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1982 AHSME Problems/Problem 11

Revision as of 12:48, 5 August 2017 by Mamis511 (talk | contribs) (Created page with "== Problem 11 Solution == Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\angleMBO = \angleOBC<...")
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Problem 11 Solution

Since $BO$ and $CO$ are angle bisectors of angles $B$ and $C$ respectively, $\angleMBO = \angleOBC$ (Error compiling LaTeX. Unknown error_msg) and similarly $\angleNCO = \angleOCB$ (Error compiling LaTeX. Unknown error_msg). Because $MN$ and $BC$ are parallel, $\angleOBC = \angleMOB$ (Error compiling LaTeX. Unknown error_msg) and $\angleNOC = \angleOCB$ (Error compiling LaTeX. Unknown error_msg) by corresponding angles. This relation makes $\bigtriangleupMOB$ (Error compiling LaTeX. Unknown error_msg) and $\bigtriangleupNOC$ (Error compiling LaTeX. Unknown error_msg) isosceles. This makes $MB = MO$ and $NO = NC$. Therefore the perimeter of $\bigtriangleupAMN$ (Error compiling LaTeX. Unknown error_msg) is $12 + 18 = 30$.

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