Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
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<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math> | <b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math> | ||
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<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. | <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. | ||
Revision as of 17:52, 25 November 2021
Solution
Claim: 
Proof: Note that 
and 
meaning that our claim is true by AA similarity.
Because of this similarity, we have that ![\[\frac{AC}{AD} = \frac{AE}{AB} \to AB \cdot AC = AD \cdot AE = AB \cdot AF\]](http://latex.artofproblemsolving.com/3/7/7/3775bac8f0539aaac0bb765d9c35e00ce55876ba.png)
by Power of a Point. Thus, 
Now, note that 
and plug into Law of Cosines to find the angle's cosine: ![\[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB = -\frac{1}{8}.\]](http://latex.artofproblemsolving.com/8/5/6/8568c0fea68d7d41da07bb5b871cb50c0bbe26d3.png)
So, we observe that we can use Law of Cosines again to find 
: ![\[AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = CF^2 \to CF = \boxed{\textbf{(C) }30}.\]](http://latex.artofproblemsolving.com/8/d/f/8dfb4002450cc3849180a76c1708168fb736fa0d.png)
- kevinmathz
