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

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* each of the <math> n </math> line segments intersects at least one of the other line segments at a point other than an endpoint,
 
* each of the <math> n </math> line segments intersects at least one of the other line segments at a point other than an endpoint,
 
* all of the angles at <math> P_1, P_2,\ldots, P_n </math> are congruent,
 
* all of the angles at <math> P_1, P_2,\ldots, P_n </math> are congruent,
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* all of the <math> n </math> line segments <math> P_2P_3,\ldots, P_nP_1 </math> are congruent, and
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* all of the <math> n </math> line segments <math>P_1P_2, P_2P_3,\ldots, P_nP_1 </math> are congruent, and
 
* the path <math> P_1P_2, P_2P_3,\ldots, P_nP_1 </math> turns counterclockwise at an angle of less than 180 degrees at each vertex.
 
* the path <math> P_1P_2, P_2P_3,\ldots, P_nP_1 </math> turns counterclockwise at an angle of less than 180 degrees at each vertex.
  
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== Solution ==
 
== Solution ==
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Uses PIE (principle of inclusion-exclusion).
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We use the [[Principle of Inclusion-Exclusion]] (PIE).
  
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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:
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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>
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<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:
 
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:
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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.
 
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.
  
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Number of non-similar 1000-pointed stars
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Therefore, the number of non-similar 1000-pointed stars is <math>\frac{1000-600-2}{2}= \boxed{199}.</math>
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<math>= \frac{1000-600-2}{2}= 199.</math>
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----
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Note that in general, the number of <math>n</math>-pointed stars is given by <math>\frac{\phi(n)}{2} - 1</math> (dividing by <math>2</math> to remove the reflectional symmetry, subtracting <math>1</math> to get rid of the <math>1</math>-step case), where <math>\phi(n)</math> is the [[Euler's totient function]]. It is well-known that <math>\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)</math>, where <math>p_1,\,p_2,\ldots,\,p_n</math> are the distinct prime factors of <math>n</math>. Thus <math>\phi(1000) = 1000\left(1 - \frac 12\right)\left(1 - \frac 15\right) = 400</math>, and the answer is <math>\frac{400}{2} - 1 = 199</math>.
  
 
== See also ==
 
== See also ==
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* [[2004 AIME I Problems/Problem 7| Previous problem]]
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{{AIME box|year=2004|n=I|num-b=7|num-a=9}}
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* [[2004 AIME I Problems/Problem 9| Next problem]]
 
  
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* [[2004 AIME I Problems]]
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[[Category:Intermediate Combinatorics Problems]]
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 20:26, 5 June 2021

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 $P_1P_2, P_2P_3,\ldots, P_nP_1$ 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

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 $\frac{1000-600-2}{2}= \boxed{199}.$

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 $\frac{400}{2} - 1 = 199$.

See also

2004 AIME I (Problems • Answer Key • Resources)
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