2021 Fall AMC 12B Problems/Problem 24
Contents
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 (Reflection)
By the Law of Cosine
As is a cyclic quadrilateral, . As is a cyclic quadrilateral, .
by
Hence,
By the Law of Cosine
Note that is 's reflection over line , quadrilateral is a kite symmetrical by line , .
Solution 2 (Olympiad Solution using Spiral Similarity)
Construct the -antipode, . Notice by spiral similarity at , thus . Let ; by symmetry as well and from Law of Cosines in , so by Law of Cosines in we have from which . Now, .
Solution 3
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,
Two solution methods follow from here.
Solution 3.1 (Stewart's theorem)
Applying Stewart's theorem on with cevian using the directed lengths and , we obtain so .
Solution 3.2 (Double Cosine Law)
Note that so we may plug into Law of Cosines to find the angle's cosine:
So, we observe that we can use Law of Cosines again to find : both ways.
- Kevinmathz
Solution 4
This solution is based on this figure:
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)
Solution 5
Denote and . Note that by Heron's formula the area of is so the -coordinate of (height of above the -axis) is easily computed by the base-height formula as .
Now, since , the -coordinate of satisfies and solving gives .
The circumcircle of has radius . We know by the perpendicular bisector rule that the circumcenter is located directly below the midpoint of (-coordinate ).
So, the negative coordinate of satisfies and solving gives .
It's also clear that point is going to be located directly below on the circle, because the angle bisector intersects the circumcircle at the midpoint of the arc (Fact 5). Since the radius of the circle is , we have the coordinates of
Solving for point (the point on the -axis between and ), we get that .
So now we know six of the critical points: ; ; ; ; ; .
We are now ready to add in the circumcircle of , which has radius . From the above information, , , and .
After a bit of simplification we end up with and .
For the area of , the altitude dropped from vertex has height , and the base has length , so its area is .
Thus, which after tons of cancellations becomes .
We know from the perpendicular bisector rule that the circumcenter of is located directly below the midpoint of (-coordinate ).
So, the negative -coordinate of satisfies , and solving gives . Thus, the equation of the circumcircle of is .
Point is the intersection of this circle and the line , which has equation . So, we substitute into the equation of the circle to get .
After simplifying, we have (the 's cancel out), whose solutions are and . The first corresponds to the origin, and the second corresponds to point . Thus the coordinates of are .
The coordinates of are , so
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 |
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