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 
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. 😦
where 
are positive real numbers and 
are given constants.
and not 
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 inequality to lesser number of variables. One method that people use quite frequently is the 
will abide by a minimum value. We consider, for the sake of simplicity that 
So, the problem that we will look into is:
we have 
Then find the conditions for which the expression 
has a minimum and also determine the minimum.
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}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Baligned%7DP%5Cequiv+%5Calpha+x%2B%5Cbeta+y%2B%5Cgamma+z%26%5Cgeq+%5Calpha+x%2B%5Cbeta+y%2B%5Cgamma+%5Ccdot+%5Cfrac%7Bax%2Bby%7D%7Bxy-c%7D%5C%5C%26%3D%5Calpha+x%2B%5Cfrac%7B%5Cbeta%7D%7Bx%7D+%5Ccdot+xy%2B%5Cfrac%7B%5Cgamma%7D%7Bx%7D%5Ccdot+%5Cfrac%7Bax%5E2%2Bbxy%7D%7Bxy-c%7D%5C%5C%26%3D%5Calpha+x%2B+%5Cfrac%7B%5Cbeta+c%2B%5Cgamma+b%7D%7Bx%7D%2B%5Cfrac+1x%5Cleft%5B%5Cbeta%28xy-c%29%2B%5Cgamma+%5Ccdot+%5Cfrac%7Bax%5E2%2Bbc%7D%7Bxy-c%7D%5Cright%5D%3B%5Cend%7Baligned%7D&bg=eeebf2&fg=3c3d47&s=0&c=20201002)
from our expression. What we get after applying AM-GM is that
inside the square root, but right now there is a 
Also, for the equality to hold till this point, we have made a hell lot of assumptions. In each step that we write a 
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.
such that 
Assume that we had been able to find some suitable 
all abide by our restrictions, then this must be the required minimum that we are seeking. So now we have to find the constraints.
but this will be true only if 
However, 
is obvious. But this will help us determine 
from 
and 
so let’s call this constraint the 
of the solution. For equality to hold, we must have 
Let’s call this the 
we can verify that the condition 
What we get is that
which I would like to call the Characteristic cubic of the problem:
of the characteristic equation must be positive; and also the 
which is verified when 
But, note that if 
so that any positive root of this cubic will always lead us to a corresponding value of 
Then the cubic is 
Now 
so 
has two real points of inflection : 
and 
Now 
and 
so this cubic equation will always have one positive root. It may have two or no negative roots. So any 
satisfy and permit such a minimum value.
which satisfies the cubic equation can be found out to be a 
points, and the difference with a complete quadrangle is that this thing has three diagonals.
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 
of these four triangles formed are concyclic and lie on a circle that passes through the Miquel Point.
Using this, working backwards of PP5, it is obvious that 
are concyclic. So, we are done.
of these triangles lie on a line, known as the Miquel-Steiner Line.
lie on the pedal line mentioned in PP12, and therefore their image after homothety wrt 
lie on a line. This line is known as the orthocentric or Miquel-Steiner line.
lies on 
because the reflection of 
in 
lies on the Miquel Circle.
is the centre of spiral similarity that maps 
and 
therefore 
and similarly it can be shown that 
Therefore the triangles 
are similar. Also, note that 
so we see that 
are collinear.
on 
such that 
and similarly define 
Then let 
be the midpoints of 
Then the points 
and 
lie on a straight line.
are harmonically separated by 
So, 
and 
and 
Since 
is a parallelogram, so the point 
Therefore 
So, we are done.
lie on a line perpendicular to the Newton-Gauss line.
are concyclic. So, 
and 
is collinear with projections of 
This line is the pedal line of 
The fact that this is orthogonal with the Newton-Gauss line also immediately follows.
where 
and 
This triangle 
. Thus this point is the pseudo-orthocentre of 
is defined as the circumcentre of 
therefore it is forced that the circle 
is congruent to the circle 
and the points 
correspond to the vertices 
meaning that the segments 
are bisected at a common point, so that 
is a parallelogram. Therefore, 
This gives us 
where the last line follows from 
in any triangle 
So this establishes that 
the centre of the Miquel circle, 
then we have 
and 
form a parallelogram.
we already have that
meet the Miquel circle again at points 
then 
and similars.
and 
note that we have 
and also 
which leads us to the fact that 
and 
are isogonal conjugates wrt 
Because 
In other words, 
leading to the fact that 
through the vertices 
the nine-point circles of triangles 
and the pedal circles of the corresponding remaining point, are concurrent.
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 
degenerate into the Simson lines of 
wrt 
and similars. A nice geometric proof is posted 
be the Miquel point for the quadruple of lines 
, 
be that for the quadruple of lines 
and 
be that of 
Let 
Let 
and 
be midpoints of the segments 
and 
as midpoints of 
and 
be that of 
Let 
be the Newton-Gauss lines for the corresponding quadruples of lines. Then,
concur at a point 
Similarly define 
are the isogonal conjugates of 
with respect to triangle 
are Miquel points for quadruples of lines joining 
are parallel.
are concurrent.
lie on a certain circle 
. Similarly define circles 
Then, 
have a common point 
is the point of concurrency of 
to 
be denoted as 
Then the lines 
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.
, then this property is true for all the four associated triangles.
be the centroid and orthocentre of a 
and let the Euler line 
meet the sideline 
Let a parallel 
to 
Let 
be the orthocentre of the 
Then 
(both perpendicular to 
). Since 
and 
is also the orthocentre of the 
Let a parallel to 
through 
meet the line 
at a point 
The 
are (centrally) similar, having the corresponding sides parallel. Their similarity coefficient is -3 (because 
and 
is the centroid of 
is the Euler line of the 
is any other line parallel to the Euler line 
), intersecting 
at 
the 
are centrally similar with center 
and their Euler lines 
are parallel, hence 
as well.
at points 
Let a parallel 
to 
meet 
at 
The 
are centrally similar with center 
hence, their Euler lines 
are parallel. In exactly the same way as above, we can show that the Euler lines 
of the 
meet on 
and that 
ie if 
then the Morley Point and the Hewer point are the same point.
and the circumcircle of the diagonal triangle 
Moreover, as each of the quadruples $latex (A, U, B’,C’),(B,V,C’,A’)$ and 
is harmonic, the circle 
is orthogonal to the three circles with diameter 
and 
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 
The Problem Solver's Paradise 