Skip to main content

2017 AMC 12B Problem 15

Problem 15 of 25IntermediateGeometry

Let ABCABC be an equilateral triangle. Extend side AB‾\overline{AB} beyond BB to a point B′B' so that BB′=3⋅AB.BB' = 3 \cdot AB. Similarly, extend side BC‾\overline{BC} beyond CC to a point C′C' so that CC′=3⋅BC,CC' = 3 \cdot BC, and extend side CA‾\overline{CA} beyond AA to a point A′A' so that AA′=3⋅CA.AA' = 3 \cdot CA. What is the ratio of the area of △A′B′C′\triangle A'B'C' to the area of △ABC?\triangle ABC?

Answer choices

Show solution

Solution

Let X=[△ABC],X = [\triangle ABC], and draw segments CB′,CB', AC′,AC', and BA′.BA'. Triangle BB′CBB'C has base BB′=3⋅ABBB' = 3 \cdot AB and the same altitude as △ABC\triangle ABC from CC to line AB,AB, so its area is 3X;3X; likewise △CC′A\triangle CC'A and △AA′B\triangle AA'B each have area 3X.3X. Next, △AA′C′\triangle AA'C' has 33 times the base and the same height as △ACC′,\triangle ACC', so its area is 9X;9X; similarly △CC′B′\triangle CC'B' and △BB′A′\triangle BB'A' each have area 9X.9X. Thus [△A′B′C′]=X+3(3X)+3(9X)=37X, \begin{aligned} &[\triangle A'B'C'] = X \\ &\quad {}+ 3(3X) + 3(9X) \\ &\quad {}= 37X, \end{aligned} so the ratio is 37:1.37 : 1. Thus, the correct answer is E.
AoPS wiki

Tagged: area ratio · area decomposition · equilateral triangle

More practice