Difference between revisions of "1990 AIME Problems/Problem 3"
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Thus, <math>\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}</math>. Cross multiplying and simplifying, we get <math>\frac{58(r-2)}{r} = \frac{59(s-2)}{s}</math>. Cross multiply and combine like terms again to yield <math>58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs</math>. Solving for <math>r</math>, we get <math>r = \frac{116s}{118 - s}</math>. | Thus, <math>\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}</math>. Cross multiplying and simplifying, we get <math>\frac{58(r-2)}{r} = \frac{59(s-2)}{s}</math>. Cross multiply and combine like terms again to yield <math>58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs</math>. Solving for <math>r</math>, we get <math>r = \frac{116s}{118 - s}</math>. | ||
| − | <math>r \ge 0</math> and <math>s \ge 0</math>, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], <math>s < 118</math>; the largest possible value of <math>s</math> is <math> | + | <math>r \ge 0</math> and <math>s \ge 0</math>, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], <math>s < 118</math>; the largest possible value of <math>s</math> is <math>117</math>. |
| + | |||
| + | This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>\boxed{117}</math> | ||
== See also == | == See also == | ||
Revision as of 21:52, 18 May 2015
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
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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
