# Concurrence of Ten Nine-Point Circles

### What Might This Be About?

### Problem

Given point $P$ in the plane of $\Delta ABC,$ let $H$ be the orthocenter of the triangles; consider three quadrilaterals $ABPH,$ $BCPH,$ and $ACPH.$ Each quadrilateral defines four triangles whose nine-point circles concur at a point, known as the Euler-Poncelet points of the quadrilateral.

Prove that the three Euler-Poncelet points coincide.

### Solution

Solution is wanting; meanwhile you may wonder why the caption says "10 circles" instead of $4\times 3=12.$

Also, it has been observed that, similar to Griffiths' theorem, that point is the same for all points $P$ on a straight line through the circumcenter of $\Delta ABC.$

### Acknowledgment

The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.

### Nine Point Circle

- Nine Point Circle: an Elementary Proof
- Feuerbach's Theorem
- Feuerbach's Theorem: a Proof
- Four 9-Point Circles in a Quadrilateral
- Four Triangles, One Circle
- Hart Circle
- Incidence in Feuerbach's Theorem
- Six Point Circle
- Nine Point Circle
- 6 to 9 Point Circle
- Six Concyclic Points II
- Bevan's Point and Theorem
- Another Property of the 9-Point Circle
- Concurrence of Ten Nine-Point Circles
- Garcia-Feuerbach Collinearity
- Nine Point Center in Square

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny71074922