1997 AIME Problems/Problem 6

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

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 $n = \boxed{42}$

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1997 AIME Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See also