AIME 2004 I · 第 8 题

Solution 1

We use the Principle of Inclusion-Exclusion (PIE).

If we join the adjacent vertices of the regular $n$-star, we get a regular $n$-gon. We number the vertices of this $n$-gon in a counterclockwise direction: $0, 1, 2, 3, \ldots, n-1.$

A regular $n$-star will be formed if we choose a vertex number $m$, where $0 \le m \le n-1$, and then form the line segments by joining the following pairs of vertex numbers: $(0 \mod{n}, m \mod{n}),$ $(m \mod{n}, 2m \mod{n}),$ $(2m \mod{n}, 3m \mod{n}),$ $\cdots,$ $((n-2)m \mod{n}, (n-1)m \mod{n}),$ $((n-1)m \mod{n}, 0 \mod{n}).$

If $\gcd(m,n) > 1$, then the star degenerates into a regular $\frac{n}{\gcd(m,n)}$-gon or a (2-vertex) line segment if $\frac{n}{\gcd(m,n)}= 2$. Therefore, we need to find all $m$ such that $\gcd(m,n) = 1$.

Note that $n = 1000 = 2^{3}5^{3}.$

Let $S = \{1,2,3,\ldots, 1000\}$, and $A_{i}= \{i \in S \mid i\, \textrm{ divides }\,1000\}$. The number of $m$'s that are not relatively prime to $1000$ is: $\mid A_{2}\cup A_{5}\mid = \mid A_{2}\mid+\mid A_{5}\mid-\mid A_{2}\cap A_{5}\mid$ $= \left\lfloor \frac{1000}{2}\right\rfloor+\left\lfloor \frac{1000}{5}\right\rfloor-\left\lfloor \frac{1000}{2 \cdot 5}\right\rfloor$ $= 500+200-100 = 600.$

Vertex numbers $1$ and $n-1=999$ must be excluded as values for $m$ since otherwise a regular n-gon, instead of an n-star, is formed.

The cases of a 1st line segment of (0, m) and (0, n-m) give the same star. Therefore we should halve the count to get non-similar stars.

Therefore, the number of non-similar 1000 pointed stars is $\frac{1000-600-2}{2}= \boxed{199}.$

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Note that in general, the number of $n$-pointed stars is given by $\frac{\phi(n)}{2} - 1$ (dividing by $2$ to remove the reflectional symmetry, subtracting $1$ to get rid of the $1$-step case), where $\phi(n)$ is the Euler's totient function. It is well-known that $\phi(n) = n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_n}\right)$, where $p_1,\,p_2,\ldots,\,p_n$ are the distinct prime factors of $n$. Thus $\phi(1000) = 1000\left(1 - \frac 12\right)\left(1 - \frac 15\right) = 400$, and the answer is $\frac{400}{2} - 1 = 199$.

Solution 2

Continue as before by cyclically constructing a star by taking every $k$-th point. We can think of this construction as the orbit of a point by a rotation $R^k\in D_{1000}$. Then, we want $|R^k|=1000$, meaning that $k\in (\mathbb Z/1000\mathbb Z)^\times$. Then, $k$ is coprime to $1000$, meaning $k$ may take on $\phi(1000)=400$ possible values. For any $R^k$, the dihedral symmetry $TR^k$ defines the figure-preserving bijection $R^k\mapsto R^{-k}$, meaning half of our possible values of $k$ are similar. Finally, the condition that each edge must intersect another rules out $k=1$, giving a total of $\frac{400}2-1=\boxed{199}$ pointed stars.

~pineconee