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2025 AMC 12A Problems/Problem 8

Problem

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 $BF$?

$\textbf{(A) } \frac{57}{11} \qquad\textbf{(B) } \frac{59}{11} \qquad\textbf{(C) } \frac{60}{11} \qquad\textbf{(D) } \frac{61}{11} \qquad\textbf{(E) } \frac{63}{11}$

Solution 1

We will scale down the diagram by a factor of $3$ so that $AB = 3$ and $AD = 8$. Because $\angle BEC = 30$, then $\angle BAC = \angle BDC = 30$ because they all subtend the same arc. Similarly, because $\angle CED = 30$, $\angle CAD = \angle CBD = 30$ as well.

Notice $\triangle ABD$, which has $\angle BAD = 60$. Applying Law of Cosines, we get: \[BD^2 = AB^2+AD^2-2AB\cdot AD \cdot\cos{60}\] \[BD^2 = 9 + 64 - 2 \cdot 3 \cdot 7 \cdot \frac{1}{2}\] \[BD^2 = 49.\] So, $BD = 7$. From here, we want $BF$. Noticing that $AF$ is the angle bisector of $\angle BAD$, we apply the Angle Bisector Theorem: \[\frac{AB}{BF} = \frac{AD}{DF}\] \[\frac{3}{BF}=\frac{8}{7-BF}.\] 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 $\frac{21}{11}\cdot 3 = \boxed{\frac{63}{11}}.$

~lprado


Solution 2 Law of (Co)Sine

From cyclic quadrilateral $CDAE$, we have $\angle CAD = \angle CED = 30^\circ.$ Since $ABDE$ is also cyclic, we have $\angle BAD = \angle BED = 60^\circ$, so, \[\angle BAC= \angle BAD - \angle CAD = 60^\circ - 30^\circ = 30^\circ.\] Using Law of Cosines on $\triangle ABD$, we get \[BD^2=9^2+24^2-2(9)(24)\cos(60^\circ).\] Solving, we get $BD=21$. Next, let $\overline{BF}=x$, and $\angle AFB = \theta$, which means $\overline{FD}=21-x$ and $\angle AFD = 180-\theta$. Using Law of Sines on $\triangle AFB$, we have \[\frac{9}{\sin \theta}=\frac{x}{\sin 30}.\] Solving for $\sin \theta$, we get $\sin \theta = \frac{9}{2x}$. Now we apply the Law of Sines to $\triangle AFD.$ We have \[\frac{24}{\sin(180-\theta)} = \frac{21-x}{\sin 30}.\] Since $\sin(180-\theta) = \sin(\theta),$ and $\sin \theta = \frac{9}{2x}$, we have \[\frac{16x}{3} = 42-2x.\] Solving for $x$ gives $\boxed{x=\frac{63}{11}}$ or $\boxed{\text{E}}$.

~evanhliu2009

Solution 3 (Ptolemy’s + Similarity)

We have $ABCDE$ cyclic, so $\angle BAC=\angle CAD=\angle BEC=30^\circ$. Hence cyclic quadrilateral $ABCD$ has $\angle BAD=60^\circ$. Law of Cosines on triangle $BAD$ gives $\overline{BD}^2=9^2+24^2-2\cdot9\cdot24\cos60^\circ$. Hence $\overline{BD}=21$. Since triangle $BCD$ is a 120-30-30 triangle, we can use law of sines or just memorize ratios to get $\overline{BC}=\overline{CD}=7\sqrt3$. Now Ptolemy’s on $ABCD$ yields $7\sqrt3(9+24)=21\overline{AC}$. Hence $\overline{AC}=11\sqrt3$. Now notice that $\angle BCF=\angle ACB$, and $\angle CBF=\angle CAB=30^\circ$. Hence triangles $CBF$ and $CAB$ are similar, and $\frac{\overline{BF}}{\overline{BC}}=\frac{\overline{AB}}{\overline{AC}}$, so $\frac{\overline{BF}}{7\sqrt3}=\frac9{11\sqrt3}$ and $\overline{BF}=\frac{63}{11}$, or $\boxed{\textit{E}}$.

~benjamintontungtungtungsahur (look guys im famous)

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