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

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== Solution 3 ==
 
== Solution 3 ==
  
As in above, we have
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As in above, we have <math>rs = 118r - 116s.</math> This means that <math>rs + 116s - 118r = 0.</math> Using SFFT we obtain <math>s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.</math> Since <math>r+116</math> is always positive, we know thta <math>s-118</math> must be negative. Therefore the maximum value of <math>s</math> must be <math>\boxed{117}</math> which indeed yields an integral value of <math>r.</math>
 
 
== Solution 3 ==
 
 
 
As in above, we have <math>rs = 118r - 116s.</math> This means that <math>rs + 116s - 118r = 0.</math> Using SFFT we obtain <math>s(r+116) - 118(r+116) = -118 \cdot 116 \implies </math>(s-118)(r+116) = -118 \cdot 116.<math> Since </math>r+116<math> is always positive, we know thta </math>s-118<math> must be negative. Therefore the maximum value of </math>s<math> must be </math>\boxed{117}<math> which indeed yields an integral value of </math>r.$
 
  
 
== See also ==
 
== See also ==

Latest revision as of 21:53, 20 September 2020

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 $r.$

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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