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

Latest revision as of 22:28, 11 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 $BF$ ?

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

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(200); real r = 7*sqrt(3); pair O, A, B, C, D, E, F; O = origin; B = r*dir(30); C = r*dir(-30); D = r*dir(-90); E = r*dir(180); A = intersectionpoints(Circle(O,r),Circle(B,9))[0]; F = intersectionpoint(A--C,B--D); draw(Circle(O,r)^^A--B--C--D--E--cycle^^D--B--E--C--A--cycle); dot("$B$",B,1.5*B/r,linewidth(4)); dot("$C$",C,1.5*C/r,linewidth(4)); dot("$D$",D,1.5*D/r,linewidth(4)); dot("$E$",E,1.5*E/r,linewidth(4)); dot("$A$",A,1.5*A/r,linewidth(4)); dot("$F$",F,1.5*F/r,linewidth(4)); label("$30^{\circ}$",E,6*(1,0),fontsize(8)); label("$30^{\circ}$",E,7*dir(-32),fontsize(8)); label("$9$",0.92*midpoint(A--B)); label("$24$",1.8*midpoint(A--D)); [/asy]$ ~MRENTHUSIASM

Solution 1

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

We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(200); real r = 7*sqrt(3); pair O, A, B, C, D, E, F; O = origin; B = r*dir(30); C = r*dir(-30); D = r*dir(-90); E = r*dir(180); A = intersectionpoints(Circle(O,r),Circle(B,9))[0]; F = intersectionpoint(A--C,B--D); draw(Circle(O,r)^^B--C--D--E--A^^B--E--C--F); draw(A--D--B--cycle^^A--F,red); dot("$B$",B,1.5*B/r,linewidth(4)); dot("$C$",C,1.5*C/r,linewidth(4)); dot("$D$",D,1.5*D/r,linewidth(4)); dot("$E$",E,1.5*E/r,linewidth(4)); dot("$A$",A,1.5*A/r,linewidth(4)); dot("$F$",F,1.5*F/r,linewidth(4)); label("$30^{\circ}$",E,6*(1,0),fontsize(8)); label("$30^{\circ}$",E,7*dir(-32),fontsize(8)); label("$30^{\circ}$",A,9*dir(-56),red+fontsize(8)); label("$30^{\circ}$",A,9*dir(-84),red+fontsize(8)); label("$3$",1.1*midpoint(A--B),red); label("$8$",0.4*midpoint(A--D),red); [/asy]$ Note that $\triangle ABD$ has $\angle BAD = 60^{\circ}.$ Applying Law of Cosines, we get $\begin{align*} BD^2 &= AB^2+AD^2-2AB\cdot AD \cdot\cos{60^{\circ}} \\ &= 9 + 64 - 2 \cdot 3 \cdot 8 \cdot \frac{1}{2} \\ &= 49. \end{align*}$ So, $BD = 7.$ From here, we want $BF.$ Noticing that $AF$ 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 $\frac{21}{11}\cdot 3 = \boxed{\textbf{(E) } \frac{63}{11}}.$

~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)

Video Solution by Power Solve

https://youtu.be/Vd_kvodRjNQ?si=ZuoUjGLXcZter8PB&t=753

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=ycwWI10M244

2025 AMC 12A (Problems • Answer Key • Resources)
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