1982 AHSME Problems/Problem 11

Revision as of 13:48, 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<...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 11 Solution

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

Invalid username
Login to AoPS