2017 AMC 10B Problem 19

Problem 19 of 25HarderGeometry

Let

ABC

be an equilateral triangle. Extend side

\overline{AB}

beyond

B

to a point

B'

so that

BB'=3 \cdot AB.

Similarly, extend side

\overline{BC}

beyond

C

to a point

C'

so that

CC'=3 \cdot BC,

and extend side

\overline{CA}

beyond

A

to a point

A'

so that

AA'=3 \cdot CA.

What is the ratio of the area of

\triangle A'B'C'

to the area of

\triangle ABC?

Show solution

Solution

Let

X

be the area of

\triangle ABC.

Each of

\triangle BB'C,

\triangle CC'A,

and

\triangle AA'B

has a base three times as long as a side of

\triangle ABC

and the same corresponding altitude. Each therefore has area

3X.

Next,

\triangle AA'C'

has three times the base and the same altitude as

\triangle ACC',

whose area is

3X.

Thus

\triangle AA'C'

has area

9X.

Similarly,

\triangle CC'B'

and

\triangle BB'A'

each have area

9X.

These seven regions partition the large triangle, so

\begin{aligned}[A'B'C']&=X+3(3X)+3(9X)\\&=37X.\end{aligned}

The requested ratio is

37:1.

Thus, the correct answer is E .