Question

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 $BF?$

$\dfrac{57}{11}$

$\dfrac{59}{11}$

$\dfrac{60}{11}$

$\dfrac{61}{11}$

$\dfrac{63}{11}$

Answer: E

Concepts:inscribed angle angle bisector theorem law of cosines

Difficulty rating: 1440

Solution:

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 Problem 8

Answer: E

Difficulty rating: 1440

Solution:

Problem 8 in Other Years