Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
Math Topics
»
geometry.college
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
More on the Kosnita point and the reflection triangle
Replies:
1
Last Post:
Apr 26, 2010 9:42 AM




More on the Kosnita point and the reflection triangle
Posted:
Aug 8, 2003 2:12 PM


In my note "On the Kosnita Point and the Reflection Triangle", Forum Geometricorum 3 (2003) pages 105111, I studied the "reflection triangle" A'B'C' of a triangle ABC, where A' is the reflection of A in BC, and similarly B' and C' are defined. I proved that the circles B'C'A, C'A'B, A'B'C concur.
What is, may be, new: If Ua, Ub, Uc are the centers of these circles, then the lines AUa, BUb, CUc concur at...
... the ninepoint center of triangle ABC.
Here is a PROOF.
Let Rc and Ra be the circumcenters of triangles ABC' and BCA', respectively, i. e. the reflections of the circumcenter O of triangle ABC in the sides AB and BC. (This is because the circumcenter of a reflected triangle is the reflection of the circumcenter of the original triangle.)
On the other hand, both circumcenters Rc and Ub lie on the perpendicular bisector of BC'. Let this perpendicular bisector meet BC at X and BC' at C" (i. e., C" is the midpoint of BC'). Analogously, the circumcenters Ra and Ub lie on the perpendicular bisector of BA'. Let this perpendicular bisector meet AB at Y and BA' at A" (i. e., A" is the midpoint of BA').
Triangles XBC" and YBA" are similar (equal angles: angle XBC" = 180ÃÂ°  2B = angle YBA", and angle XC"B = 90ÃÂ° = angle YA"B). Hence,
BX : BY = BC" : BA".
But BC" = BC' / 2 = BC / 2 and BA" = BA' / 2 = BA / 2, so that we obtain
BX : BY = BC : BA,
and the line XY is parallel to CA.
It should be wellknown that the reflection of the circumcenter of a triangle in a sideline is the reflection of the opposite vertex in the ninepoint center. Thus, the points Rc and Ra are the reflections of C and A in the ninepoint center N of triangle ABC. I. e., the lines ARa and CRc intersect at N, and we have RcRa  CA.
Together with XY  CA, we easily see that the intersections of corresponding sides of the triangles RaUbRc and ABC are collinear. By the Desargues theorem, the two triangles are perspective, and the lines ARa, BUb, CRc concur. Thus, the line BUb passes through the ninepoint center N. Analogously for the lines AUa and CUc.
Darij Grinberg



