The Incenter and Euler's LineDate: 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/ |
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