Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
Line 18: Line 18:
  
 
- kevinmathz
 
- kevinmathz
 +
 +
==Solution 2 (Stewart Bash)==
 +
By the Inscribed Angle Theorem and the definition of angle bisectors note that<cmath>\angle ABD=\angle ABC=\angle AEC\ \text{and}\ \angle BAD=\angle DAC=\angle EAC</cmath>so <math>\triangle ABD\sim\triangle AEC</math>. Therefore <math>\frac{AB}{AD}=\frac{AE}{AC}\rightarrow AB\cdot AC=AD\cdot AE</math>. By PoP, we can also express <math>AD\cdot AE</math> as <math>AB\cdot AF,</math> so <math>AB\cdot AC=AB\cdot AF\rightarrow AC=AF=20</math> and <math>BF=20-AB=20-11=9</math>. Let <math>CF=x</math>. Applying Stewart’s theorem on <math>\triangle ACF</math> with cevian <math>\overrightarrow{CB},</math> we have
 +
<cmath>\begin{align*}
 +
11\cdot 9\cdot 20+24\cdot 20\cdot 24&=11x^{2}+20\cdot 9\cdot 20 \\
 +
1980+11{,}520&=11x^{2}+3600 \\
 +
13{,}500&=11x^{2}+3600 \\
 +
11x^{2}&=9900 \\
 +
x^{2}&=900 \\
 +
x&=\boxed{\textbf{(C)} ~30}.
 +
\end{align*}</cmath>
 +
 +
~Punxsutawney Phil
  
 
== Solution 3 ==
 
== Solution 3 ==
Line 88: Line 101:
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
−
 
−
==Solution 3 (Stewart Bash)==
 
−
By the Inscribed Angle Theorem and the definition of angle bisectors note that<cmath>\angle ABD=\angle ABC=\angle AEC\ \text{and}\ \angle BAD=\angle DAC=\angle EAC</cmath>so <math>\triangle ABD\sim\triangle AEC</math>. Therefore <math>\frac{AB}{AD}=\frac{AE}{AC}\rightarrow AB\cdot AC=AD\cdot AE</math>. By PoP, we can also express <math>AD\cdot AE</math> as <math>AB\cdot AF,</math> so <math>AB\cdot AC=AB\cdot AF\rightarrow AC=AF=20</math> and <math>BF=20-AB=20-11=9</math>. Let <math>CF=x</math>. Applying Stewart’s theorem on <math>\triangle ACF</math> with cevian <math>\overrightarrow{CB},</math> we have
 
−
<cmath>\begin{align*}
 
−
11\cdot 9\cdot 20+24\cdot 20\cdot 24&=11x^{2}+20\cdot 9\cdot 20 \\
 
−
1980+11{,}520&=11x^{2}+3600 \\
 
−
13{,}500&=11x^{2}+3600 \\
 
−
11x^{2}&=9900 \\
 
−
x^{2}&=900 \\
 
−
x&=\boxed{\textbf{(C)} ~30}.
 
−
\end{align*}</cmath>
 
−
 
−
~Punxsutawney Phil
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=25|num-b=23}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=25|num-b=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:16, 28 November 2021

Problem

Triangle $ABC$ has side lengths $AB = 11, BC=24$, and $CA = 20$. The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$, and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$. The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$. What is $CF$?

$\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}$

Solution 1

Claim: $\triangle ADC \sim \triangle ABE.$

Proof: Note that $\angle CAD = \angle CAE = \angle EAB$ and $\angle DCA = \angle BCA = \angle BEA$ meaning that our claim is true by AA similarity.

Because of this similarity, we have that \[\frac{AC}{AD} = \frac{AE}{AB} \to AB \cdot AC = AD \cdot AE = AB \cdot AF\] by Power of a Point. Thus, $AC=AF=20.$

Now, note that $\angle CAF = \angle CAB$ and plug into Law of Cosines to find the angle's cosine: \[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.\]

So, we observe that we can use Law of Cosines again to find $CF$: \[AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = CF^2 \to CF = \boxed{\textbf{(C) }30}.\]

- kevinmathz

Solution 2 (Stewart Bash)

By the Inscribed Angle Theorem and the definition of angle bisectors note that\[\angle ABD=\angle ABC=\angle AEC\ \text{and}\ \angle BAD=\angle DAC=\angle EAC\]so $\triangle ABD\sim\triangle AEC$. Therefore $\frac{AB}{AD}=\frac{AE}{AC}\rightarrow AB\cdot AC=AD\cdot AE$. By PoP, we can also express $AD\cdot AE$ as $AB\cdot AF,$ so $AB\cdot AC=AB\cdot AF\rightarrow AC=AF=20$ and $BF=20-AB=20-11=9$. Let $CF=x$. Applying Stewart’s theorem on $\triangle ACF$ with cevian $\overrightarrow{CB},$ we have \begin{align*} 11\cdot 9\cdot 20+24\cdot 20\cdot 24&=11x^{2}+20\cdot 9\cdot 20 \\ 1980+11{,}520&=11x^{2}+3600 \\ 13{,}500&=11x^{2}+3600 \\ 11x^{2}&=9900 \\ x^{2}&=900 \\ x&=\boxed{\textbf{(C)} ~30}. \end{align*}

~Punxsutawney Phil

Solution 3

This solution is based on this figure: Image:2021_AMC_12B_(Nov)_Problem_24,_sol.png

Denote by $O$ the circumcenter of $\triangle BED$. Denote by $R$ the circumradius of $\triangle BED$.

In $\triangle BCF$, following from the law of cosines, we have \begin{align*} CF^2 & = BC^2 + BF^2 - 2 BC \cdot BF \cos \angle CBF \\ & = BC^2 + BF^2 + 2 BC \cdot BF \cos \angle ABC . \hspace{1cm} (1) \end{align*}

For $BF$, we have \begin{align*} BF & = 2 R \cos \angle FBO \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \angle CBO \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - \angle BOD}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BED}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BCA}{2} \right) \\ & = 2 R \cos \left( 90^\circ - \angle ABC + \angle BCA \right) \\ & = 2 R \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BED} \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BCA} \sin \left( \angle ABC - \angle BCA \right) \\ & = BD \left( \sin \angle ABC \cot \angle BCA - \cos \angle ABC \right) . \hspace{1cm} (2) \end{align*} The fourth equality follows from the property that $B$, $D$, $E$ are concyclic. The fifth and the ninth equalities follow from the property that $A$, $B$, $C$, $E$ are concyclic.

Because $AD$ bisects $\angle BAC$, following from the angle bisector theorem, we have \[ \frac{BD}{CD} = \frac{AB}{AC} . \]

Hence, $BD = \frac{24 \cdot 11}{31}$.

In $\triangle ABC$, following from the law of cosines, we have \begin{align*} \cos \angle ABC & = \frac{AB^2 + BC^2 - AC^2}{2 AB \cdot BC} \\ & = \frac{9}{16} \end{align*} and \begin{align*} \cos \angle BCA & = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} \\ & = \frac{57}{64} . \end{align*}

Hence, $\sin \angle ABC = \frac{5 \sqrt{7}}{16}$ and $\sin \angle BCA = \frac{11 \sqrt{7}}{64}$. Hence, $\cot \angle BCA = \frac{57}{11 \sqrt{7}}$.

Now, we are ready to compute $BF$ whose expression is given in Equation (2). We get $BF = 9$.

Now, we can compute $CF$ whose expression is given in Equation (1). We have $CF = 30$.


Therefore, the answer is $\boxed{\textbf{(C) }30}$.

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_24&oldid=166798"