1982 AHSME Problems/Problem 10

Problem

In the adjoining diagram, $BO$ bisects $\angle CBA$ , $CO$ bisects $\angle ACB$ , and $MN$ is parallel to $BC$ . If $AB=12, BC=24$ , and $AC=18$ , then the perimeter of $\triangle AMN$ is

$[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0)); draw(B--M--O--B--C--O--N--C^^N--A--M); label("$A$", A, dir(90)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$M$", M, dir(90)*dir(B--A)); label("$N$", N, dir(90)*dir(A--C)); label("$O$", O, dir(90));[/asy]$

$\textbf {(A)}\ 30 \qquad \textbf {(B)}\ 33 \qquad \textbf {(C)}\ 36 \qquad \textbf {(D)}\ 39 \qquad \textbf {(E)}\ 42$

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 $\boxed{A}$ .

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1982 AHSME Problems/Problem 10

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