2004 AIME I Problems/Problem 8
Define a regular -pointed star to be the union of line segments such that
- the points are coplanar and no three of them are collinear,
- each of the line segments intersects at least one of the other line segments at a point other than an endpoint,
- all of the angles at are congruent,
- all of the line segments are congruent, and
- the path turns counterclockwise at an angle of less than 180 degrees at each vertex.
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?
We use the Principle of Inclusion-Exclusion (PIE).
If we join the adjacent vertices of the regular -star, we get a regular -gon. We number the vertices of this -gon in a counterclockwise direction:
A regular -star will be formed if we choose a vertex number , where , and then form the line segments by joining the following pairs of vertex numbers:
If , then the star degenerates into a regular -gon or a (2-vertex) line segment if . Therefore, we need to find all such that .
Let , and . The number of 's that are not relatively prime to is:
Vertex numbers and must be excluded as values for 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
Note that in general, the number of -pointed stars is given by (dividing by to remove the reflectional symmetry, subtracting to get rid of the -step case), where is the Euler's totient function. It is well-known that , where are the distinct prime factors of . Thus , and the answer is .
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