2025 AMC 12A Problem 8

Attempt Problem 8 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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8.

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

Small hint:

\angle BAC

and

\angle CAD

both subtend

30^\circ

arcs, so

AC

bisects

\angle BAD

Big hint:

Find

BD

with the Law of Cosines in

\triangle ABD,

then use the Angle Bisector Theorem

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.

Problem 8 in other years