Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
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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> | 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> |
So, we observe that we can use Law of Cosines again to find <math>CF</math>: <cmath>AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = CF^2 \to CF = \boxed{\textbf{(C) }30}.</cmath> | So, we observe that we can use Law of Cosines again to find <math>CF</math>: <cmath>AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = CF^2 \to CF = \boxed{\textbf{(C) }30}.</cmath> | ||
- kevinmathz | - kevinmathz |
Revision as of 17:53, 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 by Power of a Point. Thus,
Now, note that and plug into Law of Cosines to find the angle's cosine:
So, we observe that we can use Law of Cosines again to find :
- kevinmathz