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Nagel Point

Date: 04/15/2001 at 19:54:00
From: Max Levine
Subject: Nagel Point

What relation does the Nagel Point have to the incenter, centroid, and 
Spieker point of a triangle, and why? Please give a hint at how to 
prove it, but not necessarily the proof.

Date: 04/17/2001 at 04:59:38
From: Doctor Floor
Subject: Re: Nagel Point

Hi, Max,

Thanks for writing.

First of all, I would like to point you to two answers in the Dr. Math 

   Euler Line and Nagel Point

   Incenters, Orthocenters, and the Spieker Point

The second shows that the Spieker point (which I call now S) is the 
incenter of the triangle formed by the midpoints of ABC. You can prove 
that the incenter (I), the Spieker point S, and the centroid (G) are 
collinear in the same way the Euler line is proven in the first 

The Nagel point (N) is collinear with all of these. The easiest way to 
prove it is to find barycentric coordinates for G, I, and N, and show 
that either of them is the linear combination of the others. For some 
background information, see from the dr. Math archives:

   Barycentric Calculus

From this message you find that the barycentric coordinates for I and 
G are:

     I = a:b:c
     G = 1:1:1

With the help of the very first reference about the Nagel point, it 
shouldn't be too difficult to show that the Nagel point is given by:

     N = s-a : s-b : s-c

These coordinates can are clearly the linear combination s*G - I. That 
proves that N, I, G are collinear, and thus N, I, G, S are collinear.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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