Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 52: Line 52:
  
 
-jackshi2006
 
-jackshi2006
 +
 +
 +
== Solution 3 ==
 +
 +
As in above, we have
 +
 +
== 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 ==

Revision as of 20:52, 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

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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&oldid=133902"