Difference between revisions of "1990 AIME Problems/Problem 3"

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<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>.
 
<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>
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This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>116 \cdot 117</math> and <math>s = \boxed{117}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 03:24, 7 June 2015

Problem

Let $P_1^{}$ be a regular $r~\mbox{gon}$ and $P_2^{}$ be a regular $s~\mbox{gon}$ $(r\geq s\geq 3)$ such that each interior angle of $P_1^{}$ is $\frac{59}{58}$ as large as each interior angle of $P_2^{}$. What's the largest possible value of $s_{}^{}$?

Solution

The formula for the interior angle of a regular sided polygon is $\frac{(n-2)180}{n}$.

Thus, $\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}$. Cross multiplying and simplifying, we get $\frac{58(r-2)}{r} = \frac{59(s-2)}{s}$. Cross multiply and combine like terms again to yield $58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs$. Solving for $r$, we get $r = \frac{116s}{118 - s}$.

$r \ge 0$ and $s \ge 0$, making the numerator of the fraction positive. To make the denominator positive, $s < 118$; the largest possible value of $s$ is $117$.

This is achievable because the denominator is $1$, making $r$ a positive number $116 \cdot 117$ and $s = \boxed{117}$.

See also

1990 AIME (ProblemsAnswer KeyResources)
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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