2021 Fall AMC 12B Problems/Problem 24

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

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

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2021 Fall AMC 12B Problems/Problem 24

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