1997 AIME Problems/Problem 6
Contents
[hide]Problem
Point is in the exterior of the regular -sided polygon , and is an equilateral triangle. What is the largest value of for which , , and are consecutive vertices of a regular polygon?
Solution 1
Let the other regular polygon have sides. Using the interior angle of a regular polygon formula, we have , , and . Since those three angles add up to ,
Using SFFT,
Clearly is maximized when .
Solution 2
As above, find that using the formula for the interior angle of a polygon.
Solve for to find that . Clearly, for to be positive.
With this restriction of , the larger gets, the smaller the fraction becomes. This can be proven either by calculus, by noting that is a transformed hyperbola, or by dividing out the rational function to get
Either way, minimizng will maximize , and the smallest integer such that is positive is , giving
Solution 3
From the formula for the measure for an individual angle of a regular n-gon, , the measure of . Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of (Notice that this value decreases as increases; hence, we are looking for the least possible value of ). For to be vertices of a regular polygon, must be of the form , where is a natural number greater than or equal to 3. It is obvious that . The least angle satisfying this condition is . Equating this with and solving yields
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.