1988 AMC 12 Problem 16

Question

ABC

and

A'B'C'

are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side

BC

and side

B'C'

is

\frac16

the altitude of

\triangle ABC.

The ratio of the area of

\triangle A'B'C'

to the area of

\triangle ABC

is

\frac1{36}

\frac16

\frac14

\frac{\sqrt3}{4}

\frac{9+8\sqrt3}{36}

A centroid lies one-third of the altitude above the base

Relate the two altitudes using the distance between their parallel bases

Let the outer and inner altitudes be

h

and

h'.

Their common center is

\frac{h}{3}

above

BC

and

\frac{h'}{3}

above

B'C'.

Hence

\frac{h-h'}{3}=\frac{h}{6},

so

h'=\frac{h}{2}.

Areas scale as the square of corresponding lengths, giving

(\frac{h'}{h})^2=\frac{1}{4}.

Thus the correct answer is C.

Problem 16 in other years