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1997 AIME Problems/Problem 6

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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

Image: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 - 36... Using SFFT,

\begin{eqnarray*}(m-6)(n-6) &=& 36\end{eqnarray*} Clearly n is maximized when m = 7, n = \boxed{42}.

See also

1997 AIME (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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