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### Nagel Point

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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
archives:

Euler Line and Nagel Point
http://mathforum.org/dr.math/problems/wanwipa7.20.98.html

Incenters, Orthocenters, and the Spieker Point
http://mathforum.org/dr.math/problems/wilson.2.13.00.html

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
http://mathforum.org/dr.math/problems/noble1.6.99.html

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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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