Question

五边形 $ABCDE$ 内接于一个圆，且 $\angle BEC = \angle CED = 30^\circ$ 。令 $AC$ 和 $BD$ 交于点 $F$ ，并且 $AB = 9$ 、 $AD = 24$ 。 $BF$ 是多少？

Pentagon $ABCDE$ is inscribed in a circle, and $\angle BEC = \angle CED = 30^\circ.$ Let $AC$ and $BD$ intersect at point $F,$ and suppose that $AB = 9$ and $AD = 24.$ What is $BF?$

$\dfrac{57}{11}$

$\dfrac{59}{11}$

$\dfrac{60}{11}$

$\dfrac{61}{11}$

$\dfrac{63}{11}$

答案：E

知识点：圆周角角平分线定理余弦定理

难度评级：1440

解答：

圆周角 $\angle BEC = 30^\circ$ 对着弧 $BC = 60^\circ$ ，所以同样对着弧 $BC$ 的 $\angle BAC$ 等于 $30^\circ$ 。同理， $\angle CAD = 30^\circ$ 。

因此 $AC$ 平分 $\angle BAD = 60^\circ$ 。在 $\triangle ABD$ 中，所以 $BD = 21$ 。 $\begin{aligned} BD^2 &= 9^2 + 24^2 \\ &\quad {}- 2(9)(24)\cos 60^\circ \\ &= 657 - 216 = 441, \end{aligned}$

因为 $AF$ （沿着 $AC$ ）平分 $\angle BAD$ ，由角平分线定理， $\dfrac{BF}{FD} = \dfrac{AB}{AD} = \dfrac{9}{24} = \dfrac{3}{8}$ 。因此 $BF = \dfrac{3}{11}\cdot 21 = \dfrac{63}{11}$ 。

因此，正确答案是 E。

The inscribed angle $\angle BEC = 30^\circ$ subtends arc $BC = 60^\circ,$ so $\angle BAC,$ which also subtends arc $BC,$ equals $30^\circ.$ Likewise $\angle CAD = 30^\circ.$

Thus $AC$ bisects $\angle BAD = 60^\circ.$ In $\triangle ABD,$ $\begin{aligned} BD^2 &= 9^2 + 24^2 \\ &\quad {}- 2(9)(24)\cos 60^\circ \\ &= 657 - 216 = 441, \end{aligned}$ so $BD = 21.$

Since $AF$ (along $AC$ ) bisects $\angle BAD,$ the Angle Bisector Theorem gives $\dfrac{BF}{FD} = \dfrac{AB}{AD} = \dfrac{9}{24} = \dfrac{3}{8}.$ Hence $BF = \dfrac{3}{11}\cdot 21 = \dfrac{63}{11}.$

Thus, the correct answer is E.

2025 AMC 12A 第 8 题

答案：E

难度评级：1440

解答：

其他年份的第 8 题