2017 AMC 12B Problem 15

Question

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?

9 : 1

16 : 1

25 : 1

36 : 1

37 : 1

Answer: E

Concepts:area ratio area decomposition equilateral triangle

Difficulty rating: 1660

Small hint:

Let

X

be the area of

\triangle ABC.

Draw

CB',

AC',

BA'

to split

\triangle A'B'C'

into pieces

Big hint:

A triangle with

k

times the base and the same height has

k

times the area; find each piece as a multiple of

X

Solution:

Let

X = [\triangle ABC],

and draw segments

CB',

AC',

and

BA'.

Triangle

BB'C

has base

BB' = 3 \cdot AB

and the same altitude as

\triangle ABC

from

C

to line

AB,

so its area is

3X;

likewise

\triangle CC'A

and

\triangle AA'B

each have area

3X.

Next,

\triangle AA'C'

has

3

times the base and the same height as

\triangle ACC',

so its area is

9X;

similarly

\triangle CC'B'

and

\triangle BB'A'

each have area

9X.

Thus

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

so the ratio is

37 : 1.

Thus, the correct answer is E.

2017 AMC 12B Problem 15

Answer: E

Difficulty rating: 1660

Small hint:

Big hint:

Solution:

Problem 15 in other years