Question

在边长均为整数的 $\triangle ABC$ 中，

$\begin{aligned} \cos A &= \dfrac{11}{16}, \\ \cos B &= \dfrac{7}{8}, \\ \cos C &= -\dfrac{1}{4}. \end{aligned}$

$\triangle ABC$ 的最小可能周长是多少？

In $\triangle ABC$ with integer side lengths,

$\begin{aligned} \cos A &= \dfrac{11}{16}, \\ \cos B &= \dfrac{7}{8}, \\ \cos C &= -\dfrac{1}{4}. \end{aligned}$

What is the least possible perimeter for $\triangle ABC?$

$9$

$12$

$23$

$27$

$44$

每个正弦值为 $\sqrt{1 - \cos^2}$ ： $\sin A = \dfrac{3\sqrt{15}}{16}$ ， $\sin B = \dfrac{2\sqrt{15}}{16}$ ， $\sin C = \dfrac{4\sqrt{15}}{16}$ 。

由正弦定理，边长比为 $3 : 2 : 4$ 。最小的整数边长为 $3, 2, 4$ ，它们满足三角形不等式。

最小周长为 $3 + 2 + 4 = 9$ 。

所以正确答案是 A。

Each sine is $\sqrt{1 - \cos^2}:$ $\sin A = \dfrac{3\sqrt{15}}{16},$ $\sin B = \dfrac{2\sqrt{15}}{16},$ $\sin C = \dfrac{4\sqrt{15}}{16}.$

By the Law of Sines the sides are in ratio $3 : 2 : 4.$ The smallest integer sides are $3, 2, 4,$ which satisfy the triangle inequality.

The least perimeter is $3 + 2 + 4 = 9.$

Thus, the correct answer is A.