2021 Fall AMC 12B Problems/Problem 24
Contents
[hide]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 1
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
Solution 2
By the Inscribed Angle Theorem and the definition of angle bisectors note thatso . Therefore . By PoP, we can also express as so and . Let . Applying Stewart’s theorem on with cevian we have ~Punxsutawney Phil
Solution 3
This solution is based on this figure: Image:2021_AMC_12B_(Nov)_Problem_24,_sol.png
Denote by the circumcenter of . Denote by the circumradius of .
In , following from the law of cosines, we have For , we have The fourth equality follows from the property that , , are concyclic. The fifth and the ninth equalities follow from the property that , , , are concyclic.
Because bisects , following from the angle bisector theorem, we have Hence, .
In , following from the law of cosines, we have and Hence, and . Hence, .
Now, we are ready to compute whose expression is given in Equation (2). We get .
Now, we can compute whose expression is given in Equation (1). We have .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Power of Logic(Trig and Power of a point)
~math2718281828459
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.