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Difference between revisions of "2004 AIME I Problems/Problem 8"

(→‎Solution)
(→‎See also)
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== See also ==
 
== See also ==
 
* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]
 +
Euler <math>\phi</math> function gives the number of integers less than n that do not have a factor in common with n besides 1.
 +
<math>\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3}) \cdots,</math>
 +
where the <math>{p_i}</math>'s are distinct prime factors of n.
 +
 +
<math>\phi (1000) = 1000 \cdot (1-\frac{1}{2})(1-\frac{1}{5}) = 400.</math>

Revision as of 22:15, 20 August 2006

Problem

Define a regular $n$-pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that

  • the points $P_1, P_2,\ldots, P_n$ are coplanar and no three of them are collinear,
  • each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint,
  • all of the angles at $P_1, P_2,\ldots, P_n$ are congruent,
  • all of the $n$ line segments $P_2P_3,\ldots, P_nP_1$ are congruent, and
  • the path $P_1P_2, P_2P_3,\ldots, P_nP_1$ 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?

Solution

Uses PIE (principle of inclusion-exclusion).

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 half the count to get non-similar stars.

Number of non-similar 1000-pointed stars $= \frac{1000-600-2}{2}= 199.$

See also

Euler $\phi$ function gives the number of integers less than n that do not have a factor in common with n besides 1. $\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3}) \cdots,$ where the ${p_i}$'s are distinct prime factors of n.

$\phi (1000) = 1000 \cdot (1-\frac{1}{2})(1-\frac{1}{5}) = 400.$

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