# For every three points on a line, does there exist a triangle such that the three points are the orthocenter, circumcenter and centroid?

The Euler line states that the orthocenter, circumcenter and centroid of a given triangle are on one line. This made me wondering whether the following is true:

For every three points on a line (not necessarily different), does there exist a triangle such that the three points are the orthocenter, circumcenter and centroid?

I already eliminated the case that two points coincide and the third not, because if two points coincide, the the third point is also on that point. I assume that if two points are very close and the third is far away, it is not possible, but I'm not able to show it.

I thought of this problem myself.

• No. the distance between the orthocenter and the centroid must be twice the distance between the centroid and the circumcenter: jwilson.coe.uga.edu/EMAT6680Fa09/Rosonet/Rosonet_Assignment4/…
– lulu
Nov 22, 2015 at 19:07
• @lulu You might want to post that as an answer, to keep the question form the unanswered list. Nov 22, 2015 at 19:32

To repeat the contents of an earlier comment: there is a very tight relation between these points. Indeed, the distance between the orthocenter and the centroid is always twice that between the centroid and the circumcenter. This is not obvious, but neither is the argument terribly difficult. A good reference for it can be found here.

For every three points on a line $($not necessarily different$),$ does there exist a triangle,
such that the three points are the orthocenter, circumcenter and centroid ?

No, because the three are not independent$^1$ of one another !

You already know that the centroid G splits all three medians in a ratio of $1:2$, right ?
Well, it does the same with the segment determined by the other two centers, OH.

$^1$ “Well, how come ?”, you might ask. Because, when you think about it, the circumcenter
O is nothing more than a “combination” between the centroid and the orthocenter, since
the perpendicular bisectors which determine the former are themselves a “combination”
between the medians and altitudes which determine the latter two. Just like the altitudes,
the perpendicular bisectors are perpendicular on the three sides; and, just like the medians,
the perpendicular bisectors pass through the midpoint of each segment. I used the term
“combination” deliberately, since there is just such a concept in mathematics called linear
combination
, whose geometrical interpretation is, quite unsurprisingly, a line. In this case,
the Euler line.