# “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.

# 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.

Introduction.

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.

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.

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.

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 $M.$

Proof.
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!

$\Box$

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.

Proof.
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.

$\Box$

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.

Proof.
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.

$\Box$

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

Proof.
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.

$\Box$

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

Proof.
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 $CAB.$

$\Box$

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

Proof.

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.

$\Box$

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.

Proof.
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.

$\Box$

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

Proof.
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.

Corollary.
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 $ABCD.$

$\Box$

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 $O_1O_2O_3O_4.$

Proof.
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 $O_1O_2O_3O_4.$

$\Box$

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.

Proof.
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.

$\Box$

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.

Proof.
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.

Proof.
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.

$\Box$

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 $XP_X, YP_Y, ZP_Z.$

Proof.
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.

Proof.
AWAITING PROOFS WITHOUT BARYCENTRICS (Unsolved for me as well)

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.

Proof.(Yetti)
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.

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 $G_1'H_1' \parallel B'E \parallel BC.$

$\Box$

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.

Proof.
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.

$\Box$

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]

REFERENCES

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 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.

# 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.

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.

$\cdots$