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Difference between revisions of "1997 AIME Problems/Problem 6"
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(m-6)(n-6) &=& 36
 
(m-6)(n-6) &=& 36
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
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Clearly <math>n</math> is maximized when <math>m = 7, n = \boxed{42}</math>.
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Clearly <math>n</math> is maximized when <math>m = 7, n = \boxed{042}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
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With this restriction of <math>m>6</math>, the larger <math>m</math> gets, the smaller the fraction <math>\frac{6m}{m-6}</math> becomes.  This can be proven either by calculus, by noting that <math>n = \frac{6m}{m-6}</math> is a  transformed hyperbola, or by dividing out the rational function to get <math>n = 6 + \frac{36}{m - 6}.</math>
 
With this restriction of <math>m>6</math>, the larger <math>m</math> gets, the smaller the fraction <math>\frac{6m}{m-6}</math> becomes.  This can be proven either by calculus, by noting that <math>n = \frac{6m}{m-6}</math> is a  transformed hyperbola, or by dividing out the rational function to get <math>n = 6 + \frac{36}{m - 6}.</math>
  
−
Either way, minimizng <math>m</math> will maximize <math>n</math>, and the smallest integer <math>m</math> such that <math>n</math> is positive is <math>m=7</math>, giving <math>n = \boxed{42}</math>
+
Either way, minimizng <math>m</math> will maximize <math>n</math>, and the smallest integer <math>m</math> such that <math>n</math> is positive is <math>m=7</math>, giving <math>n = \boxed{042}</math>
  
 
== Solution 3 ==
 
== Solution 3 ==
  
−
From the formula for the measure for an individual angle of a regular n-gon, <math>180 - \frac{360}{n}</math>, the measure of <math>\angle A_2A_1A_n = 180 - \frac{360}{n}</math>. Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of <math>\angle A_nA_1B = 120 + \frac{360}{n}</math> (Notice that this value decreases as <math>n</math> increases; hence, we are looking for the least possible value of <math>\angle A_nA_1B</math>). For <math>A_n, A_1, B</math> to be vertices of a regular polygon, <math>\angle A_nA_1B</math> must be of the form <math>180 - \frac{360}{n}</math>, where <math>n</math> is a natural number greater than or equal to 3. It is obvious that <math>\angle A_nA_1B > 120</math>. The least angle satisfying this condition is <math>180 - \frac{360}{7}</math>. Equating this with <math>120 + \frac{360}{n}</math> and solving yields <math>n = \boxed{42}</math>
+
From the formula for the measure for an individual angle of a regular n-gon, <math>180 - \frac{360}{n}</math>, the measure of <math>\angle A_2A_1A_n = 180 - \frac{360}{n}</math>. Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of <math>\angle A_nA_1B = 120 + \frac{360}{n}</math> (Notice that this value decreases as <math>n</math> increases; hence, we are looking for the least possible value of <math>\angle A_nA_1B</math>). For <math>A_n, A_1, B</math> to be vertices of a regular polygon, <math>\angle A_nA_1B</math> must be of the form <math>180 - \frac{360}{n}</math>, where <math>n</math> is a natural number greater than or equal to 3. It is obvious that <math>\angle A_nA_1B > 120</math>. The least angle satisfying this condition is <math>180 - \frac{360}{7}</math>. Equating this with <math>120 + \frac{360}{n}</math> and solving yields <math>n = \boxed{042}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 23:39, 21 December 2018

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 1

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 $m = 7, n = \boxed{042}$.

Solution 2

As above, find that $mn - 6m - 6n = 0$ using the formula for the interior angle of a polygon.

Solve for $n$ to find that $n = \frac{6m}{m-6}$. Clearly, $m>6$ for $n$ to be positive.

With this restriction of $m>6$, the larger $m$ gets, the smaller the fraction $\frac{6m}{m-6}$ becomes. This can be proven either by calculus, by noting that $n = \frac{6m}{m-6}$ is a transformed hyperbola, or by dividing out the rational function to get $n = 6 + \frac{36}{m - 6}.$

Either way, minimizng $m$ will maximize $n$, and the smallest integer $m$ such that $n$ is positive is $m=7$, giving $n = \boxed{042}$

Solution 3

From the formula for the measure for an individual angle of a regular n-gon, $180 - \frac{360}{n}$, the measure of $\angle A_2A_1A_n = 180 - \frac{360}{n}$. Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of $\angle A_nA_1B = 120 + \frac{360}{n}$ (Notice that this value decreases as $n$ increases; hence, we are looking for the least possible value of $\angle A_nA_1B$). For $A_n, A_1, B$ to be vertices of a regular polygon, $\angle A_nA_1B$ must be of the form $180 - \frac{360}{n}$, where $n$ is a natural number greater than or equal to 3. It is obvious that $\angle A_nA_1B > 120$. The least angle satisfying this condition is $180 - \frac{360}{7}$. Equating this with $120 + \frac{360}{n}$ and solving yields $n = \boxed{042}$

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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