Question

非退化 $\triangle ABC$ 的边长都是整数， $\overline{BD}$ 是角平分线， $AD = 3$ ， $DC = 8$ 。周长的最小可能值是多少？

Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3,$ and $DC = 8.$ What is the smallest possible value of the perimeter?

$30$

$33$

$35$

$36$

$37$

答案：B

知识点：角平分线定理三角不等式最优化

难度评级：1600

解答：

由角平分线定理， $\dfrac{AB}{3} = \dfrac{BC}{8}$ $AB = \dfrac{3}{8} BC.$

为使 $AB$ 和 $BC$ 都是整数， $BC$ 必须是 $8$ 的倍数。

若为了最小化周长取 $BC = 8$ 、 $AB = 3$ ，三角形会退化。

因此 $BC$ 必须取 $16$ ，此时 $AB = 6$ 。又因为 $AC = AD + DC = 11$ ，所以周长为 $16 + 6 + 11 = 33.$

所以正确答案是 B。

Using the Angle Bisector Theorem, we have that $\dfrac{AB}{3} = \dfrac{BC}{8}$ $AB = \dfrac{3}{8} BC.$

For $AB$ and $BC$ to be integers, we must have that $BC$ is a multiple of $8.$

To minimize the perimeter, we can set $BC = 8$ and $AB = 3.$ This, however, makes the triangle degenerate.

$BC$ must then be $16$ and $AB = 6.$ Since $AC = AD + DC = 11,$ the perimeter is $16 + 6 + 11 = 33.$

Thus, B is the correct answer.

2010 AMC 10A 第 16 题

答案：B

难度评级：1600

解答：

其他年份的第 16 题