AMC12 2003 A

AMC12 2003 A · Q7

AMC12 2003 A · Q7. It mainly tests Triangles (properties), Area & perimeter.

How many non-congruent triangles with perimeter $7$ have integer side lengths?

周长为 $7$ 的非全等整数边三角形有多少个？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (B)

正确答案：（B）

Solution

By the triangle inequality, no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$. Since all sides must be integers, the largest possible length of a side is $3$. Therefore, all such triangles must have all sides of length $1$, $2$, or $3$. Since $2+2+2=6<7$, at least one side must have a length of $3$. Thus, the remaining two sides have a combined length of $7-3=4$. So, the remaining sides must be either $3$ and $1$ or $2$ and $2$. Therefore, the number of triangles is $\boxed{\mathrm{(B)}\ 2}$.

由三角形不等式，任一边长都不能大于半周长，即 $\frac{1}{2}\cdot7=3.5$。由于三边都必须是整数，最大可能边长为 $3$。因此所有这样的三角形的边长只能是 $1$、$2$ 或 $3$。因为 $2+2+2=6<7$，至少有一条边长为 $3$。于是另外两边之和为 $7-3=4$，所以另外两边只能是 $3$ 和 $1$ 或 $2$ 和 $2$。因此三角形的个数为 $\boxed{\mathrm{(B)}\ 2}$。

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