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 (s-118)(r+116) = -118 \cdot 116.</math> Since <math>r+116</math> is always positive, we know | + | 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 that <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> |
− | == Video Solution | + | == Video Solution == |
https://www.youtube.com/watch?v=9YwQlFAJqvc | https://www.youtube.com/watch?v=9YwQlFAJqvc |
Latest revision as of 19:18, 16 April 2024
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 This means that
Using SFFT we obtain
Since
is always positive, we know that
must be negative. Therefore the maximum value of
must be
which indeed yields an integral value of
Video Solution
https://www.youtube.com/watch?v=9YwQlFAJqvc
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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