AIME 1997 · 第 6 题

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 $360^{\circ}$,

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

$$\begin{aligned} (m-6)(n-6) &=& 36 \end{aligned}$$ 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 $n = \boxed{042}$