AMC12 2019 A

AMC12 2019 A · Q19

AMC12 2019 A · Q19. It mainly tests Triangles (properties), Trigonometry (basic).

In $\triangle ABC$ with integer side lengths, $$\cos A = \frac{11}{16}, \quad \cos B = \frac{7}{8}, \quad \text{and} \quad \cos C = -\frac{1}{4}.$$ What is the least possible perimeter for $\triangle ABC$?

在具有整数边长的$\triangle ABC$中，$$\cos A = \frac{11}{16}, \quad \cos B = \frac{7}{8}, \quad \text{and} \quad \cos C = -\frac{1}{4}.$$$\triangle ABC$的最小可能周长是多少？

(A) 9 9

(B) 12 12

(D) 27 27

(E) 44 44

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The sines of the angles can be calculated using the Pythagorean Identity: $\sin A=\frac{3\sqrt{15}}{16},\quad \sin B=\frac{\sqrt{15}}{8}=\frac{2\sqrt{15}}{16},\quad \sin C=\frac{\sqrt{15}}{4}=\frac{4\sqrt{15}}{16}.$ The ratio of these values is $3:2:4$. By the Law of Sines, the side lengths are in the same ratio. The smallest triangle having this ratio has sides of lengths $3,2,$ and $4$; the requested least possible perimeter is $9$.

答案（A）：角的正弦值可以用勾股恒等式计算： $\sin A=\frac{3\sqrt{15}}{16},\quad \sin B=\frac{\sqrt{15}}{8}=\frac{2\sqrt{15}}{16},\quad \sin C=\frac{\sqrt{15}}{4}=\frac{4\sqrt{15}}{16}.$ 这些数值的比为 $3:2:4$。由正弦定理可知，边长之比相同。满足该比例的最小三角形边长为 $3、2、4$；所求的最小周长为 $9$。

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