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Difference between revisions of "1982 AHSME Problems/Problem 10"

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== Problem 11 Solution ==
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== Problem 10 Solution ==
  
 
Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\angle MBO = \angle OBC</math> and similarly <math>\angle NCO = \angle OCB</math>. Because <math>MN</math> and <math>BC</math> are parallel, <math>\angle OBC = \angle MOB</math> and <math>\angle NOC = \angle OCB</math> by corresponding angles. This relation makes <math>\triangle MOB</math> and <math>\triangle NOC</math> isosceles. This makes <math>MB = MO</math> and <math>NO = NC</math>. Therefore the perimeter of <math>\triangle AMN</math> is <math>12 + 18 = 30</math>.
 
Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\angle MBO = \angle OBC</math> and similarly <math>\angle NCO = \angle OCB</math>. Because <math>MN</math> and <math>BC</math> are parallel, <math>\angle OBC = \angle MOB</math> and <math>\angle NOC = \angle OCB</math> by corresponding angles. This relation makes <math>\triangle MOB</math> and <math>\triangle NOC</math> isosceles. This makes <math>MB = MO</math> and <math>NO = NC</math>. Therefore the perimeter of <math>\triangle AMN</math> is <math>12 + 18 = 30</math>.

Revision as of 11:43, 11 October 2020

Problem 10 Solution

Since $BO$ and $CO$ are angle bisectors of angles $B$ and $C$ respectively, $\angle MBO = \angle OBC$ and similarly $\angle NCO = \angle OCB$. Because $MN$ and $BC$ are parallel, $\angle OBC = \angle MOB$ and $\angle NOC = \angle OCB$ by corresponding angles. This relation makes $\triangle MOB$ and $\triangle NOC$ isosceles. This makes $MB = MO$ and $NO = NC$. Therefore the perimeter of $\triangle AMN$ is $12 + 18 = 30$.

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