1982 AHSME Problems/Problem 10

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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1982 AHSME Problems/Problem 10

Problem 10 Solution