HYACINTHOS 3841

 

Antreas P. Hatzipolakis

 

We are given three collinear points A, A', A" (in that order).

Which is the locus of the centroids of the triangles ABC, which have:

AA' = A-cevian of H (ie AA' is the A-altitude)

AA" = A-circumcevian of H (ie A" is the [other than A] intersection of
AA' and the circumcircle).

Analysis:

A
/\
/ \
/ \
/ \
/ \
/ H \
/ \
/ G G' \
/ O O' \
B----------A'------C

 

A"

The orthocenter H is the reflection of A" in A', therefore it is known.

Let O', G' be the orth. projections of O, G (resp.) on AA".

OO' is the perp. bisector of AA" ==> O' is the midpoint of AA": known point.

Now, HG / HO = HG' / HO' = 2/3 ==> G' is known point.

Therefore, the locus of G is the perpendicular to AA" at the
known point point G'.

Variations: Replace H with I, O. Namely:

1. AA' = A-cevian of I (ie AA' is the A-bisector)

AA" = A-circumcevian of I (ie A" is the [other than A] intersection
of AA' and the circumcircle).

2. AA' = A-cevian of O

AA" = diameter of the circumcircle (through A').

Which are the loci of G's in these cases?

Note: The variation #1 is a problem in I. Panakis' collection
of 2500 geom. loci [in Greek]

APH