1990 AIME Problems/Problem 3
The formula for the interior angle of a regular sided polygon is .
Thus, . Cross multiplying and simplifying, we get . Cross multiply and combine like terms again to yield . Solving for , we get .
This is achievable because the denominator is , making a positive number and .
Like above, use the formula for the interior angles of a regular sided polygon.
This equation tells us divides . If specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is , which does give a solution: . Although, the problem asks for , not . The only conceivable reasoning behind this is that is greater than 1000. This prompts us to look into the second case, where divides . Make . Rewrite the equation using this new information.
Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.
As in above, we have This means that Using SFFT we obtain Since is always positive, we know thta must be negative. Therefore the maximum value of must be which indeed yields an integral value of
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