2025 AMC 12A Problems/Problem 8: Difference between revisions

Revision as of 01:42, 7 November 2025

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)

@@ Line 29: / Line 29: @@
 ==Solution 3 (Ptolemy’s + Similarity)==
-We have <imath>ABCDE</imath> cyclic, so <imath>\angle BAC=\angle CAD=\angle BEC=30^\circ </imath>. Hence cyclic quadrilateral <imath>ABCD</imath> has <imath>\angle BAD=60\circ </imath>. Law of Cosines on triangle <imath>BAD</imath> gives <imath>\overline{BD}^2=9^2+24^2-2\cdot9\cdot24\cos60^\circ </imath>. Hence <imath>\overline{BD}=21</imath>. Since triangle <imath>BCD</imath> is a 120-30-30 triangle, we can use law of sines or just memorize ratios to get <imath>\overline{BC}=\overline{CD}=7\sqrt3</imath>. Now Ptolemy’s on <imath>ABCD</imath> yields <imath>7\sqrt3(9+24)=21\overline{AC}</imath>. Hence <imath>\overline{AC}=11\sqrt3</imath>. Now notice that <imath>\angle BCF=\angle ACB</imath>, and <imath>\angle CBF=\angle CAB=30^\circ</imath>. Hence triangles <imath>CBF</imath> and <imath>CAB</imath> are similar, and <imath>\frac{\overline{BF}}{\overline{BC}}=\frac{\overline{AB}}{\overline{AC}}</imath>, so <imath>\frac{\overline{BF}}{7\sqrt3}=\frac9{11\sqrt3}</imath> and <imath>\overline{BF}=\frac{63}{11}</imath>, or <imath>\boxed{\textit{E}}</imath>.
+We have <imath>ABCDE</imath> cyclic, so <imath>\angle BAC=\angle CAD=\angle BEC=30^\circ </imath>. Hence cyclic quadrilateral <imath>ABCD</imath> has <imath>\angle BAD=60^\circ </imath>. Law of Cosines on triangle <imath>BAD</imath> gives <imath>\overline{BD}^2=9^2+24^2-2\cdot9\cdot24\cos60^\circ </imath>. Hence <imath>\overline{BD}=21</imath>. Since triangle <imath>BCD</imath> is a 120-30-30 triangle, we can use law of sines or just memorize ratios to get <imath>\overline{BC}=\overline{CD}=7\sqrt3</imath>. Now Ptolemy’s on <imath>ABCD</imath> yields <imath>7\sqrt3(9+24)=21\overline{AC}</imath>. Hence <imath>\overline{AC}=11\sqrt3</imath>. Now notice that <imath>\angle BCF=\angle ACB</imath>, and <imath>\angle CBF=\angle CAB=30^\circ</imath>. Hence triangles <imath>CBF</imath> and <imath>CAB</imath> are similar, and <imath>\frac{\overline{BF}}{\overline{BC}}=\frac{\overline{AB}}{\overline{AC}}</imath>, so <imath>\frac{\overline{BF}}{7\sqrt3}=\frac9{11\sqrt3}</imath> and <imath>\overline{BF}=\frac{63}{11}</imath>, or <imath>\boxed{\textit{E}}</imath>.
 ~benjamintontungtungtungsahur (look guys im famous)

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