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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
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<math>\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}</math>
 
<math>\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}</math>
  
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==Solution==
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==Solution 1==
  
 
<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math>
 
<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math>
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- kevinmathz
 
- kevinmathz
  
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==Solution 2==
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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>
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==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 22:29, 25 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*}

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

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