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

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== Problem 10 Solution ==
 
== Problem 10 Solution ==
  
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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>. <math>AM</math> + <math>MB</math> = 12, and <math>AN</math> + <math>NC</math> = 18. So, <math>AM</math> + <math>MO</math> = 12, and <math>AN</math> + <math>NO</math> = 18, and those are all of the lengths that make up <math>\triangle AMN</math>. Therefore, the perimeter of <math>\triangle AMN</math> is <math>12 + 18 = 30</math>.
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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>. <math>AM</math> + <math>MB</math> = 12, and <math>AN</math> + <math>NC</math> = 18. So, <math>AM</math> + <math>MO</math> = 12, and <math>AN</math> + <math>NO</math> = 18, and those are all of the lengths that make up <math>\triangle AMN</math>. Therefore, the perimeter of <math>\triangle AMN</math> is <math>12 + 18 = 30</math>. The answer is <math>\boxed{A}</math>.

Revision as of 23:24, 1 March 2023

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 $NO = NC$. $AM$ + $MB$ = 12, and $AN$ + $NC$ = 18. So, $AM$ + $MO$ = 12, and $AN$ + $NO$ = 18, and those are all of the lengths that make up $\triangle AMN$. Therefore, the perimeter of $\triangle AMN$ is $12 + 18 = 30$. The answer is $\boxed{A}$.

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