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The Incenter and Euler's Line


Date: 11/27/2001 at 01:34:49
From: Madison Kaplan
Subject: The Incenter and Euler's Line

Dear Dr. Math,

I am having problems proving why the incenter is not included in 
Euler's line.  I have already proved Euler's line; I just need help 
understanding the logic behind why the Incenter is not included.

-Madison


Date: 11/27/2001 at 06:50:41
From: Doctor Floor
Subject: Re: The Incenter and Euler's Line

Hi, Madison,

Thanks for writing.

We will give a counterexample in which indeed the incenter is not on 
the Euler line. In most triangles the incenter is not on the Euler 
line; the exceptions are isosceles triangles.

Let us consider a simple triangle in a rectangular grid:

 A(0,0)
 B(4,0)
 C(0,3)

This is a right triangle, so the vertex at the right angle is also the 
orthocenter and the midpoint of the hypotenuse is the circumcenter. 
The Euler line is the line through (0,0) and (2,1.5).

The incenter is at point (1,1). This can be seen from the fact that 
the inradius is 1. See for instance from the Dr. Math archives:

   Radius of a Circle Inscribed in a Triangle
   http://mathforum.com/dr.math/problems/michael6.02.99.html   

This point (1,1) is not on the line through (0,0) and (2,1.5). We 
conclude that - in this triangle - the incenter is not on the Euler 
line.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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