a turn, the dot describes another circular arc, congruent to the first one, and moves another $18$ units linearly around the large triangle. It is now $12$ units from the lower lefthand vertex of the large triangle, so its distance from the lower lefthand axle is $$\sqrt{3^2+\left(3\sqrt3\right)^2}=6$$ units. On the next third of a turn of the triangles, therefore, it moves through a circular arc comprising one-third the circumference of a circle of radius $6$, so it moves a circular distance of $4\pi$ units. It still progresses $18$ units linearly around the large triangle, however, so it is now $6$ units from the lower lefthand vertex. The next third of a turn by the small triangles again moves it through a circular arc of length $4\pi\sqrt3$ units and takes it to the midpoint of the bottom edge of the large triangle. At this point it has moved a total of

$$3\left(4\pi\sqrt3\right)+4\pi=4\pi\left(1+3\sqrt3\right)$$

units.

The other half of its trip is a mirror image of the first half, so the total distance covered by the dot is $8\pi(1+3\sqrt3)$ units, or about $155.7263$ units.

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What do you mean by circular distance? And how are you getting 4pi/3 units? –  Xuan Huang Oct 1 '12 at 17:26
    
Never mind... made an extraordinarily stupid calculation error. –  Xuan Huang Oct 1 '12 at 18:56
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