Properties of the Neuberg Locus and Brocard Problem

June 27, 2012 by Potla

Our article is finally complete. Although RSM did the major stuff, we were able to crash it all in two days and one night.

The pdf file has been uploaded to Mediafire here. (2.05 MB)

Note that the file will be updated as soon as we get some new stuff to add there. In other words, we welcome people to proofread the file for possible typos and other errors.

Well, nothing else to say. Explore the properties of Neuberg Cubic with your own eyes, and comments, suggestions, feedback etc are obviously welcome. \o/

Synthetic Proof of the Neuberg Locus Properties (without Barycentrics)

June 18, 2012 by Potla

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.

On The Maximization of Prod_{cyc}|(a-b)|

June 2, 2012 by Potla

This is my third time submitting an article to the University of Computer Science, Romania. This time, I wrote the article on something new. Although the inequalities are somewhat classical-looking, I have tried to explain the logic behind maximization of the product, for nonnegative reals $a,b,c.$ I have attached the full article with this post, for those who are interested in seeing the whole article.

This article is currently at v1.0, so please contact me in case any changes are necessary. 🙂 The first part of the article:

1. Introduction

Most inequalities that we come across have a simple type of equality case, ie $a=b=c$ or $a=b,c=0.$ But when we think of maximizing the product $(a-b)(b-c)(c-a)$ or its magnitude (both mean the same), the equality case cannot be as simple as $(1,1,1)$ because any two equal values of $a$ and $b$ can change the value of this expression directly to zero. Of course, considering the same thing over nonnegative reals and whole of reals is much different. So these types of inequalities usually have different equality cases which are not so obvious to determine. However, we will demonstrate how most of these inequalities can be easily solved with the help of AM-GM Inequality. After that, we will try to determine some general forms of this maximum, for different fixed parameters.

1.1 Pre-requisites

Here we assume that the reader has a good idea of the different ways of applying the AM-GM inequality, and the Cauchy-Schwarz inequality. These are the most basic and classical inequalities having monstrous applications. Also some basic knowledge about the $pqr$ and $uvw$ methods might come in handy, so that the reader can relate between these stuff easily. Basic knowledge on calculus, derivatives and solving polynomial equations is also required.

2. The $|(a-b)(b-c)(c-a)|$ problem

We will discuss this method by considering a well-known problem of Tran Quoc Anh(forum name, Nguoivn) which runs as follows.

If $a,b,c\geq 0,$ then we have $\displaystyle (a+b+c)^3\geq 6\sqrt 3 |(a-b)(b-c)(c-a)|.$

2.1. Determining the Equality Case

Firstly, we have to order $a,b,c$ so as to remove the modulus sign. Also, note that in these inequalities, it’s important to figure out an equality case somehow. In most of the cases, the equality occurs when one of the variables is set to zero. This is due to the nature of $(c-a)(c-b)(b-a)$ (assuming $c\geq b\geq a.$ ) When $a$ is the minimum of the three numbers, we see that the expression itself is a decreasing function in terms of $a,$ if we keep $b,c$ fixed. So this expression will be able to reach its maximum for a given value of $b,c$ if and only if $a$ is as small as possible, which is zero itself. Plugging in $a=0$ and $c=bt$ leads us to the cubic equation

$(t+1)^3=6\sqrt 3 t(t-1).$

A factorisation of this cubic expression gives us

$\displaystyle \left(t-\sqrt 3 -2\right)^2(t+7-4\sqrt{3})=0;$

Which leads us to the positive double root $t=\sqrt 3 +2.$

2.2. Significance of the Double Root

The inequality that we get after putting $a=0$ under the assumption $c\geq b\geq a$ is

$\displaystyle (b+c)^3\geq 6\sqrt{3}bc(c-b).$

Note that in this inequality, when the equality occurs for some value of $\dfrac cb,$ the curve $(t+1)^3-6\sqrt 3 t(t-1)$ must be just touching the $t$ axis at that point, so as to satisfy the inequality. Another consequence of this concept is that the other root will be on the negative side of the $t$ axis. Now, we can differentiate this equation to get the double root easily.

$\displaystyle \left[(t+1)^3-6\sqrt 3 t(t-1)\right]'=3(t+1)^2-12\sqrt 3 t-1=3(t-2-\sqrt 3)(t+4-3\sqrt 3).$

Now, we can check that which one of these is our double root by just plugging in these values into the previous equation.

2.3. The Equality Case

If we see that $c=\left(\sqrt{3}+2\right)b$ satisfies the conditions of equality, then the actual equality case turns out to be

$\displaystyle \boxed{(a,b,c)=\left(0,1,(2+\sqrt 3)\right)\equiv \left(0,\sqrt 3 -1 , \sqrt 3 +1\right)}$

The writing as conjugate surds is for aesthetic satisfaction.

2.4. Determining the Coefficients for AM-GM

Once we have determined the equality case, the rest is somewhat easy. Assuming that $c\geq b\geq a$ helps us writing $\displaystyle |(a-b)(b-c)(c-a)|\leq bc(c-b);$

And $(a+b+c)\geq (b+c);$ So that it is sufficient to check that

$\displaystyle (b+c)^3\geq 6\sqrt 3 bc(c-b).$

For this, we need to apply AM-GM in such a way that the three terms we choose on the right hand side are balanced. Using the equality case, we see that $(\sqrt 3+1)b=(\sqrt 3-1)c=c-b=2,$

And therefore the terms have been determined.

$\displaystyle 2bc(c-b)=\left[\left(\sqrt 3+1\right)b\right]\left[\left(\sqrt3-1\right)c\right]\left[(c-b)\right]\leq \frac{\left\{\sqrt 3(b+c)\right\}^3}{27}=\frac{(b+c)^3}{3\sqrt 3}.$

Since we have determined the equality case, therefore we need not worry about the right hand side. This method works mostly without fail, and is easy to implement. So, the method has worked out for $6\sqrt 3.$ We will discuss some more applications of this method. However, note that higher the degree of the inequality, the more unusable this method becomes.

3. Determination of Best Constants

In cases, these $|(a-b)(b-c)(c-a)|$ problems come with some determination of the “best constant” phrases. What it actually means is, say, we were talking of the inequality

$\displaystyle (a+b+c)^3\geq k|(a-b)(b-c)(c-a)|.$

Where $k$ is some positive real number, and $a,b,c\geq 0.$ In such a problem, they want us to figure out the largest value that $k$ can attain, because the inequality implies $k\leq\dfrac{(a+b+c)^3}{|(a-b)(b-c)(c-a)|}.$ If this is satisfied for all $a,b,c\geq 0,$ then we might try to minimize the right side first by minimizing $a,$ ie putting $a=0$ and $c=tb$ where $t\geq 1.$ Then this rewrites into

$\displaystyle k\leq \frac{(t+1)^3}{t(t-1)}.$

This is a single-variable inequality, so we may just want to differentiate this to find the minimum of $f(t)=\dfrac{(t+1)^3}{t(t+1)}.$ Simple calculus gives us the value $k=6\sqrt 3$ for $t=2+\sqrt 3.$ This way can also help us determine the equality case as we saw in the previous section. For some applications, we can get our hands on some problems. Note that this is, in cases, not complete enough to prove the original problem. There may be some other paramatrization in the problem, for example, an extra $(ab+bc+ca)$ term may be there, thus affecting the nature of the $|(a-b)(b-c)(c-a)|$ term, which was always decreasing in $\min\{a,b,c\}.$ But, this can always fetch us the best constants, and the equality cases in a packed fashion, so that the application of AM-GM becomes easier.