Q:

(Taken from the 2019-11-01 Riddler)

Six snails are situated at the corners of a regular hexagon whose sides are all 10 meters. Each snail is determined to reach its clockwise neighbor, and they all travel at the same speed. But keep in mind that as each snail moves along, slowly but surely, the snail it’s traveling toward is also moving toward another snail.

The result is six snails — and six trails of slime — that spiral inward toward the center of the hexagon. How long is each snail’s slimy trail?

A:

Hello again! By the willpower of Franklin Lu, we are once again back in the problem solving math blog business. It’s good to be back.

This little problem was sent to me by Mr. Lu himself, and its solution is very neat and not at all calc-y, despite what one might expect.

Remark 1

By the rotational symmetry of the problem, we know that any given time, the snails always remain in a regular hexagonal arrangement.

Remark 2

Because the snails always remain in regular hexagonal arrangement, it is easy to show that each snail’s radial velocity (that is, the component of each snail’s velocity pointing towards the center of the hexagon) is constant.

Consider one snail. Suppose it travels with velocity $v$. At any moment, the snail’s velocity points towards the next snail. Since the snails are always arranged regular hexagonally, the snail’s velocity vector makes a 60 degree angle with the vector pointing from the snail towards the center of the hexagon. The snail’s radial velocity can be computed to be

$$\lvert v \rvert \cos 60^{\circ} = \frac{1}{2} \lvert v \rvert$$

Note that the above argument is time-independent. Therefore each snail’s radial velocity is a constant $\frac{1}{2} \lvert v \rvert$.

And Thus the Answer Is

Knowing the radial velocity, we can then compute (assuming $\lvert v \rvert$ is in m/s) that each snail takes

$$\frac{10}{\frac{1}{2} \lvert v \rvert} = \frac{20}{\lvert v \rvert}$$

seconds to reach the center of the hexagon (where they all finally meet).

In that time, each snail will have traveled

$$\lvert v \rvert \cdot \frac{20}{\lvert v \rvert} = 20$$

meters.