1990 AIME Problems/Problem 3
Contents
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) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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