Triangles on HO

Source

Triangles on HO, source

The problem has been shared by Miguel Ochoa Sanchez at the Peru Geometrico facebook group.

Remark and Reformulation

As stated, the problem is incorrect: It is not always true that $[\Delta AHO]=[\Delta BHO]+[\Delta CHO].\,$ The suspicion may be aroused by the lack of symmetry in the formulation: both the orthocenter $H\,$ and the circumcenter $O\,$ are symmetric functions of the vertices of $\Delta ABC.\,$ There is no reason to single out one of the vertices. The improved formulation is as follows:

Assume $[XYZ]\,$ denotes the signed area of $\Delta XYZ.\,$ Then, with $H\,$ the orthocenter and $O\,$ the circumcenter of $\Delta ABC,\,$

$[AHO]+[BHO]+[CHO]=0.$

A Better Formulation

The problem, as stated above, appears somewhat artificial. Let $h_A,\,$ $h_B,\,$ $h_C,\,$ be the altitudes of the three triangle involved to the line $HO.\,$ Since all three have the same base $(HO)\,$ the problem reduce to proving that

$h_A+h_B+h_C=0.$

(This in fact explains the idea of signed areas: depending on the position of the vertices relative to the line $HO,\,$ the altitudes may be either positive or negative.)

Remarkably, the above condition does not depend on the specific common base of the three triangles. As the line $HO\,$ is the Euler line of $\Delta ABC,\,$ it is obvious that for any two points $P,Q\,$ on the Euler line we'll have

$[APQ]+[BPQ]+[CPQ]=0.$

Thus, inclusion of $H\,$ and $O\,$ in the formulation is a red herring. But this is not the only red herring in the formulation. What makes the Euler line so special as to possess such a nice property? To make sure, the Euler line is a very special construct in every triangle, but the condition $h_A+h_B+h_C=0\,$ seems too general to be strictly and rigidly associated with the Euler line. Indeed, if we imagine unit masses placed on the vertices of the triangles, the equation $h_A+h_B+h_C=0\,$ will simply assert that the moments of the system of three masses with respect to $HO\,$ is zero, i.e., the system is in equilibrium. This observation removes the mystery in the appearance of the Euler line. The Euler line passes through $H,\,$ $O,\,$ but - in this case - more importantly - through $G,\,$ the centroid, i.e., the barycenter of the triangle. For every line though the centroid of $\Delta ABC,\,$ the required condition will hold!

Assume line $\ell\,$ passes through the centroid $G\,$ of $\Delta ABC.\,$ Let $h_A,\,$ $h_B,\,$ $h_C,\,$ denote the signed distance from the vertices $A,\,$ $B,\,$ and $C\,$ to the line $\ell.\,$ Then

$h_A+h_B+h_C=0.$

 

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