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 ABC and a point P, suppose, A',B',C' are the reflections of P on BC,CA,AB. Prove that AA',BB',CC' are concurrnt iff PP^* is parallel to the Euler line of ABC where P^* is the isogonal conjugate of P wrt ABC.

The locus of such point P is the Neuberg Cubic of ABC. But in our proof we will not use any property of cubics. So let’s call the locus of points P 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 ABC, if a point P lies on the Neuberg locus of ABC, then isogonal conjugate of P wrt ABC also lies on the Neuberg locus of ABC.

Proof.

Suppose, A',B',C' are the reflections of P on BC,CA,AB. Suppose, A_1=PA'\cap \odot A'B'C'. Similarly define B_1,C_1. Under inversion wrt P with power PA'.PA_1, A goes to the reflection of P on B_1C_1 and similar for others. Now note that, if A_2,B_2,C_2 are the reflections of P^* on BC,CA,AB, then \Delta A_2B_2C_2 is homothetic to \Delta A_1B_1C_1. So if A_3,B_3,C_3 are the reflections of P^* on B_2C_2, C_2A_2,A_2B_2, then \odot P^*A_2A_3, \odot P^*B_2B_3, \odot P^*C_2C_3 are co-axial. But the center of \odot P^*A_2A_3 is the intersection point of B_2C_2 and BC and similar for others. So B_2C_2\cap BC,C_2A_2\cap CA,A_2B_2\cap AB are collinear. So by Desargues’ Theorem, \Delta ABC and \Delta A_2B_2C_2 are perspective. So P^* lies on the Neuberg locus of ABC.

\Box
Property 2.

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

Proof.

Suppose, P^* is the isogonal conjugate of P wrt ABC. A',B',C' be the reflections of P^* on BC,CA,AB. Note that, P is the circumcenter of \Delta A'B'C'. P' be the isogonal conjugate of P^* wrt A'B'C'. A_1,B_1,C_1 be the circumcenters of \Delta B'P'C', \Delta C'P'A', \Delta A'P'B'. By some simple angle-chasing we get that PA_1.PA=PA'^2. So A'A_1 is the isogonal conjugate of A'A wrt \angle B'A'C' and similar for others. But from property 1, if P lies on the Neuberg locus of \Delta ABC iff so does P^*. So A'A,B'B,C'C concur. So A'A_1,B'B_1,C'C_1 concur. So P' lies on the Neuberg locus of \Delta A_1B_1C_1. Now note that, if A_2B_2C_2 is the pedal triangle of P wrt ABC and Q is the isogonal conjugate of P wrt A_2B_2C_2, then the configuration A_2B_2C_2Q is homothetic to the configuration A_1B_1C_1P'. So Q lies on the Neuberg locus of \Delta A_2B_2C_2. So again using property 1, we get that P lies on the Neuberg locus of \Delta A_2B_2C_2.

Corollary.

Given a triangle ABC and a point P on its Neuberg Locus, suppose, P_a,P_b,P_c are the circumcenters of \Delta BPC,\Delta CPA,\Delta APB. Prove that, AP_a,BP_b,CP_c are concurrent. Call that concurrency point as Q. If P^* is the isogonal conjugate of P wrt ABC, then define Q^* for P^* similarly. Then Q and Q^* are isogonal conjugates wrt ABC.

\Box
Property 3.

Given a triangle ABC and a point P, prove that P lies on the Neuberg locus of ABC iff the Euler lines of \Delta BPC,\Delta CPA,\Delta APB are concurrent.

Proof.

Let G_a, G_b, G_c be the centroids of the triangles BPC,CPA,APB. and O_a,O_b,O_c are the their circumcenters. Then we have G_bG_c\parallel BC and BG_c\cap CG_b=\text{midpoint}(AP). So, the homothety that maps G_bG_c to BC maps O_bO_c\cap G_bG_c\equiv K_1 to O_bO_c\cap BC\equiv K_1'. Thus, \frac{G_cK_1}{G_bK_1}=\frac{BK'_1}{CK'_1}.
Since the triangles G_aG_bG_c and O_aO_bO_c are perspective, so we get
\prod_{cyc}\frac{G_cK_1}{G_bK_1}=-1,

Which leads to
\prod_{cyc}\frac{BK_1'}{CK_1'}=-1,
so that ABC and O_aO_bO_c are perspective. So using the corollary of Property 2 we get that P lies on the Neuberg cubic of \Delta ABC.

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

Proof.

On the sides of BC,CA,AB take points C_1,A_1,B_1 so that \Delta A_1B_1C_1 is homothetic to \Delta A'B'C'. Through C_1 draw parallel to AC to intersect AB at A_2. Cyclically define B_2,C_2 on BC,AC. Applying the converse of Pascal’s theorem on the hexagon A_1C_2B_1A_2C_1B_2 we get that A_1,B_1,C_1,A_2,B_2,C_2 lie on a conic(call this conic \Gamma(A_1B_1C_1)). A_3=A_2C_1\cap B_2A_1, B_3=B_2A_1\cap B_1C_2, C_3=C_2B_1\cap A_2C_1. Clearly, ABC and A_3,B_3,C_3 are homothetic. So AA_3,BB_3,CC_3 concur. Since A_2A_1 is the harmonic conjugate of AA_3 wrt AB,AC, so if we draw parallels to A_1A_2,B_1B_2,C_1C_2 through A,B,C, then they intersect BC,CA,AB at three collinear points. So there exists a conic passing through A,B,C so that the tangent at A is parallel to A_1A_2 and similar for B,C. Call this conic \Gamma(ABC). Take any point X on \Gamma(ABC). Through A_1 draw a line l_a parallel to AX and similarly define l_b,l_c. Note that (l_a,A_1A_2;A_1C_2,A_1B_2)=(AX,A_1A_2;AC,AB) =(BX,BA;BC,B_1B_2)=(l_b,B_1A_2;B_1C_2,B_1B_2). So l_a\cap l_b lies on \Gamma(A_1B_1C_1) and similar for others. So l_a,l_b,l_c concur on \Gamma(A_1B_1C_1). Its easy to prove that its not true if X doesn’t lie on \Gamma(ABC).

\Box
Lemma 2(Sondat’s Theorem).

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

Proof.

Suppose, AP\perp B'C', BP\perp C'A', CP\perp A'B'. Similarly define P' such that A'P'\perp BC, B'P'\perp CA, C'P'\perp AB. Suppose, AA',BB',CC' concur at some point Q.
Note that, if we define \Gamma(ABC) wrt \Delta A'B'C' using Lemma 1, then we get that, its the rectangular hyperbola passing through A,B,C,P and the orthocenter of ABC. Now note that, BPC and B'P'C' are orthologic and A,A' are the two centers of orthology. Note that, \Gamma(ABC) passes through the orthocenter of \Delta BPC, but \Gamma(BPC) wrt \Delta B'P'C' is the rectangular hyperbola passing through B,C,P,A and the orthocenter of \Delta BPC. So \Gamma(BPC) wrt \Delta B'P'C' is same as \Gamma(ABC) wrt \Delta A'B'C'. But note that, Q lies on \Gamma(ABC). So Q lies on \Gamma(BPC) wrt B'P'C'. So PQ\parallel P'Q. So P,P',Q are collinear.

\Box
Lemma 3.

Given two points P,Q and their isogonal conjugates wrt ABC be P',Q', then suppose, X=PQ'\cap QP' and Y=PQ\cap P'Q', then Y is the isogonal conjugate of X wrt ABC.

Proof.

Consider the projective transformation followed by an affine transformation that takes a point P to Q keeping the triangle ABC fixed. This takes Q' to P' where Q',P' are the isogonal conjugates of Q,P wrt ABC. Because, suppose, AP,AQ,AP',AQ' intersect BC at P_a,Q_a,P'_a,Q'_a. So after the transformation, P_a goes to Q_a. Suppose, Q'_a goes to K, Then (BC;P_aQ'_a)=(BC;Q_aK). So K\equiv P'_a. Similar for other sides, so Q' goes to P'.

So if the aforementioned transformation sends P' to Q, then it sends Q' to P. So (P'A,P'B;P'C,P'Q')=(QA,QB;QC,QP) \implies (P'A,P'B;P'C,P'Y)=(QA,QB;QC,QY). So A,B,C,Q,Y,P' lie on the same conic, which is the isogonal conjugate of PQ' wrt ABC. So isogonal conjugate of Y lies on PQ' and similarly it lies on QP'. So X is the isogonal conjugate of Y.
Also note that, if two points X,Y are on PQ' and PQ, such that X,Y are isogonal conjugates, then X=PQ'\cap QP' and Y=PQ\cap P'Q'.

\Box
Some Observations.

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

Proof.

Suppose, O_a,O_b,O_c are the circumcenters of \Delta BPC, \Delta CPA,\Delta APB. G_a,G_b,G_c be the centroids of \Delta BPC,\Delta CPA,\Delta APB. Suppose, O_aG_a,O_bG_b,O_cG_c concur at some point X. Note that, \Delta O_aO_bO_c and \Delta G_aG_bG_c are perspective. Also note that, the perpendiculars from O_a,O_b,O_c to the sides of \Delta G_aG_bG_c concur at the circumcenter of ABC and the perpendiculars from G_a,G_b,G_c to the sides of O_aO_bO_c concur at the centroid of \Delta ABC. So by Sondat’s Theorem O_aG_a,O_bG_b,O_cG_c concur on the Euler line of ABC.

2. In the corollary of Property 2, we mentioned two points Q,Q^*. Using Sondat’s theorem we get that, Q lies on OP and Q^* lies on OP^*. But since Q and Q^* are isogonal conjugates, so using Lemma 3, we get that Q=OP\cap HP^* and Q^*=OP^*\cap HP where H is the orthocenter of ABC.

\Box

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

Given a triangle ABC and a point P, suppose, P^* is the isogonal copnjugate of P wrt ABC. P_a,P_b,P_c be the circumcenters of \Delta BPC,\Delta CPA,\Delta APB. P^*_a,P^*_b,P^*_c be the circumcenters of \Delta BP^*C,\Delta CP^*A,\Delta AP^*B. Suppose, P lies on the Neuberg locus of \Delta P_aP_bP_c. So AP_a,BP_b,CP_c are concurrent. Clearly, it implies AP^*_a,BP^*_b,CP^*_c are concurrent. Suppose, O is the circumcenter of \Delta ABC. Note that, the triangle formed by the circumcenters of \Delta OP_bP_c,\Delta OP_cP_a,\Delta OP_aP_b is homothetic to \Delta ABC. So if we draw parallels to the Euler lines of \Delta OP_bP_c,\Delta OP_cP_a,\Delta OP_aP_b through A,B,C they will meet at some point Q. Similarly define Q^* for P^*. Now note that, P_bP_c is anti-parallel to P^*_b,P^*_c wrt \angle P_bOP_c. So the Euler line of \Delta OP_bP_c is anti-parallel to the Euler line of \Delta OP^*_bP^*_c wrt \angle P_bOP_c. Similar for others. So Q^* is the isogonal conjugate of Q wrt ABC. Note that, the triangle formed by the centroids of \Delta OP_bP_c,\Delta OP_cP_a,\Delta OP_aP_b is homothetic to \Delta P_aP_bP_c. So if we draw parallels to AQ,BQ,CQ through P_a,P_b,P_c, then they will concur. So using lemma 1 we get that, Q lies on \Gamma(ABC) wrt \Delta P_aP_bP_c which is the rectangular hyperbola passing through A,B,C,H,P where H is the orthocenter of \Delta ABC. So Q^* lies on OP^*. Similarly, Q lies on OP. So using lemma 3, we get that, Q=OP\cap HP^* and Q^*=OP^*\cap HP. Now note that, if R is the concurreny point of AP_a,BP_b,CP_c and R^* is the concurrency point of AP^*_a,BP^*_b,CP^*_c, then from Observation 2, we have R=OP\cap HP^* and R^*=OP^*\cap HP. So R\equiv Q and R^*\equiv Q^*. From Observation 1, we get that the PQ is parallel to the Euler line of \Delta P_aP_bP_c. But now we have, Q lies on OP. O is the isogonal conjugate of P wrt P_aP_bP_c. So the line joining P and isogonal conjugate of P wrt P_aP_bP_c is parallel to the Euler line of P_aP_bP_c. 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 P lies on the Neuberg locus of ABC, then A lies on the Neuberg Locus of BPC and similarly.

Proof.

Instead of ABC, take P_aP_bP_c and the notations in the previous proof. Note that, A is the isogonal conjugate of P_a wrt \Delta OP_bP_c. But AP_a passes through Q, so AP_a is parallel to the Euler line of \Delta OP_bP_c. 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.

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

    • 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.

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

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