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

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== Problem ==
 
== Problem ==
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Let <math>P_1^{}</math> be a regular <math>n~\mbox{gon}</math> and <math>P_2^{}</math> be a regular <math>s~\mbox{gon}</math> <math>(r\geq s\geq 3)</math> such that each interior angle of <math>P_1^{}</math> is <math>\frac{59}{58}</math> as large as each interior angle of <math>P_2^{}</math>. What's the largest possible value of <math>s_{}^{}</math>?
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Let <math>P_1^{}</math> be a [[Regular polygon|regular]] <math>r~\mbox{gon}</math> and <math>P_2^{}</math> be a regular <math>s~\mbox{gon}</math> <math>(r\geq s\geq 3)</math> such that each [[interior angle]] of <math>P_1^{}</math> is <math>\frac{59}{58}</math> as large as each interior angle of <math>P_2^{}</math>. What's the largest possible value of <math>s_{}^{}</math>?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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The formula for the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>.
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Thus, <math>\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}</math>. Cross multiplying and simplifying, we get <math>\frac{58(r-2)}{r} = \frac{59(s-2}{s}</math>. Cross multiply and combine like terms again to yield <math>58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs</math>. Solving for <math>r</math>, we get <math>r = \frac{116s}{118 - s}</math>.
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<math>r \ge 0</math> and <math>s \ge 0</math>, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], <math>s \le 118</math>; the largest possible value of <math>s</math> is <math>117</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=2|num-a=4}}
 
{{AIME box|year=1990|num-b=2|num-a=4}}

Revision as of 20:32, 2 March 2007

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, $s \le 118$; the largest possible value of $s$ is $117$.

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
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