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Difference between revisions of "1997 AIME Problems/Problem 6"

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== Problem ==
 
== Problem ==
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Point <math>B</math> is in the exterior of the regular <math>n</math>-sided polygon <math>A_1A_2\cdots A_n</math>, and <math>A_1A_2B</math> is an equilateral triangle. What is the largest value of <math>n</math> for which <math>A_1</math>, <math>A_n</math>, and <math>B</math> are consecutive vertices of a regular polygon?
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[[Point]] <math>B</math> is in the exterior of the [[regular polygon|regular]] <math>n</math>-sided polygon <math>A_1A_2\cdots A_n</math>, and <math>A_1A_2B</math> is an [[equilateral triangle]]. What is the largest value of <math>n</math> for which <math>A_1</math>, <math>A_n</math>, and <math>B</math> are consecutive vertices of a regular polygon?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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[[Image:1997_AIME-6.png]]
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Let the other regular polygon have <math>m</math> sides. Using the interior angle of a regular polygon formula, we have <math>\angle A_2A_1A_n = \frac{(n-2)180}{n}</math>, <math>\angle A_nA_1B = \frac{(m-2)180}{m}</math>, and <math>\angle A_2A_1B = 60^{\circ}</math>. Since those three angles add up to <math>360^{\circ}</math>,
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<cmath>\begin{eqnarray*}
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\frac{(n-2)180}{n} + \frac{(m-2)180}{m} &=& 300\\
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m(n-2)180 + n(m-2)180 &=& 300mn\\
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360mn - 360m - 360n &=& 300mn\\
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mn - 6m - 6n &=& 0
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\end{eqnarray*}</cmath>
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Using [[Simon's Favorite Factoring Trick|SFFT]],
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<cmath>\begin{eqnarray*}
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(m-6)(n-6) &=& 36
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\end{eqnarray*}</cmath>
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Clearly <math>n</math> is maximized when <math>m = 7, n = \boxed{42}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1997|num-b=5|num-a=7}}
 
{{AIME box|year=1997|num-b=5|num-a=7}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 22:21, 21 November 2007

Problem

Point $B$ is in the exterior of the regular $n$-sided polygon $A_1A_2\cdots A_n$, and $A_1A_2B$ is an equilateral triangle. What is the largest value of $n$ for which $A_1$, $A_n$, and $B$ are consecutive vertices of a regular polygon?

Solution

1997 AIME-6.png

Let the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\angle A_2A_1A_n = \frac{(n-2)180}{n}$, $\angle A_nA_1B = \frac{(m-2)180}{m}$, and $\angle A_2A_1B = 60^{\circ}$. Since those three angles add up to $360^{\circ}$,

\begin{eqnarray*} \frac{(n-2)180}{n} + \frac{(m-2)180}{m} &=& 300\\ m(n-2)180 + n(m-2)180 &=& 300mn\\ 360mn - 360m - 360n &=& 300mn\\ mn - 6m - 6n &=& 0 \end{eqnarray*} Using SFFT,

\begin{eqnarray*} (m-6)(n-6) &=& 36 \end{eqnarray*} Clearly $n$ is maximized when $m = 7, n = \boxed{42}$.

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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