What type of polygon will fit this description?

Question

A particle moving with constant speed turns left by an angle of 74° after travelling every 1m distance. It returns back to the starting point in 18s and we are required to find out the speed of the particle. My attempt: I know that the particle will return back to its initial position if the line joining it to the orgin rotates by a multiple of 360°,therefore after it changes it's direction 180 times,it will have rotated an angle of 13320° which is 37 times 360.Therfore,the time taken for one rotation will be 18/180= 0.1 second and hence the speed is distance divided by time which is 1/0.1=10m/s.How will the motion of the particle look like?After solving the problem independently, I tried to look up the solution where it was mentioned that the motion will describe a regular polygon which seems incorrect.

Answer 1

Because $74$ is not of the form $\frac{n-2}{n}*180$ for some integer $n\geq 3$, this will not trace a convex regular polygon. It will however trace a star polygon. Let $n$ be the number of edges of this star polygon. Then we have that $74*n$ is divisible by $360$ and $n$ is the smallest such number. After some calculations, you get that $n = 180$ and thus that the speed $180/18=10$ meters per second. Below is a picture of the polygon.

Answer 2

I interpret the question as saying that the particle travels along a circular path, not a polygonal one. In that case the answer would be:

$1$ metre is $74/360$ths of a full circle.
Therefore a full circle is $360/74$ metres.
This is travelled in $18$ seconds, so it has a speed of $\frac{360}{74\cdot18}=\frac{10}{37}=0.27027$ metres per second.

As for the polygon - it would be the regular star polygon $\{180/37\}$.