Find the distance covered by a corner of the plate in one revolution round the fixed plate.

Question

An equilateral triangular plate of side ‘a’ is rolling without slipping on the periphery of another identical fixed equilateral triangular plate as shown.

Find the distance covered by a corner of the plate in one revolution around the fixed plate.

I was able to get a rough idea of the path but I can't express it mathematically as I can't find a point which is at a fixed distance throughout the revolution.

Can someone suggest a way? Any help would be appreciated.

Answer 1

The diagram below shows the first half of the rolling trip. By symmetry, the total distance travelled will be twice the sum of thick arcs $\sum r_i \theta_i$. \begin{align} 2\cdot \left( |FB| \frac{\pi}3 + |HC| \frac{2\pi}3 + |OI| \frac{2\pi}3 \right) &= 2\cdot \left( \frac{a\sqrt 3}2 \frac{\pi}3 + a \frac{2\pi}3 + \frac{a}2 \frac{2\pi}3 \right) \\ &=2a\pi\left( \frac{\sqrt 3}6 + \frac13 + \frac16\right) \\ &= a\pi \left(1 + \frac{\sqrt 3}3 \right) \end{align}

Point $F$ travels to $H$, rotating about $B$ with radius $|FB| = |HB| = \frac{a\sqrt 3}2$.

Point $H$ travels to $I$ (through $J$), rotating about $C$ with radius $|HC| = |JC| = a$.

Point $I$ travels to $K$, rotating about $O$ with radius $|OI| = |OK| = \frac{a}2$.

Answer 2

We have $6$ rotations through $\frac{2\pi}3$ each, $3$ about the corners of the moving plate and $3$ about its midpoints. In each such rotation, a given corner is rotated through an arc of length $\frac{2\pi}3\cdot d$, where $d$ is the distance of the corner from the centre of rotation.

If we assume that the diagram is meant to imply that the corners are always on the midpoints of the sides, the total distance travelled is

$$ \frac{2\pi}3\left(0+\frac a2+a+\frac{\sqrt3}2a+a+\frac a2\right)=2\pi a\left(1+\frac{\sqrt3}6\right)\;. $$

For the general case, assume that when a corner lies on a side, its distance from the nearest corner of the other triangle is $x\le\frac a2$. Then the total distance travelled is

$$ \frac{2\pi}3\left(0+d+a+\sqrt{\left(\frac a2-d\right)^2+\frac34a^2}+a+(a-d)\right)=2\pi a\left(1+\frac13\sqrt{1-\frac da+\left(\frac da\right)^2}\right)\;, $$

which is minimal for $d=\frac12$, the case above, and maximal for $d=0$, in which case it’s $\frac{8\pi a}3$ (and you can check that that’s the result you get if you just rotate about each of the $3$ corners through $\frac{4\pi}3$.

Answer 3

Here is a diagram of the paths, which consists three kinds of arcs.