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1997 AIME Problems/Problem 6: Difference between revisions

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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?
[[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 1==
[[Image:1997_AIME-6.png]]
[[Image:1997_AIME-6.png]]


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\end{eqnarray*}</cmath>
\end{eqnarray*}</cmath>
Clearly <math>n</math> is maximized when <math>m = 7, n = \boxed{42}</math>.
Clearly <math>n</math> is maximized when <math>m = 7, n = \boxed{42}</math>.
== Solution 2 ==
As above, find that <math>mn - 6m - 6n = 0</math> using the formula for the interior angle of a polygon. 
Solve for <math>n</math> to find that <math>n = \frac{6m}{m-6}</math>.  Clearly, <math>m>6</math> for <math>n</math> to be positive.
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>


== See also ==
== See also ==

Revision as of 16:48, 10 June 2011

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

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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 $m = 7, n = \boxed{42}$.

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 $n = \boxed{42}$

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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