Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Friday, April 24, 2020
Dynamic Geometry 1473: Kosnita's Theorem, Triangle, Four Circumcenters, Concurrent Line
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My solution at https://stanfulger.blogspot.com/2021/06/problem-1473-gogeometry-cosnita-theorem.html
ReplyDeleteLet ABC have nine point center N. We will show that BO_B and BN are isogonal in angle B, and then the desired concurrency would follow since the isogonal conjugate of N would lie on all the lines AO_A, BO_B, CO_C.
ReplyDeleteConsider the transformation f which consists of an inversion of center B and ratio sqrt(BA*BC) and a reflection about the angle bisector of angle B. It is not hard to see that f(O) is B', the reflection of B about AC. So, circle (AOC) is sent onto (CB'A) under f. If we call H the orthocenter of ABC, a famous lemma tells us that H is on (CB'A), so f((AOC)) = (AHC). Let D be the foot of the B-altitude in ABC. The inversion centered at B which swaps H and D swaps (AHC) with the nine point circle of ABC, so in particular BN passes through the center of (AHC).
So, BO_B swaps with BN under f. It follows that these two lines are isogonal.
QED