AMC8 2005

AMC8 2005 · Q21

AMC8 2005 · Q21. It mainly tests Basic counting (rules of product/sum), Counting in geometry (lattice points).

How many distinct triangles can be drawn using three of the dots below as vertices?

使用以下点中的三个作为顶点，可以画出多少个不同的三角形？

(A) 9 9

(B) 12 12

(D) 20 20

(E) 24 24

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) To make a triangle, select as vertices two dots from one row and one from the other row. To select two dots in the top row, decide which dot is not used. This can be done in three ways. There are also three ways to choose one dot to use from the bottom row. So there are $3 \times 3 = 9$ triangles with two vertices in the top row and one in the bottom. Similarly, there are nine triangles with one vertex in the top row and two in the bottom. This gives a total of $9 + 9 = 18$ triangles. Note: Can you find the four noncongruent triangles?

（C）要组成一个三角形，从一行中选取两个点作为顶点，并从另一行中选取一个点作为顶点。要在上面一行中选两个点，只需决定哪个点不被使用，这有三种方式。下面一行中选择一个点也有三种方式。因此，上面一行取两个顶点、下面一行取一个顶点的三角形共有 $3 \times 3 = 9$ 个。同理，上面一行取一个顶点、下面一行取两个顶点的三角形也有 9 个。总数为 $9 + 9 = 18$ 个三角形。注：你能找出四个互不全等的三角形吗？

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