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

Revision as of 02:24, 7 June 2015

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, $s < 118$ ; 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 $s = \boxed{117}$ .

@@ Line 9: / Line 9: @@
 <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 < 118</math>; the largest possible value of <math>s</math> is <math>117</math>.
-This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>\boxed{117}</math>
+This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>116 \cdot 117</math> and <math>s = \boxed{117}</math>.
 == 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
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Difference between revisions of "1990 AIME Problems/Problem 3"

Revision as of 02:24, 7 June 2015

Problem

Solution

See also