1990 AIME Problems/Problem 3
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[hide]Problem
Let be a regular and be a regular such that each interior angle of is as large as each interior angle of . What's the largest possible value of ?
Solution 1
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 .
and , making the numerator of the fraction positive. To make the denominator positive, ; the largest possible value of is .
This is achievable because the denominator is , making a positive number and .
Solution 2
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.
-jackshi2006
See also
1990 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 4 | |
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