2023 USAMO Problems/Problem 6
Problem
Let ABC be a triangle with incenter and excenters , , opposite , , and , respectively. Given an arbitrary point on the circumcircle of that does not lie on any of the lines , , or , suppose the circumcircles of and intersect at two distinct points and . If is the intersection of lines and , prove that .
Video Solution by mop 2024
https://youtube.com/watch?v=LAuyU2OuVzE
~r00tsOfUnity
Solution 1
Consider points and such that the intersections of the circumcircle of with the circumcircle of are and , the intersections of the circumcircle of with the circumcircle of are and , the intersections of the circumcircle of with line are and , the intersections of the circumcircle of with line are and , the intersection of lines and is , and the intersection of lines and is .
Since is cyclic, the pairwise radical axes of the circumcircles of and concur. The pairwise radical axes of these circles are and , so and are collinear. Similarly, since is cyclic, the pairwise radical axes of the cirucmcircles of and concur. The pairwise radical axes of these circles are and , so and are collinear. This means that , so the tangents to the circumcircle of at and intersect on . Let this intersection be . Also, let the intersection of the tangents to the circumcircle of at and be a point at infinity on called and let the intersection of lines and be . Then, let the intersection of lines and be . By Pascal's Theorem on and , we get that and are collinear and that and are collinear, so and are collinear, meaning that lies on since both and lie on .
Consider the transformation which is the composition of an inversion centered at and a reflection over the angle bisector of that sends to and to . We claim that this sends to and to . It is sufficient to prove that if the transformation sends to , then is cyclic. Notice that since and . Therefore, we get that , so is cyclic, proving the claim. This means that .
We claim that . Construct to be the intersection of line and the circumcircle of and let and be the intersections of lines and with the circumcircle of . Since and are the reflections of and over , it is sufficient to prove that are concyclic. Since and concur and and are concyclic, we have that are concyclic, so , so are concyclic, proving the claim. We can similarly get that .
Let line intersect the circumcircle of at and . Notice that is the midpoint of and , so is a parallelogram with center , so . Similarly, we get that if line intersects the circumcircle of at and , we have that , so , so , so are concyclic. Then, the pairwise radical axes of the circumcircles of and are and , so and concur, so and concur, so . We are then done since .
~Zhaom
Solution 2
Set as the reference triangle, and let with homogenized coordinates. To find the equation of circle , we note that (not homogenized) and . Thus, for this circle with equation , we compute that For circle with equation ,, we find that The equation of the radical axis is with , , and . We want to consider the intersection of this line with line , so set . The equation reduces to . We see that and , so which is the required condition for and to be isogonal.
~MathIsFun286
See Also
2023 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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