Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "<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 = \ang...")
 
m
Line 1: Line 1:
−
<b>Claim:</b> <math>\triangle ADC \sim \triangle ABE.</math>
+
Claim: <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.
+
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.
  
 
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>
Line 7: Line 7:
  
 
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

Revision as of 18:51, 25 November 2021

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

- kevinmathz

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_24&oldid=166337"