“Neuberg Locus and its Properties” published in JCGeometry

So, well, after a lot of re-edits and excerpting, and changes, the joint article of Debdyuti and mine on Neuberg Locus has found its place in the Second Volume of the Journal of Classical Geometry. I had to change and redraw all the diagrams in metapost, and now I feel that all the tedious work was worth it. 🙂

I’m personally thankful to A.V.Akopyan for helping out with the \LaTeX and metapost issues of the article. The full article, as appeared in the Volume 2, is uploaded here.

And well, a small update on my current status… I am currently burdened with Board Exams and stuff (Chemistry exam tomorrow), and all these Exams Business will end on 2nd June. Before that, I’m sorry that I can’t update the blog at all. 😦


2012 in review

With nothing better to post, and because I am totally burdened down with different ranges of exams and olympiads, you can be rest assured that this blog won’t get any updates before 19th Jan.

The WordPress.com stats helper monkeys prepared a 2012 annual report for this blog.

Here’s an excerpt:

600 people reached the top of Mt. Everest in 2012. This blog got about 5,100 views in 2012. If every person who reached the top of Mt. Everest viewed this blog, it would have taken 9 years to get that many views.

Click here to see the complete report.

Condition ax+by+cz≤xyz, expression αx+βy+γz

Haha, so I am back after a long time, it feels so nice to write something again on my good old blog. I was kind of busy with school exams and different mock exams for IIT-JEE in my coaching institute (NOT FIITJEE) and also a maths talent exam (which had a total shi**y question paper mostly full of calculations). So after all this business, I am back to maths. It’s not as if I was slacking off too much, but I really got into playing osu!

Today is the first day of Durga Puja, known as Mahalaya; and therefore a holiday. The month of November of a 12th grader in India is the busiest time of the year, full of different exams – competitive and school. So maybe it will be another long wait for this blog before it gets another update.

kk, I guess that concludes my set of excuses for not updating the blog for a long time. However, the title and the topic is totally random. Obviously it’s on inequalities(because it’s random), and it is on a method for solving problems of the type ax+by+cz\leq xyz\implies \alpha x+\beta y+\gamma z\geq k where x,y,z are positive real numbers and a,b,c,\alpha,\beta, \gamma are given constants.


I don’t consider this necessary, but probably I should show one inequality of this kind. This is an excerpt from my “Topic of Tran Quoc Anh” file, which is under development. This inequality probably inspired me to find out the technique of which I am talking now.

Hmm, now you may ask some questions, like how the hell did I come up with the exact coefficients and why did I use 10=9+1 and not 10=1+9. Well, when I was trying to solve it in a way other than how the author came up with the problem, I was trying different ways of reducing the variables. The strategy that I used was to boil down the inequality on 3 variables to 1 variable. The same procedure will work for our general case.

Variable Elimination.

On a totally general note, this is quite useful in non-symmetric inequalities. In fact, this is one definite method which should work always. We try to reduce a n variable inequality to lesser number of variables. One method that people use quite frequently is the Entirely Mixing Variables Method(EMV), and its stronger version: The Stronger Mixing Variables Method(SMV). Now, this one is quite unrelated to MV and is quite a raw form of elimination, but still quite useful.

The General Problem.

Well, it’s obvious that not all a,b,c,\alpha, \beta,\gamma will abide by a minimum value. We consider, for the sake of simplicity that x,y,z,a,b,c,\alpha,\beta,\gamma>0. So, the problem that we will look into is:

Let x,y,z be positive real numbers such that for some constants a,b,c>0 we have ax+by+cz\leq xyz. Then find the conditions for which the expression P(x,y,z)\equiv \alpha x+\beta y+\gamma z has a minimum and also determine the minimum.

The “General” Solution.

Again, we follow the same method as what we used for the example problem.
Note that z\geq \dfrac{ax+by}{xy-c}. Then,

\displaystyle \begin{aligned}P\equiv \alpha x+\beta y+\gamma z&\geq \alpha x+\beta y+\gamma \cdot \frac{ax+by}{xy-c}\\&=\alpha x+\frac{\beta}{x} \cdot xy+\frac{\gamma}{x}\cdot \frac{ax^2+bxy}{xy-c}\\&=\alpha x+ \frac{\beta c+\gamma b}{x}+\frac 1x\left[\beta(xy-c)+\gamma \cdot \frac{ax^2+bc}{xy-c}\right];\end{aligned}

And now we know how to eliminate y from our expression. What we get after applying AM-GM is that

\displaystyle \begin{aligned}P&\geq \alpha x+\frac{\beta c+\gamma b}{x}+2\sqrt{\beta\gamma\left(\frac{ax^2+bc}{x^2}\right)}\\&=\alpha x+\frac{\beta c+\gamma b}{x}+2\sqrt{\beta\gamma\left(a+\frac{bc}{x^2}\right)}.\end{aligned}

Hmm, this is a one-variable inequality. We have been able to generalise our method till this line, but after this I was just lucky to have stumbled upon 10 inside the square root, but right now there is a \beta \gamma. Also, for the equality to hold till this point, we have made a hell lot of assumptions. In each step that we write a \geq, we need to verify for the equality case. So for people who are too lazy to push this thing into an elegant solution, might as well differentiate the expression. But, the AM-GM solution is way more spectacular than a calculus bash; also Cauchy-Schwarz and AM-GM are my favourite inequalities – so anyway let’s continue.

So, we want to find k,l such that ak^2+bcl^2=4\beta \gamma. Assume that we had been able to find some suitable k,l (I am not explaining what “suitable” means right now). Then we can get

\displaystyle \begin{aligned}P&\geq \alpha x+\frac{\beta c+\gamma b}{x}+\sqrt{(ak^2+bcl^2)\left(a+\frac{bc}{x^2}\right)}\\&\geq \alpha x+\frac{\beta c+\gamma b}{x}+ak+\frac{bcl}{x}\\&=ak+\alpha x+\frac{\beta c+\gamma b+ bcl}{x}\\&\geq ak+2\sqrt{\alpha\left(\beta c+\gamma b+bcl\right)}.\end{aligned}

So one thing is for sure. If we can find k,l and if the a,b,c,\alpha,\beta,\gamma all abide by our restrictions, then this must be the required minimum that we are seeking. So now we have to find the constraints.


Start with the first line of the solution. I suddenly wrote z\geq \dfrac{ax+by}{xy-c}, but this will be true only if xy>c. However, xyz\geq ax+by+cz>cz is obvious. But this will help us determine z from x and y, so let’s call this constraint the z– determination constraint.
Now come to the second \geq of the solution. For equality to hold, we must have (xy-c)^2=\dfrac{\gamma}{\beta}\cdot (ax^2+bc). Let’s call this the y– determination constraint.
Now the third prey. For equality in the Cauchy-Schwarz step; ie \displaystyle (ak^2+bcl^2)\left(a+\frac{bc}{x^2}\right)\geq \left(ak+\frac{bcl}{x}\right)^2, we can verify that the condition x must follow is x=\dfrac{k}{l}.
Fourth condition. In the final step, for equality to hold we should have x=\sqrt{\dfrac{\beta c+\gamma b+bcl}{\alpha}}.

The last two constraints will let us solve for k,l. What we get is that
\dfrac{k^2}{l^2}=\dfrac{\beta c+\gamma b +bc l}{\alpha};
Which simplifies to the following cubic in l; which I would like to call the Characteristic cubic of the problem:

\displaystyle l^3+\left(\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}\right)l^2-\frac{4\alpha\beta\gamma}{abc}=0.

Conditions for Minima.

So we have obtained some necessary conditions for the minimum to exist. For example, the real root l_1 of the characteristic equation must be positive; and also the l_1 that we obtain this way must give a corresponding k, which is verified when \dfrac{4\beta\gamma}{bc}\geq l^2. But, note that if l_1 is a positive root of the characteristic cubic then

\displaystyle l_1^2\cdot \frac{\alpha}{a}-\frac{4\alpha\beta\gamma}{abc}< l_1^3+l_1^2\sum_{cyc}\frac{\alpha}{a}-\frac{4\alpha\beta\gamma}{abc}=0; so that any positive root of this cubic will always lead us to a corresponding value of k, thus fixing the x,y,z for which the minimum is attained. So finally what we have to check for is the existence of a positive real root of the characteristic cubic.

Take r=\frac{\alpha}{a}, s=\frac{\beta}{b}, t=\frac{\gamma}{c}. Then the cubic is f(x)=x^3+(r+s+t)x^2-4rst. Now f'(x)=3x^2+2x(r+s+t) so f has two real points of inflection : x=0 and x=-\frac{2}{3}(r+s+t). Now f(0)=-4rst<0 and f is increasing for x>0, so this cubic equation will always have one positive root. It may have two or no negative roots. So any a,b,c,\alpha,\beta, \gamma>0 satisfy and permit such a minimum value.

The value of l which satisfies the cubic equation can be found out to be a huge expression.

So we can take any a,b,c,\alpha,\beta,\gamma we like and then make a new inequality using the characteristic cubic. \Box

This was not too tl;dr, but the method is awesome, so I wrote down this thing in an hour. The next post should be on some other more interesting topic. 🙂

On the Complete Quadrilateral Configurations

Another interesting configuration, this time associated with a more tl;dr post. Firstly, a complete quadrilateral is the figure formed by four lines intersecting at 6 points, and the difference with a complete quadrangle is that this thing has three diagonals.

Before starting, I’d like to present a set of disclaimers.


1. This post is around 40% compilation and 60% observations. I have used geogebra for experimenting with the different configurations, and also searched a lot on the internet for different properties of this configuration. In the meanwhile, there are a lot of properties that I have independently figured out, but some I cannot prove myself.
2. I have mentioned “Awaiting proofs” to whichever property I have not been able to solve. For example, in Property 13, I was not able to prove all of the results thanks to the complexity. If I am able to come up with solutions, I will add them later. However, anyone eager to try is free to try and prove these results. Discussions regarding these are welcome as well.
3. Nothing else. Go ahead and enjoy reading. 🙂

Main Post.
Whenever we talk about complete quadrilaterals, we firstly think of Miquel’s theorem. A really well-known and useful theorem as it is, we will check it out along with some other basic results. I will refer to the following configuration always, ’cause it’s annoying to name each of the points every time I state a new theorem/result. We will just analyse the different properties of this configutation:

Property 1.(Miquel)
Let the lines \ell_1,\ell_2,\ell_3,\ell_4 meet in six points (named or unnamed 😛 ). Then the circumcircles of the four triangles that are formed out of these six vertices concur at a point, known as the Miquel Point, which I will designate as M.

Surely this is well-known and simple angle-chasing. I won’t give a proof for this one, since the post is getting more and more tl;dr everyday!


Property 2.(Miquel Circle Theorem)
The circumcentres O_1,O_2,O_3,O_4 of these four triangles formed are concyclic and lie on a circle that passes through the Miquel Point.

Refer to the diagram in Property 5. This Miquel circle doesn’t really look too hard, just note that \angle O_3MO_2=\pi-\angle MO_2O_3-\angle MO_3O_2=\pi-\angle MDE-\angle MFE=\angle DMF. Using this, working backwards of PP5, it is obvious that M,O_2,O_3,O_4 are concyclic. So, we are done.


Property 3.(Miquel-Steiner Line)
The orthocentres H_1, H_2, H_3, H_4 of these triangles lie on a line, known as the Miquel-Steiner Line.

For the proof of this, we make use of PP12. So maybe you should first scroll down a little, read what it is all about, then come back. 🙂
Obviously the midpoints of MH_1, MH_2, MH_3, MH_4 lie on the pedal line mentioned in PP12, and therefore their image after homothety wrt (M,2) lie on a line. This line is known as the orthocentric or Miquel-Steiner line.


Property 4.
The orthocentre of O_2O_3O_4 lies on \ell _1.

I don’t really care about the configurations much, so I’d like you to refer to the diagram in the next Property. Now, according to the notations of PP5, we need to show that H_1'\in DE.
Note that \angle DH_1'F=\angle DH_1'O_2+\angle O_3H_1'F+\angle O_2H_1'O_3=\angle O_2O_4O_3+\angle O_2H_1'O_3=180^{\circ}, because the reflection of H_1' in O_2O_3 lies on the Miquel Circle.


Property 5.
M is the centre of spiral similarity that maps O_2O_3O_4 to CAB.

Since O_4O_2\perp DM and O_2O_3\perp ME, therefore \angle O_4O_2O_3=\angle DCE and similarly it can be shown that \angle O_2O_3O_4=\angle CBA. Therefore the triangles O_2O_3O_4 and CAB are similar. Also, note that \angle O_{3}MO_{4}=\angle O_{3}O_{2}O_{4}=\angle BCA=\angle BMA, so we see that M is the centre of spiral similarity that maps O_2O_3O_4 to CAB.
import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsi...

Property 6.(Newton-Gauss Line)
The midpoints of AC, BD, EF are collinear.


Not really too eager to write it all down. The following stronger version of this theorem holds, which is known as Gauss-Bodenmiller theorem.

Extension of property 6:
The circles drawn on the diagonals of a complete quadrilateral taking them as diameters, are coaxal.
For a proof, see here.


Property 7.
Define the unique point K_1 on AD such that BC\parallel GK_1 and similarly define K_2\in BC, K_3\in DC, K_4\in AB. Then let M_1, M_2 be the midpoints of FG, DG. Then the points K_i and M_i lie on a straight line.

Note that the sides DA, BC are harmonically separated by FG, FE. So, K_2 and K_1 are harmonically divided by FG\cap K_2K_1 and FG\cap K_2K_1. Since GK_1FK_2 is a parallelogram, so the point FG\cap K_2K_1 is the point at infinity of K_1K_1. Therefore FE\parallel K_1K_2. So, we are done.


Property 8.
The projections of Miquel point onto the sidelines AB,BC,CD,DA lie on a line perpendicular to the Newton-Gauss line.

Let the points K_i be defined as in our previous property. Then it can be seen that \displaystyle \frac{\tau(K_1,\odot(FCD))}{\tau(K_1, \odot(FAB))}=\frac{\tau(K_2, \odot(FCD))}{\tau(K_2, \odot(FAB))}\Longrightarrow K_1,K_2,F,M are concyclic. So, M is the Miquel point of both ABK_2K_1 and CDK_3K_4.
Using this, we see that projection of M on K_1K_2 is collinear with projections of M on AB,BC,AD. This line is the pedal line of ABCD. The fact that this is orthogonal with the Newton-Gauss line also immediately follows.

Using this property, it can be shown that the Miquel point lies on the Nine Point Circle of the triangle GYZ where Y=AC\cap EF and Z=BD\cap EF. This triangle GYZ is the diagonal triangle of ABCD.


Property 9.

We define the Hewer Point as the centre of the circle that passes through the orthocentres of O_iO_jO_k. Thus this point is the pseudo-orthocentre of O_1O_2O_3O_4.

Since the point H_P is defined as the circumcentre of H_1'H_2'H_3', therefore it is forced that the circle \odot(H_1'H_2'H_3') is congruent to the circle \odot(O_1O_2O_3), and the points H_i' correspond to the vertices O_i, meaning that the segments O_iH_i' are bisected at a common point, so that O_1OH_1'H_P is a parallelogram. Therefore, \vec{OO_1}=\vec{H_1'H_P}. This gives us \vec{OH_P}=\vec{OH_1'}+\vec{OO_1} =\vec{OO_1}+\vec{OO_2}+\vec{OO_3}+\vec{OO_4}, where the last line follows from \vec{OH}=\vec{OA}+\vec{OB}+\vec{OC} in any triangle ABC. So this establishes that H_P is the pseudo-orthocentre of O_1O_2O_3O_4.

import graph; size(18.52cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real...


Property 10.
If H_P is the Hewer point, O the centre of the Miquel circle, H_1' the orthocentre of O_2O_3O_4; then we have O_1,H_P,O,H_1' and H_1', H_2', O_1, O_2 form a parallelogram.

Since H_P is the pseudo-orthocentre of O_1O_2O_3O_4, we already have that
\displaystyle \overrightarrow{OH_P}=\overrightarrow{OO_1}+\overrightarrow{OO_2}+\overrightarrow{OO_3}+\overrightarrow{OO_4}=\overrightarrow{OO_1}+\overrightarrow{OH_1'}\Longrightarrow\overrightarrow{O_1H_P}=\overrightarrow{OH_1'};
And all these similar properties give us our desired results.

import graph; size(18.52cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real...


Property 11.
If the Euler lines of the four triangles formed, ie FAB, FCD, EBC, EDA meet the Miquel circle again at points S_i, then O_1H_P\perp MS_1 and similars.

Due to the spiral similarity that maps O_1 and H_1' to O and H_1, note that we have \angle (MO, MO_1) =\angle (OH_1' , O_1H_1) , and also \vec{OH_1'}=\vec{O_1H_P}, which leads us to the fact that O_1H_P and O_1O are isogonal conjugates wrt \angle MO_1S_1. Because O_1O passes through the circumcentre, so its isogonal passes through the orthocentre of MO_1S_1. In other words, O_1H_P\perp MS_1, leading to the fact that H_P is the point of concurrence of the four altitudes of MO_iS_i through the vertices O_i .


Property 12.
For an arbitrary quadrilateral ABCD, the nine-point circles of triangles BCD, CDA, DAB, ABC and the pedal circles of the corresponding remaining point, are concurrent.

The proof makes use of a known property of conics, ie the pedal circle of a fixed point chosen on a conic wrt all the triangles inscribed in it pass through a fixed point. For rectangular hyperbola, this fixed point coincides with the centre. Thus every conic through the orthocentre and vertices of a triangle is a rectangular hyperbola, and because the centres of all these conics lie on the nine-point circle, therefore we are done. Refer to Ref. 4 for details. (P123.)

Special Case.
When the quadrilateral ABCD becomes cyclic, then the point of concurrency is termed as the Euler Point of the Cyclic Quadrilateral or the anticentre, in which case, the pedal circles of A,B,C degenerate into the Simson lines of A wrt BCD and similars. A nice geometric proof is posted here.



Family of Properties 13.
Let X be the Miquel point for the quadruple of lines AB, AC, BD, CD., Y be that for the quadruple of lines AB, AD, BC, CD, and Z be that of BC, AC, BD, AD. Let P_X=AD\cap BC, P_Y =AC\cap BD, P_Z=AB\cap CD. Let K_X and L_X be midpoints of the segments BC and AD respectively. Similarly, define K_Y , L_Y as midpoints of AC, BD, and K_Z, L_Z be that of AB, CD. Let \Gamma _X=K_X L_X, \Gamma _Y =K_Y L_Y, \Gamma _Z=K_Z L_Z be the Newton-Gauss lines for the corresponding quadruples of lines. Then,
Property 13(a)
The lines AX, BY, CZ concur at a point D_0.  Similarly define A_0, B_0, C_0.
Property 13(b)
A_0, B_0, C_0, D_0 are the isogonal conjugates of A, B, C, D with respect to triangle XYZ.
Property 13(c)
X, Y, Z are Miquel points for quadruples of lines joining A_0, B_0, C_0, D_0.
Property 13(d)
The lines AA_0, BB_0, CC_0, DD_0 are parallel.
Property 13(e)
The lines AD, A_0D_0, YZ are concurrent.
Property 13(f)
The points X,Z, P_Y, K_Y lie on a certain circle \omega_Y. Similarly define circles \omega _X, \omega _Z. Then, \omega_X, \omega_Y, \omega_Z have a common point T.
Property 13(g)
T is the point of concurrency of XP_X, YP_Y, ZP_Z.

AWAITING FOOL-PROOF COMPLETE PROOFS (I have not been able to solve this yet).

Property 14.
Let the perpendicular from the midpoint of O_1H_1 to CD be denoted as m_1. Then the lines m_1, m_2, m_3, m_4 concur at a point known as the Morley point of the complete quadrilateral, and this point lies on the Miquel-Steiner line. Refer to Ref. 2 for more details.


Property 15.
If the Euler line of one associated triangle is parallel to the corresponding line \ell_n, then this property is true for all the four associated triangles.

Let G, H be the centroid and orthocentre of a \triangle ABC and let the Euler line GH meet the sideline BC at a point D. Let a parallel b_1 to GH through $ latexA$ meet BC at C_1. Let H_1 \in AH be the orthocentre of the \triangle HDB. Then DH_1 \parallel CA (both perpendicular to BH). Since (AH_1 \equiv AH) \perp (BC \equiv BC_1) and BH_1 \perp (DH \parallel C_1A), H_1 is also the orthocentre of the \triangle ABC_1. Let a parallel to BC \equiv BC_1 through G meet the line DH_1 at a point G_1. The \triangle AC_1C \sim \triangle DGG_1 are (centrally) similar, having the corresponding sides parallel. Their similarity coefficient is -3 (because G is the centroid of the \triangle ABC and their corresponding sides are oppositely oriented). Therefore, \overline{C_1C} = -3\cdot \overline{GG_1}=3 \cdot \overline{G_1G} and G_1 is the centroid of \triangle ABC_1. As a result, (G_1H_1 \equiv DH_1) \parallel CA is the Euler line of the \triangle ABC_1.

If b_2 is any other line parallel to the Euler line GH of the \triangle ABC (different from GH, b_1 \equiv C_1A), intersecting AB, BC at A_2, C_2, the \triangle A_2BC_2 \sim \triangle ABC_1 are centrally similar with center B and their Euler lines G_2H_2 \parallel G_1H_1 are parallel, hence G_2H_2 \parallel CA as well.
Euler_line GH.GIF
Let the Euler line GH of \triangle ABC meet CA, AB at points E, F. Let a parallel a' to BC through E meet AB at B'. The \triangle AB'E \sim \triangle ABC are centrally similar with center A, hence, their Euler lines G'H' \parallel GH are parallel. In exactly the same way as above, we can show that the Euler lines G'H', G_1'H_1' of the \triangle AB'E, \triangle AFE meet on AB' and that G_1'H_1' \parallel B'E \parallel BC.


Property 16.
If one of the Euler lines is parallel to the corresponding line \ell_n, ie if O_1H_1\parallel AB, then the Morley Point and the Hewer point are the same point.

We already know that the perpendiculars through the perpendicular bisectors of O_1H_1 and similars concur at the Hewer point. And if O_1H_1\parallel AB, then by PP15, we already have that these lines are the same as the lines perpendicular to the corresponding sides, and passing through the nine-point centres of the corresponding triangles, so that Morley point and Hewer points coincide for these quadrilaterals.


Property 17.
The polar circles of the four triangles formed by \ell_1, \ell_2, \ell_3,\ell_4  and the circumcircle of the diagonal triangle GYZ are coaxal.

Proof. (taken from Ref. 3)
Obvious from some harmonic divisions. Observe that each of the four polar circles is orthogonal to the three circles with diameter AU, BV, CW. Moreover, as each of the quadruples $latex  (A, U, B’,C’),(B,V,C’,A’)$ and (C, W, A',B') is harmonic, the circle \odot(A'B'C') is orthogonal to the three circles with diameter AU, BV and CW. So this gives a shorter proof of the PP5, and also, helps us in figuring out that the orthocentric line and the pedal line are both perpendicular to the line A'B'C'.

Family of Property 18.
We have some additional results when ABCD is a cyclic quadrilateral. We will just refer to the file “Cyclic Quadrilaterals – The Big Picture” by Yufei Zhao. [Ref. 1]


1. Cyclic Quadrilaterals – The Big Picture by Yufei Zhao.
2. Orthocentric Properties of the Plane n-Line by F. Morley.
3. Steiner’s Theorems on the Complete Quadrilateral by Jean-Pierre Ehrmann.
4. A Treatise on the Circle and Sphere by Julian Lowell Coolidge.
5. Some Properties of The Newton-Gauss Line by Catalin Barbu and Ion Patrascu.
6. Cut-The-Knot Page on Complete Quadrilaterals.
7. ENCYCLOPEDIA  OF  QUADRI-FIGURES by Chris Van Tienhoven (The ultimate set of properties everyone searches for).

Properties of the Neuberg Locus and Brocard Problem

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)

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.


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.

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.


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.


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.

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.


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

Which leads to
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.

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.


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).

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.


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.

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.


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'.

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.


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.


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.

Property 4.

If P lies on the Neuberg locus of ABC, then A lies on the Neuberg Locus of BPC and similarly.


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.


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. 🙂

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)|

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.

Caution: In an exam, we never mention the differentiations etc, just scribble the AM-GM step. All of this should be rough work.


For checking out the full article, see here.