AMC12 2010 A

AMC12 2010 A · Q14

AMC12 2010 A · Q14. It mainly tests Triangles (properties).

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?

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

(A) 30 30

(B) 33 33

(D) 36 36

(E) 37 37

Answer

Correct choice: (B)

正确答案：（B）

Solution

By the Angle Bisector Theorem, we know that $\frac{AB}{BC} = \frac{3}{8}$. If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$, respectively), then $AB + BC = AD + DC = AC$, contradicting the Triangle Inequality. If we use the next lowest values ($AB = 6$ and $BC = 16$), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$, or choice $\textbf{(B)}$.

由角平分线定理可知 $\frac{AB}{BC} = \frac{3}{8}$。如果取 $AB$ 和 $BC$ 的最小整数值（分别为 $AD$ 和 $DC$ 的长度），则 $AB + BC = AD + DC = AC$，与三角形不等式矛盾。若取下一组最小值（$AB = 6$ 且 $BC = 16$），则满足三角形不等式。因此答案为 $6 + 16 + 3 + 8 = \boxed{33}$，对应选项 $\textbf{(B)}$。

Topics

GE.002 Triangles (properties)