Selection tests for the Gulf Mathematical Olympiad 2013

Because all angles are congruent and all line segments are congruent, the regular $n$-pointed star is cyclic. Indeed, consider any four consecutive vertices $P_i,P_{i+1},P_{i+2},P_{i+3}$, for some $i\in \{1,2,\dots ,n\}$. They form an isosceles trapezoid. So they are cocyclic. By induction, all vertices $P_1,P_2,\dots ,P_n$ are on the same circle. Let $O$ be the center of this circle.

Because all the line segments are congruent and the path turns counterclockwise, there exists a positive real number $0^{\circ }<\theta <180^{\circ }$ such that $\angle P_{i}O{P}_{i+1}=\theta$ for all $i=1,2,\dots ,n$. Because $P_{n+1}=P_1$, there exists a positive integer $k$ such that $n\theta =360k^{\circ }$. But $\theta <180^{\circ }$. Therefore $k<\frac{n}{2}$. Because each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint, $k>1$. Because all the vertices $P_1,P_2,\dots ,P_n$ are different, the integers $k$ and $n$ have no common divisors.

Conversely, for a given circumradius, if $1<k<\frac{n}{2}$ is an integer relatively prime to $n$, there exists a unique regular $n$-pointed star of angle $\theta =\frac{360k^{\circ }}{n}$. Hence, the number of non-similar regular $n$-pointed stars is the number of such integers $k$, that is $\frac{\varphi (n)}{2}-1$, where $\varphi$ is the Euler totient function. Therefore, it remains to solve the equation $\varphi (n)=60$.

There are three cases. The first case is when there exists a prime $p$ divisor of $n$ such that $3\times 5$ divides $p-1$. Here, the only possibilities are $p=61$ or $31$.

If $p=61$, there are 2 solutions for $n$. Either $n=61$ or $n=2\times 61=122$. If $p=31$, there are 3 solutions for $n$. Either $n=3\times 31=93$, or $n=2\times 3\times 31=186$, or $n=2^2\times 31=124$.

The second case is when there exist two prime numbers $p$ and $q$ such that $3$ divides $p-1$ and $5$ divides $q-1$. In this case, the only possibility is $p=7$ and $q=11$. This gives only two solutions for $n$. Either $n=7\times 11=77$ or $n=2\times 7\times 11=154$.

The third case is when there exists an odd prime number $p$ such that $p^2$ divides $n$. Then $p=3$ or $5$. If $n=5^{2}m$ and $m$ is relatively prime to $5$, we obtain $\varphi (m)=3$ which is impossible since $3$ is an odd number. If $n=3^{2}m$ and $m$ is relatively prime to $3$, we obtain $\varphi (m)=10$. Therefore, $m=11$ or $22$. Hence $n=99$ or $198$.

In conclusion, all possible values for $n$ are $61,77,93,99,122,124,154,186$ and $198$.