1990 AIME Problems/Problem 3

Revision as of 03:24, 7 June 2015 by Jabberwock2 (talk | contribs) (Solution)

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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