Nagel PointDate: 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 answer. 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/ |
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