Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
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+ | ==Problem== | ||
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+ | Triangle <math>ABC</math> has side lengths <math>AB = 11, BC=24</math>, and <math>CA = 20</math>. The bisector of <math>\angle{BAC}</math> intersects <math>\overline{BC}</math> in point <math>D</math>, and intersects the circumcircle of <math>\triangle{ABC}</math> in point <math>E \ne A</math>. The circumcircle of <math>\triangle{BED}</math> intersects the line <math>AB</math> in points <math>B</math> and <math>F \ne B</math>. What is <math>CF</math>? | ||
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+ | <math>\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}</math> | ||
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==Solution== | ==Solution== | ||
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+ | {{AMC12 box|year=2021 Fall|ab=B|num-a=25|num-b=23}} | ||
+ | {{MAA Notice}} |
Revision as of 18:55, 25 November 2021
Problem
Triangle has side lengths
, and
. The bisector of
intersects
in point
, and intersects the circumcircle of
in point
. The circumcircle of
intersects the line
in points
and
. What is
?
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
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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