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

Revision as of 20:57, 24 November 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 $117$ .

@@ Line 11: / Line 11: @@
 == See also ==
 {{AIME box|year=1990|num-b=2|num-a=4}}
+[[Category:Intermediate Geometry Problems]]
+[[Category:Intermediate Algebra Problems]]

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

Revision as of 20:57, 24 November 2007

Problem

Solution

See also