Gábor Pete, my old school friend
(even if we never attended the same class—except A.
Krámli's lectures on statistics)
was very active in geometry during his high school years. One of his experiments
resulted in discovering a beautiful property of the joining segment of the orthocenters
in two special triangles in a general triangle.

Precisely, given an arbitrary value \(\lambda\in[0,1]\) and we consider
the two triangles \(GH'I\) and \(G'HI'\), where \[BG:GC=CH':H'A=AI:IB=BI':I'A=AH:HC=CG':G'B=\lambda.\]
For these two triangles we create their orthocenters \(J\) and \(K\).

Gábor conjectured that the midpoint \(N\) of \(JK\) always lies on the
Euler line
of the triangle \(ABC\). Moreover, the segments \(JK\) designate a set of parallel segments.
By dragging point \(G\) you can visually verify these statements.

By using GeoGebra's *LocusEquation* command it is a simple 21st century experiment
to identify that the locus of point \(N\) is usually a line. Here we highlight that
point \(G'\) was created by using the length preserving property of point reflection (about
midpoint \(D\)), and point
\(H\) was constructed by exploiting the intercept
theorem. Similar considerations can
lead to points \(H'\), \(I\) and \(I'\). These tricks are important to ensure that
GeoGebra's symbolic features can be used in a later explained proof that is performed
by the machine. Yes, *your* machine will perform the automated proof!

But first, let us double check that the locus is indeed a line:

\[x-y=1.\]

By dragging point \(A\), \(B\) or \(C\) you can do a short experiment and see how
the equation changes, but it remains linear (unless you are unfair and make the
triangle degenerate). In fact, the geometric locus is not a complete line, but the underlying
theory can only deliver a so-called Zariski
closure of the sought set, since here always an algebraic equation is computed.

To actually *prove* the first part of the statement you need to click on
the *Relation*
tool, then select point \(N\) and the Euler line.
A numerical check will be quickly performed. Then, by clicking on **More...**,
a symbolic process will be started. The computation may take several seconds
on your computer (even 1 minute is quite usual on a modern PC),
making this page unresponsive for that time.
But if you are
lucky (and maybe patient enough) your computer will indeed perform a symbolic
verification that literally means that you obtained a proof. The proof is not
shown: it consists of several thousands of very complicated algebraic steps,
that are just additions, subtractions, multiplications and divisions, but
with very long rational numbers and several terms of high degree of \(x\) and \(y\).
Clearly this is not what you want! (You only get a notification that the proof
was successfully performed and the statement holds in general. In this particular
case generality means that points \(A\) and \(B\) should differ.)
(Note: As of May 2021, the proof does not work because of an out of memory exception (OOM),
in Google Chrome. You can perform a numerical check only.)

Here comes Gábor's classmate and my old friend, András
Dőtsch, into the picture.
He proved that Gábor's conjectures are correct. His proof was
published when
we all three were between 22 and 23 years old, in the Hungarian journal
Polygon.
At the time of writing this blog entry, we are 22-23 years *after* that event—András will celebrate
his 45th birthday very soon. I have just found a
DVI version
of András's original proof, and a
plain \(\TeX\) version
can also be accessed, both in Hungarian.

Let me mention Gábor's and András's high school mathematics teacher,
Tamás Tarcsay as well,
a now retired supporter of gifted students.
His open minded style was always a great influence for young learners to do and love
mathematics. His 2002
contribution (on a Hungarian education website) summarizes the background of the above statement,
also called Pete-Dőtsch theorem (by me, at least).

The locus of points \(J\) and \(K\) are also interesting. It turns out that it can be described
by a cubic algebraic curve.

The corresponding algebraic curve is

\[112x^{3} + 102x^{2} y - 719x^{2} + 84x y^{2} - 478x y + 1518x + 40y^{3} - 272y^{2} + 636y = 1080.\]

Feel free to drag point \(A\), \(B\) or \(C\) to get a different output. Note that
here we obtain the algebraic closure of the locus, that is, the whole cubic, not just
those points that play a geometric role. (In fact, if we allow \(\lambda\in\mathbb{R}\),
all points of the cubic will be covered by the locus.)

Above (if you were lucky and patient enough) you managed to prove Gábor Pete's first statement.
The second statement can also be proven by automated means, but here you need a bit more
computational time.

Segment \(TU\) represents the segment \(JK\) for a certain \(\lambda\) value, here \(\lambda=1/4\).
Now it is possible to compare all other segments \(JK\) to this particular segment.
When comparing \(JK\) and \(TU\) about *parallelism*, after a couple of minutes you should be
able to get the symbolic answer. (Important: Do not try the other comparison that is shown
in the first appearing window, namely,
the one between the lengths of \(JK\) and \(TU\). This computation is much more difficult
and will not result in any reasonable answer.)
(Note: As of May 2021, the proof does not work because of an out of memory exception (OOM),
in Google Chrome. You can perform a numerical check only.)

- 18 November 2020—Ellipsograph of Archimedes as a simple LEGO construction
- 17 November 2020—Offsets of a trifolium
- 11 November 2020—Explore envelopes easily!
- 31 October 2020—Points attached to an algebraic curve…
- 19 October 2020—Detection of perpendicular lines…
- 6 October 2020—Better language support…
- 29 September 2020—A new GeoGebra version with better angle bisectors…
- 28 September 2020—I restart my blog…
- …newest entry (as of 16 April 2023)

Zoltán KovácsLinz School of Education Johannes Kepler University Altenberger Strasse 69 A-4040 Linz |