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

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Claim: <math>\triangle ADC \sim \triangle ABE.</math>
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==Solution==
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Proof: Note that <math>\angle CAD = \angle CAE = \angle EAB</math> and <math>\angle DCA = \angle BCA = \angle BEA</math> meaning that our claim is true by AA similarity.
 
  
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Because of this similarity, we have that <cmath>\frac{AC}{AD} = \frac{AE}{AB}} \to AB \cdot AC = AD \cdot AE = AB \cdot AF</cmath> by Power of a Point. Thus, <math>AC=AF=20.</math>
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<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math>
 +
<b>Proof:</b> Note that <math>\angle CAD = \angle CAE = \angle EAB</math> and <math>\angle DCA = \angle BCA = \angle BEA</math> meaning that our claim is true by AA similarity.
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Because of this similarity, we have that <cmath>\frac{AC}{AD} = \frac{AE}{AB} \to AB \cdot AC = AD \cdot AE = AB \cdot AF</cmath> by Power of a Point. Thus, <math>AC=AF=20.</math>
  
 
Now, note that <math>\angle CAF = \angle CAB</math> and plug into Law of Cosines to find the angle's cosine: <cmath>AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB = -\frac{1}{8}.</cmath>
 
Now, note that <math>\angle CAF = \angle CAB</math> and plug into Law of Cosines to find the angle's cosine: <cmath>AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB = -\frac{1}{8}.</cmath>

Revision as of 17:52, 25 November 2021

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

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