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

## 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 1

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 $s = \boxed{117}$.

## Solution 2

Like above, use the formula for the interior angles of a regular sided polygon. $\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}$ $59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s$ $59 * (rs - 2r) = 58 * (rs - 2s)$ $rs - 118r = -116s$ $rs = 118r-116s$

This equation tells us $s$ divides $118r$. If $s$ 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 $s=59$, which does give a solution: $s=59, r=116$. Although, the problem asks for $s$, not $r$. The only conceivable reasoning behind this is that $r$ is greater than 1000. This prompts us to look into the second case, where $s$ divides $r$. Make $r = s * k$. Rewrite the equation using this new information. $s * s * k = 118 * s * k - 116 * s$ $s * k = 118 * k - 116$

Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum, 116. $s * 116 = 118 * 116 - 116$ $s = 118 - 1$ $s = \boxed{117}$

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