Difference between revisions of "1990 AIME Problems/Problem 3"
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== Solution 3 == | == Solution 3 == | ||
− | As in above, we have <math>rs = 118r - 116s.</math> This means that <math>rs + 116s - 118r = 0.</math> Using SFFT we obtain <math>s(r+116) - 118(r+116) = -118 \cdot 116 \implies | + | As in above, we have <math>rs = 118r - 116s.</math> This means that <math>rs + 116s - 118r = 0.</math> Using SFFT we obtain <math>s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.</math> Since <math>r+116</math> is always positive, we know thta <math>s-118</math> must be negative. Therefore the maximum value of <math>s</math> must be <math>\boxed{117}</math> which indeed yields an integral value of <math>r.</math> |
== See also == | == See also == |
Revision as of 21:53, 20 September 2020
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
Solution 3
As in above, we have
Solution 3
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
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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