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.

$s * 116 = 118 * 116 - 116$

$s = 118 - 1$

$s = \boxed{117}$

-jackshi2006

Solution 3

As in above, we have $rs = 118r - 116s.$ This means that $rs + 116s - 118r = 0.$ Using SFFT we obtain $s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.$ Since $r+116$ is always positive, we know thta $s-118$ must be negative. Therefore the maximum value of $s$ must be $\boxed{117}$ which indeed yields an integral value of $r.$

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1990 AIME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

Solution 3

See also