Synthetic Proof of the Neuberg Locus Properties (without Barycentrics)

I had been trying this problem jointly with Chandan Banerjee, better known as RSM in the Forum, for the past month. The different properties of the Neuberg Cubic/Locus are intriguing, and last night, finally we were able to prove this problem, though I would say that the key idea is due to RSM.

Shortly we plan to write an article on this problem, and for the time being, here’s the proof, complete with the problem statement etc. Thanks to RSM for posting this in his blog, and Hyacinthos. Also, I am sorry for just copy-pasting the post from RSM’s blog. 😀

Neuberg Locus Problem.

Given a triangle and a point , suppose, are the reflections of on . Prove that are concurrnt iff is parallel to the Euler line of where is the isogonal conjugate of wrt .

The locus of such point is the Neuberg Cubic of . But in our proof we will not use any property of cubics. So let’s call the locus of points which satisfies the above concurrency fact as Neuberg locus(we are avoiding the term cubic).

At first we will prove 3 properties of the Neuberg locus. The properties are as follows:-

Property 1.

Given a triangle , if a point lies on the Neuberg locus of , then isogonal conjugate of wrt also lies on the Neuberg locus of .

Proof.

Suppose, are the reflections of on . Suppose, . Similarly define . Under inversion wrt with power , goes to the reflection of on and similar for others. Now note that, if are the reflections of on , then is homothetic to . So if are the reflections of on , then are co-axial. But the center of is the intersection point of and and similar for others. So are collinear. So by Desargues’ Theorem, and are perspective. So lies on the Neuberg locus of .

$\Box$
Property 2.

Given a triangle and a point , prove that lies on the Neuberg locus of iff lies on the Neuberg locus of the pedal triangle of wrt .

Proof.

Suppose, is the isogonal conjugate of wrt . be the reflections of on . Note that, is the circumcenter of . be the isogonal conjugate of wrt . be the circumcenters of . By some simple angle-chasing we get that . So is the isogonal conjugate of wrt and similar for others. But from property 1, if lies on the Neuberg locus of iff so does . So concur. So concur. So lies on the Neuberg locus of . Now note that, if is the pedal triangle of wrt and is the isogonal conjugate of wrt , then the configuration is homothetic to the configuration . So lies on the Neuberg locus of . So again using property 1, we get that lies on the Neuberg locus of .

Corollary.

Given a triangle and a point on its Neuberg Locus, suppose, are the circumcenters of . Prove that, are concurrent. Call that concurrency point as . If is the isogonal conjugate of wrt , then define for similarly. Then and are isogonal conjugates wrt .

$\Box$
Property 3.

Given a triangle and a point , prove that lies on the Neuberg locus of iff the Euler lines of are concurrent.

Proof.

Let be the centroids of the triangles and are the their circumcenters. Then we have and So, the homothety that maps to maps to Thus,
Since the triangles and are perspective, so we get

so that and are perspective. So using the corollary of Property 2 we get that lies on the Neuberg cubic of .

$\Box$
Before going to the proof of the main problem we will mention some lemmas that we will use in the proof.

Lemma 1.

Given two triangles and (not homothetic), consider the set of all triangles(no two of them homothetic), so that both and are orthologic to them. Then the locus of the center of orthology of and this set of triangles lie on a conic passing through or the line at infinity.

Proof.

On the sides of take points so that is homothetic to . Through draw parallel to to intersect at . Cyclically define on . Applying the converse of Pascal’s theorem on the hexagon we get that lie on a conic(call this conic ). . Clearly, and are homothetic. So concur. Since is the harmonic conjugate of wrt , so if we draw parallels to through , then they intersect at three collinear points. So there exists a conic passing through so that the tangent at is parallel to and similar for . Call this conic . Take any point on . Through draw a line parallel to and similarly define . Note that . So lies on and similar for others. So concur on . Its easy to prove that its not true if doesn’t lie on .

$\Box$
Lemma 2(Sondat’s Theorem).

Given two triangles and , such that they are orthologic and perspective, prove that the two centers of orthology are collinear with the center of perspectivity.

Proof.

Suppose, . Similarly define such that . Suppose, concur at some point .
Note that, if we define wrt using Lemma 1, then we get that, its the rectangular hyperbola passing through and the orthocenter of . Now note that, and are orthologic and are the two centers of orthology. Note that, passes through the orthocenter of , but wrt is the rectangular hyperbola passing through and the orthocenter of . So wrt is same as wrt . But note that, lies on . So lies on wrt . So . So are collinear.

$\Box$
Lemma 3.

Given two points and their isogonal conjugates wrt be , then suppose, and , then is the isogonal conjugate of wrt .

Proof.

Consider the projective transformation followed by an affine transformation that takes a point to keeping the triangle fixed. This takes to where are the isogonal conjugates of wrt . Because, suppose, intersect at . So after the transformation, goes to . Suppose, goes to , Then . So . Similar for other sides, so goes to .

So if the aforementioned transformation sends to , then it sends to . So . So lie on the same conic, which is the isogonal conjugate of wrt . So isogonal conjugate of lies on and similarly it lies on . So is the isogonal conjugate of .
Also note that, if two points are on and , such that are isogonal conjugates, then and .

$\Box$
Some Observations.

1. Given a triangle and a point , if the Euler lines of are concurrent, then they concur on the Euler line of .

Proof.

Suppose, are the circumcenters of . be the centroids of . Suppose, concur at some point . Note that, and are perspective. Also note that, the perpendiculars from to the sides of concur at the circumcenter of and the perpendiculars from to the sides of concur at the centroid of . So by Sondat’s Theorem concur on the Euler line of .

2. In the corollary of Property 2, we mentioned two points . Using Sondat’s theorem we get that, lies on and lies on . But since and are isogonal conjugates, so using Lemma 3, we get that and where is the orthocenter of .

$\Box$

Let us finally come to the crux of the problem.
Proof of The Neuberg Problem:-

Given a triangle and a point , suppose, is the isogonal copnjugate of wrt . be the circumcenters of . be the circumcenters of . Suppose, lies on the Neuberg locus of . So are concurrent. Clearly, it implies are concurrent. Suppose, is the circumcenter of . Note that, the triangle formed by the circumcenters of is homothetic to . So if we draw parallels to the Euler lines of through they will meet at some point . Similarly define for . Now note that, is anti-parallel to wrt . So the Euler line of is anti-parallel to the Euler line of wrt . Similar for others. So is the isogonal conjugate of wrt . Note that, the triangle formed by the centroids of is homothetic to . So if we draw parallels to through , then they will concur. So using lemma 1 we get that, lies on wrt which is the rectangular hyperbola passing through where is the orthocenter of . So lies on . Similarly, lies on . So using lemma 3, we get that, and . Now note that, if is the concurreny point of and is the concurrency point of , then from Observation 2, we have and . So and . From Observation 1, we get that the is parallel to the Euler line of . But now we have, lies on . is the isogonal conjugate of wrt . So the line joining and isogonal conjugate of wrt is parallel to the Euler line of . So done.

Comment by RSM: The proof may look long, but the idea is very small. You can easily realize that after reading the proof.

$\Box$
Property 4.

If lies on the Neuberg locus of , then lies on the Neuberg Locus of and similarly.

Proof.

Instead of , take and the notations in the previous proof. Note that, is the isogonal conjugate of wrt . But passes through , so is parallel to the Euler line of . So done.

$\Box$

We will come back with more properties of the Neuberg Locus after we start writing the article. Till then, rejoice! An open problem has finally been solved. 🙂

REFERENCES.
1.  The Neuberg Cubic in Locus Problems by Zvonko Cerin.
2. Bernard Gilbert’s Page on Neuberg Cubic.
3. Concurrency of four Euler lines by Antreas P. Hatzipolakis, Floor van Lamoen, Barry Wolk, and Paul Yiu.

6 thoughts on “Synthetic Proof of the Neuberg Locus Properties (without Barycentrics)”

1. Wow!! Such elementary proofs are unheard of in this realm. Truly nice and wonderful. 🙂

• Potla says:

Thanks, but the main idea was of RSM, and I made a lot of useless conjectures (using SCT, ofc :P), which will come up in the article.

2. I learned lots about the alphabet, specifically the letters A, Q, and P. Many thanks for the tl;dr wonderfulness.

• Potla says:

You’re welcome, but the main point was to show the conics through $A,Q,P$ etc and not the alphabets themselves. 😛 LOL

3. Goutham says:

Yes, it was fun scrolling down the page. 😀

• Potla says:

Yep. And it was fun copy-pasting from RSM’s blog. 😀