2021 Fall AMC 12B Problems/Problem 24

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

\[\frac{AC}{AD} = \frac{AE}{AB}} \to AB \cdot AC = AD \cdot AE = AB \cdot AF\] (Error compiling LaTeX. Unknown error_msg)

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 $CF$ : $AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = CF^2 \to CF = \boxed{\textbf{(C) }30}.$

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2021 Fall AMC 12B Problems/Problem 24