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

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

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 $n = \boxed{042}$

@@ Line 2: / Line 2: @@
 [[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]]
@@ Line 18: / Line 18: @@
 (m-6)(n-6) &=& 36
 \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{042}</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{042}</math>
+== 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{042}</math>
 == See also ==
@@ Line 25: / Line 39: @@
 [[Category:Intermediate Algebra Problems]]
 [[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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