29 December 2020

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.)