+0

Hard Inversion Question

0
300
2

Let PQ be the diameter of a circle. Let A and B be points on the circle, and let the tangents to the circle through A and B intersect at C. Let the tangent to the circle at Q intersect PA, PB, and PC at A', B', and C', respectively. Show that C' is the midpoint of A'B'.

While I have done a solution using similar triangles and symmedian properties, it is encouraged to solve this problem using inversion and I am having a lot of trouble doing that. I think I should draw a circle with radius PQ centered at P  and possibly invert a circle centered at C with radius CA but I haven't been able to do much other than that. Could someone please explain the inversion solution?

Aug 21, 2020

#1
0

You can use coordinates.  It's ez with coordinates.

Aug 21, 2020
#2
0

Thank you for the hint. I am actually looking for a inversion solution but I am going to try coordinate bashing. Is there any special things I should do to do this like having a certain point as the origin or using a certain theorem? Thanks!

Guest Aug 22, 2020