AMC12 2025 A

AMC12 2025 A · Q8

AMC12 2025 A · Q8. It mainly tests Angle chasing, Triangles (properties).

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

五边形$ABCDE$内接于圆中，且$\angle BEC = \angle CED = 30^\circ$。直线$AC$与直线$BD$相交于点$F$，已知$AB = 9$，$AD = 24$。求$BF$？

(A) \frac{57}{11} \frac{57}{11}

(B) \frac{59}{11} \frac{59}{11}

(D) \frac{61}{11} \frac{61}{11}

(E) \frac{63}{11} \frac{63}{11}

Answer

Correct choice: (E)

正确答案：（E）

Solution

We will scale down the diagram by a factor of $3$ so that $AB = 3$ and $AD = 8.$ Since $\angle BEC = 30^{\circ},$ it follows that $\angle BAC = \angle BDC = 30^{\circ}$ as they all subtend the same arc. Similarly, since $\angle CED = 30^{\circ},$ it follows that $\angle CAD = \angle CBD = 30^{\circ}$ as well. We obtain the following diagram: Note that $\triangle ABD$ has $\angle BAD = 60^{\circ}.$ Applying Law of Cosines, we get \begin{align*} BD^2 &= AB^2+AD^2-2AB\cdot AD \cdot\cos{60^{\circ}} \\ &= 9 + 64 - 2 \cdot 3 \cdot 8 \cdot \frac{1}{2} \\ &= 49, \end{align*} from which $BD = 7.$ From here, we wish to find $BF.$ As $AF$ is the angle bisector of $\angle BAD,$ we apply the Angle Bisector Theorem: \begin{align*} \frac{AB}{BF} &= \frac{AD}{DF} \\ \frac{3}{BF} &= \frac{8}{7-BF}. \end{align*} Solving for $BF,$ we get $BF = \frac{21}{11}.$ Remember to scale the figure back up by a factor of $3,$ so our answer is $\frac{21}{11}\cdot 3 = \boxed{\textbf{(E) } \frac{63}{11}}.$

我们将图按比例缩小3倍，使$AB = 3$，$AD = 8$。由于$\angle BEC = 30^{\circ}$，因此$\angle BAC = \angle BDC = 30^{\circ}$，因为它们都对应该弓。同理，由于$\angle CED = 30^{\circ}$，有$\angle CAD = \angle CBD = 30^{\circ}$。注意$\triangle ABD$有$\angle BAD = 60^{\circ}$。应用余弦定律，得 \begin{align*} BD^2 &= AB^2+AD^2-2AB\cdot AD \cdot\cos{60^{\circ}} \\ &= 9 + 64 - 2 \cdot 3 \cdot 8 \cdot \frac{1}{2} \\ &= 49, \end{align*} 故$BD = 7$。现在求$BF$。由于$AF$是$\angle BAD$的角度平分线，应用角度平分线定理： \begin{align*} \frac{AB}{BF} &= \frac{AD}{DF} \\ \frac{3}{BF} &= \frac{8}{7-BF}. \end{align*} 解得$BF = \frac{21}{11}$。记得将图形按3倍放大，因此答案是$\frac{21}{11}\cdot 3 = \boxed{\textbf{(E) } \frac{63}{11}}$。

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