AMC10 2013 A

AMC10 2013 A · Q16

AMC10 2013 A · Q16. It mainly tests Area & perimeter, Coordinate geometry.

A triangle with vertices $(6, 5)$, $(8, -3)$, and $(9, 1)$ is reflected about the line $x = 8$ to create a second triangle. What is the area of the union of the two triangles?

有一个顶点为$(6, 5)$、$(8, -3)$和$(9, 1)$的三角形，关于直线$x = 8$反射得到第二个三角形。两个三角形的并集面积是多少？

(A) 9 9

(B) $\frac{28}{3}$ $\frac{28}{3}$

(D) $\frac{31}{3}$ $\frac{31}{3}$

(E) $\frac{32}{3}$ $\frac{32}{3}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The reflected triangle has vertices $(7,1)$, $(8,-3)$, and $(10,5)$. The point $(9,1)$ is on the line segment from $(10,5)$ to $(8,-3)$. The line segment from $(6,5)$ to $(9,1)$ contains the point $\left(8,\frac{7}{3}\right)$, which must be on both triangles, and by symmetry the point $(7,1)$ is on the line segment from $(6,5)$ to $(8,-3)$. Therefore the union of the two triangles is also the union of two congruent triangles with disjoint interiors, each having the line segment from $(8,-3)$ to $\left(8,\frac{7}{3}\right)$ as a base. The altitude of one of the two triangles is the distance from the line $x=8$ to the point $(10,5)$, which is $2$. Hence the union of the triangles has area $2\cdot\left(\frac{1}{2}\cdot 2\cdot\left(\frac{7}{3}+3\right)\right)=\frac{32}{3}$.

答案（E）：反射后的三角形的顶点为 $(7,1)$、$(8,-3)$ 和 $(10,5)$。点 $(9,1)$ 在线段 $(10,5)$ 到 $(8,-3)$ 上。线段 $(6,5)$ 到 $(9,1)$ 包含点 $\left(8,\frac{7}{3}\right)$，该点必须同时在两个三角形上；并且由于对称性，点 $(7,1)$ 在线段 $(6,5)$ 到 $(8,-3)$ 上。因此，这两个三角形的并集也等于两个全等且内部不相交的三角形的并集，每个三角形都以从 $(8,-3)$ 到 $\left(8,\frac{7}{3}\right)$ 的线段为底边。两个三角形中任意一个的高等于直线 $x=8$ 到点 $(10,5)$ 的距离，即 $2$。因此并集面积为 $2\cdot\left(\frac{1}{2}\cdot 2\cdot\left(\frac{7}{3}+3\right)\right)=\frac{32}{3}$。

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