In this recent post, the following figure was formed by considering two triangular regions in three-dimensional space, where :
The line joining and satisfies
Similarly, the line joining and satisfies
Equating the two expressions gives and from which . The point of intersection is therefore at . By symmetry of this expression the line joining and also passes through this point. This is the point on the plane that is equi-distant from the xy-, yz- and xz- coordinate planes. It is also the central projection of the origin onto the plane along the vector parallel to .
In terms of the original two triangles this point is neither the centroid, incentre, orthocentre, circumcentre nor other commonly encountered triangle centre. Let us find the barycentric coordinates of this point (call it ) in terms of the triangle with vertices at .
The first barycentric coordinate will be the ratio of the area of to the area of . Since and have the same x-coordinate, this will be the ratio of the x-coordinates of to , which is . By symmetry it follows that the barycentric coordinates have the attractive form
Let the side lengths of be . Then by Pythagoras’ theorem, . Hence
By the cosine rule, (where ) which equals from the above expression. Therefore and similarly we obtain . Then
By the sine rule, ( being the circumradius of ) from which . Hence the barycentric coordinates of may be written in non-normalised form as
Comparing this with the coordinates of the orthocentre , the point is known as the square root of the orthocentre (see Theorem 1 of ). Note that the real existence of the point requires to be acute, which it is when . A geometric construction of the square root of a point is given in Section 8.1.2 of .
 Miklós Hoffmann, Paul Yiu. “Moving Central Axonometric Reference Systems”, Journal for Geometry and Graphics, Volume 9 (2005), No. 2, 127–134.