1982 AMC 12 Problem 10

Question

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

30

33

36

39

42

Because

O

is the incenter, its distance from

BC

is the inradius

Compare the altitude of

\triangle AMN

with the altitude of

\triangle ABC

The sides

12,

18,

24

have semiperimeter

27

and area

\sqrt{27\cdot15\cdot9\cdot3}=27\sqrt{15},

so the inradius is

r=\sqrt{15}.

Taking

BC

as base, the altitude is

h=\frac{2(27\sqrt{15})}{24}=\frac{9\sqrt{15}}{4}.

Thus the similarity scale from

\triangle ABC

to

\triangle AMN

is

1-\frac{r}{h}=1-\frac{4}{9}=\frac{5}{9}.

Its perimeter is

(\frac{5}{9})(12+18+24)=30.

Therefore, the correct answer is A.

Problem 10 in other years