AMC10 2017 B

AMC10 2017 B · Q19

AMC10 2017 B · Q19. It mainly tests Triangles (properties), Area & perimeter.

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 ?

设ABC为正三角形。将边AB向B外延至点B′，使BB′=3AB。类似地，将边BC向C外延至点C′，使CC′=3BC，将边CA向A外延至点A′，使AA′=3CA。△A′B′C′的面积与△ABC的面积之比是多少？

(A) 9 : 1 9 : 1

(B) 16 : 1 16 : 1

(D) 36 : 1 36 : 1

(E) 37 : 1 37 : 1

Answer

Correct choice: (E)

正确答案：（E）

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.

画线段CB′、AC′和BA′。设△ABC面积为X。因为△BB′C底边是3倍且高度相同，其面积为3X。类似地，△AA′B和△CC′A面积也为3X。此外，△AA′C′底边是△ACC′的3倍且高度相同，其面积为9X。△CC′B′和△BB′A′面积也为9X。因此△A′B′C′面积为X+3(3X)+3(9X)=37X，所求比为37:1。

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