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 $\triangle ABC?$

$9 : 1$

$16 : 1$

$25 : 1$

$36 : 1$

$37 : 1$

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 $[\triangle A'B'C'] = X + 3(3X) + 3(9X) = 37X,$ so the ratio is $37 : 1.$

Thus, the correct answer is E.

2017 AMC 12B Problem 15

Solution:

Problem 15 in Other Years