AMC12 2023 B

AMC12 2023 B · Q17

AMC12 2023 B · Q17. It mainly tests Triangles (properties), Trigonometry (basic).

Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\circ,$ what is the area of $ABC$?

三角形$ABC$的三边长成等差数列，最小边长为$6$。若三角形有一个$120^\circ$的角，求$ABC$的面积。

(A) 12\sqrt{3} 12\sqrt{3}

(B) 8\sqrt{6} 8\sqrt{6}

(D) 20\sqrt{2} 20\sqrt{2}

(E) 15\sqrt{3} 15\sqrt{3}

Answer

Correct choice: (E)

正确答案：（E）

Solution

The length of the side opposite to the angle with $120^\circ$ is longest. We denote its value as $x$. Because three side lengths form an arithmetic sequence, the middle-valued side length is $\frac{x + 6}{2}$. Following from the law of cosines, we have \begin{align*} 6^2 + \left( \frac{x + 6}{2} \right)^2 - 2 \cdot 6 \cdot \frac{x + 6}{2} \cdot \cos 120^\circ = x^2 . \end{align*} By solving this equation, we get $x = 14$. Thus, $\frac{x + 6}{2} = 10$. Therefore, the area of the triangle is \begin{align*} \frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ = \boxed{\textbf{(E) } 15 \sqrt{3}} . \end{align*}

与$120^\circ$角相对的边最长。我们记其值为$x$。因为三边长成等差数列，中间值的边长为$\frac{x + 6}{2}$。根据余弦定律，有 \begin{align*} 6^2 + \left( \frac{x + 6}{2} \right)^2 - 2 \cdot 6 \cdot \frac{x + 6}{2} \cdot \cos 120^\circ = x^2 . \end{align*} 解此方程，得$x = 14$。于是，$\frac{x + 6}{2} = 10$。因此，三角形的面积为 \begin{align*} \frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ = \boxed{\textbf{(E) } 15 \sqrt{3}} . \end{align*}

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