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

 
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== Solution ==
 
== Solution ==
 +
Uses PIE (principle of inclusion-exclusion).
 +
 +
If we join the adjacent vertices of the  regular <math>n</math>-star, we get a regular <math>n</math>-gon. We number the vertices of this <math>n</math>-gon in a counterclockwise direction:
 +
<math>0, 1, 2, 3, \ldots, n-1.</math>
 +
 +
A regular <math>n</math>-star will be formed if we choose a vertex number <math>m</math>, where <math>0 \le m \le n-1</math>, and then form the line segments by joining the following pairs of vertex numbers:
 +
<math>(0 \mod{n}, m \mod{n}),</math>
 +
<math>(m \mod{n}, 2m \mod{n}),</math>
 +
<math>(2m \mod{n}, 3m \mod{n}),</math>
 +
<math>\cdots,</math>
 +
<math>((n-2)m \mod{n}, (n-1)m \mod{n}),</math>
 +
<math>((n-1)m \mod{n}, 0 \mod{n}).</math>
 +
 +
If <math>\gcd(m,n) > 1</math>, then the star degenerates into a regular <math>\frac{n}{\gcd(m,n)}</math>-gon or a (2-vertex) line segment if
 +
<math>\frac{n}{\gcd(m,n)}= 2</math>. Therefore, we need to find all <math>m</math> such that <math>\gcd(m,n) = 1</math>.
 +
 +
Note that <math>n = 1000 = 2^{3}5^{3}.</math>
 +
 +
Let <math>S = \{1,2,3,\ldots, 1000\}</math>, and <math>A_{i}= \{i \in S \mid i \textrm{ divides }1000\}</math>. The number of <math>m</math>'s that are not relatively prime to <math>1000</math> is:
 +
<math>\mid A_{2}\cup A_{5}\mid = \mid A_{2}\mid+\mid A_{5}\mid-\mid A_{2}\cap A_{5}\mid</math>
 +
<math>= \left\lfloor \frac{1000}{2}\right\rfloor+\left\lfloor \frac{1000}{5}\right\rfloor-\left\lfloor \frac{1000}{2 \cdot 5}\right\rfloor</math>
 +
<math>= 500+200-100 = 600.</math>
 +
 +
Vertex number <math>1</math> and <math>n-1=999</math> must be excluded 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) gives the same star. Therefore we should half the count to get non-similar stars.
 +
 +
Number of non-similar 1000-pointed stars
 +
<math>= \frac{1000-600-2}{2}= 199.</math>
  
 
== See also ==
 
== See also ==
 
* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]

Revision as of 21:46, 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 number $1$ and $n-1=999$ must be excluded 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) gives 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

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