Problem:
Let ABC be an equilateral triangle. Extend side AB beyond B to a point B′ so that BB′=3AB. Similarly, extend side BC beyond C to a point C′ so that CC′=3BC, and extend side CA beyond A to a point A′ so that AA′=3CA. What is the ratio of the area of △A′B′C′ to the area of △ABC ?
Answer Choices:
A. 9:1
B. 16:1
C. 25:1
D. 36:1
E. 37:1 Solution:
Draw segments CB′,AC′, and BA′. Let X be the area of △ABC. Because △BB′C has a base 3 times as long and the same altitude, its area is 3X. Similarly, the areas of △AA′B and △CC′A are also 3X. Furthermore, △AA′C′ has 3 times the base and the same height as △ACC′, so its area is 9X. The areas of △CC′B′ and △BB′A′ are also 9X by the same reasoning. Therefore the area of △A′B′C′ is X+3(3X)+3(9X)=37X, and the requested ratio is 37:1. Note that nothing in this argument requires △ABC to be equilateral.
OR
Let s=AB. Applying the Law of Cosines to △B′BC′ gives
