AMC12 2010 A

AMC12 2010 A · Q8

AMC12 2010 A · Q8. It mainly tests Angle chasing, Triangles (properties).

Triangle $ABC$ has $AB=2 \cdot AC$. Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$, respectively, such that $\angle BAE = \angle ACD$. Let $F$ be the intersection of segments $AE$ and $CD$, and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$?

三角形 $ABC$ 有 $AB=2 \cdot AC$。点 $D$ 和 $E$ 分别在 $\overline{AB}$ 和 $\overline{BC}$ 上，使得 $\angle BAE = \angle ACD$。$F$ 是线段 $AE$ 和 $CD$ 的交点，且 $\triangle CFE$ 是等边三角形。$\angle ACB$ 是多少？

(A) 60^\circ 60^\circ

(B) 75^\circ 75^\circ

(D) 105^\circ 105^\circ

(E) 120^\circ 120^\circ

Answer

Correct choice: (C)

正确答案：（C）

Solution

Error creating thumbnail: Unable to save thumbnail to destination Let $\angle BAE = \angle ACD = x$. \begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*} Since $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$, triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90^\circ\,\textbf{(C)}}$.

Error creating thumbnail: Unable to save thumbnail to destination 设 $\angle BAE = \angle ACD = x$。 \begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*} 由于 $\frac{AC}{AB} = \frac{1}{2}$，且斜边与较短边之间的夹角为 $60^\circ$，所以三角形 $ABC$ 是 $30-60-90$ 三角形，因此 $\angle BCA = \boxed{90^\circ\,\textbf{(C)}}$。

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