Centroid, Circumcenter, Orthocenter Generalized

Centers Inside and Outside the Triangle

Here you can see things are going wrong. While the average or centroid is within the triangle the other centers are not. However they are all meeting at a unique point!

The Circumcenter Generalized

The circumcenter is defined as the point where the other vertexes are of equal distance away. So in 2D its the center of a circle, where the circle goes through three points. In 3D its the center of a sphere which goes through four points.

An easy way to calculate this is in 2D is with vectors. Make two lines at right angles and in same plane as the triangle, find their intersection where the lines go through the midpoints. The vectors work in 3D with the same reasoning and using crossproducts to get perpendicular vectors.

To show that this is unique look at the equations. Since there are as many equations as unknowns there is always one and only one unique solution.

Unknown variables $r$ , $x$ , $y$
$(a_{0} - x)^{2} + (a_{1} - x)^{2} = r^{2}$
$(b_{0} - x)^{2} + (b_{1} - x)^{2} = r^{2}$
$(c_{0} - x)^{2} + (c_{1} - x)^{2} = r^{2}$
for the points
$(x, y)$ , $(a_{0}, a_{1})$ , $(b_{0}, b_{1})$ , $(c_{0}, c_{1})$

Sphere Through Four Points

In 3D the situation is similar with a sphere through the four points of the tetrahedron.

I was having difficulty solving this situation when finally a really simple idea occurred (after much garbage). Translate all the points so that one of them is at the origin. Solve the system and translate the solution back.

Unknown variables $r$ , $a$ , $b$ , $c$
$(x_{i} - a)^{2} + (y_{i} - b)^{2} + (z_{i} - c)^{2} = r^{2}$
$i = 0, 1, 2, 3$

Translate points by X₀.

$X_{i} = X_{i} - X_{0}$
$i = 1, 2, 3$

Expand the general equation out and simplify

$x_{i}^{2} - 2 x_{i} a + a^{2} + y_{i}^{2} - 2 y_{i} b + b^{2} + z_{i}^{2} - 2 z_{i} c + c^{2} = r^{2}$

$\frac{1}{2} (x_{i}^{2} + y_{i}^{2} + z_{i}^{2}) = x_{i} a + y_{i} b + z_{i} c$
$i = 1, 2, 3$

Solve this linear system of equations for a, b and c. The center is found by translating the solution back.

$(a, b, c) + X_{0}$

The Inner Circle

Another measure of a triangles center is the point where the smallest circle inside the triangle fits. This was not put in with the others because in classical geometry(from my very limited knowledge) it does not go through the famous straight line like the others do. However I will at a later date read up on it and the centroids relationships.

The inner circles center was found by bisecting the angles. As you can see these lines meet at a unique point too. See Bisecting an Angle with Ratios.

Tetrahedron Centers

Clearly the centroid is a tetrahedron center.

Here are some other possible centers.

Equilateral Tetrahedron Derived Centers

Firstly I am defining an equilateral tetrahedron as taking a triangle and extruding it into 3D as a tetrahedron, the lengths of the new sides are the lengths of the opposite sides.

Construct a line from the equilateral triangles opposite vertex to the newely constructed point. Repeat of all tetrahedron's triangular faces. Find the point which minimized the distance to these four lines. This is similar in spirit to 2D Fermat point.
Construct a line from the quilateral's centroid to the opposite vertex. Repeat of all tetrahedron's triangular faces. Find the point which minimized the distance to these four lines. This is similar in spirit to 2D Napolean point.