Euler's Nine-point Circle

Date: 02/21/99 at 23:59:43
From: asit bhattacharyya
Subject: Nine-point Circle

Could you give me some information on "nine-point circle"?

Thanks.

Date: 02/28/99 at 06:18:05
From: Doctor Floor
Subject: Re: Nine-point Circle

This is a very famous triangle problem that was presented by the Swiss 
mathematician Euler in 1765. 

Let ABC be a triangle. Let A~ be the midpoint of BC, B~ of AC, and C~ 
of AB. Also let A' be the foot of the altitude from A on BC, and let 
B' and C' be the feet of the other two altitudes. The three altitudes 
intersect at a common point, the orthocenter H. Finally, let A' be the 
midpoint of AH, B" of BH and C" of CH.

The nine-point circle is the circle passing through the nine points 
A~, B~, C~, A', B', C', A", B" and C":
 


Two arguments are needed to prove that these nine points are indeed 
concyclic:

1. First consider quadrilateral A'A~B~C~. This is a trapezoid, because 
A'A~ is parallel to B~C~. Note that A'C~ = AC~ = AB/2 and that A~B~ = 
AB/2 too, so A'A~B~C~ is an isosceles trapezoid. An isosceles trapezoid 
can be circumscribed by a circle, so the circle circumscribing A~B~C~ 
passes through A'. In the same way, it passes through B' and C'.

2. Observe that C~A" is parallel to BH and thus parallel to BB' too. 
Also C~A~ is parallel to AC. But this gives us that angle A~C~A" is 
right. Since angle A"A'A~ is right too, we know that the circle with 
diameter A"A~ passes through  both C~ and A'. This must be the same 
circle as in our first argument, because this circle passes through 
A~, C~, and A'. So our circle from argument 1. passes through A", and 
in the same way through B" and C".

We can see that triangle A"B"C" is the product figure of ABC with 
factor 0.5 over H. This means that the center of the nine point circle 
N is the midpoint of the circumcenter and the orthocenter. This also 
means that N must be on the Euler line.

You can read about the Euler line in our archives:

  Euler Line
  http://mathforum.org/dr.math/problems/christen6.8.98.html   

A very nice result on the nine point circle is proven by Feuerbach in 
1822: it is tangent to the incircle and the excircles of a triangle.

The center of the nine point circle, also called the nine point center, 
is a very well-known triangle center. A lot of triangle centers are 
presented in a site by Professor Clark Kimberling of the University of 
Evansville:

  Triangle Centers
  http://cedar.evansville.edu/~ck6/tcenters/   

I hope this helps!

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/