Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"

Revision as of 00:17, 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

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)

@@ Line 19: / Line 19: @@
 - kevinmathz
-==Solution 2 (Stewart Bash)==
+==Solution 2==
 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*}
@@ Line 29: / Line 29: @@
 x&=\boxed{\textbf{(C)} ~30}.
 \end{align*}</cmath>
 ~Punxsutawney Phil

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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