Difference between revisions of "1997 AIME Problems/Problem 6"
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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{42}</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{42}</math> | ||
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+ | == Solution 3 == | ||
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+ | 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{42}</math> | ||
== See also == | == See also == |
Revision as of 02:43, 29 June 2011
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 |