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represented by tokens to players 
to 
so that they all have the same number of tokens in their urns.
beats 
when, when 
.
What is 
such that the following equations are both true:
and 
be positive integers with 
such that 
divides 
.
and 
;
and 
.
be reals such that 
Prove that
Y by narutomath96, parmenides51, Adventure10, Mango247, and 3 other users
I apologize if I posted to a wrong area. Can any one tell me if the bellow theorem is well known?
THEOREM. A cyclic quadrilateral
is given and let 
be, the intersection point of it's diagonals 
. Prove that the Euler lines of the triangles 
, are concurrent.
I am still searching about a reference of this theorem ( which is also true for non convex quandrilateral ) and I will with pleasure post here the proof I have in mind.
Kostas Vittas.
THEOREM. A cyclic quadrilateral
is given and let 
be, the intersection point of it's diagonals 
. Prove that the Euler lines of the triangles 
, are concurrent.I am still searching about a reference of this theorem ( which is also true for non convex quandrilateral ) and I will with pleasure post here the proof I have in mind.
Kostas Vittas.
This post has been edited 2 times. Last edited by vittasko, Nov 18, 2011, 10:18 AM
be 3 arbitrary points on a normal hyperbola 
(with perpendicular asymptotes). Let a line parallel to AB meet the hyperbola chords OA, OB at C, D. Let C', D' be reflections of C, D in a line through O parallel to one of the hyperbola asymptotes. Then the lines AC', BD' also meet on the hyperbola 
and 
be the coordinates of 
Since 
coordinates of C, D are 
and 
while coordinates of their reflections C', D' in the line 
are 
and 
Equations of the lines AC', BD' are then
i
both satisfy the hyperbola equation. Substituting 
and 
![$(x_{b}y_{a}-x_{a}y_{b})[(x^{2}-2px)-(y^{2}-2qy)] = 0$](http://latex.artofproblemsolving.com/5/6/f/56f79163f2497eb04a94fc9a71599daeb6156de5.png)
so that the 1st factor in non-zero, which means that the locus of Q is the same normal hyperbola, 
Then P lies on a unique normal hyperbola passing through the parallelogram vertices A, B, C, D, centered at the parallelogram diagonal intersection, and with asymptotes parallel to the parallelogram angle bisectors. 
circumcircles 
of 
are congruent (this is quite sufficient for our proof). Likewise, circumcircles 
of 
are also congruent. If we translate 
into 
so that A'B' coincides with DC, the quadrilateral CPDP' is cyclic with the circumcircle 
Triangles 
are congruent by SAS, hence PP' = DA. Since 
are circumcircles of these 2 congruent triangles, they are also congruent and consequently, all 4 circumcircles 
are congruent. Denote a = AB = CD, b = BC = DA and 
Let A be the coordinate origin and the positive x-axis identical with the ray (AB. Let 
be center of 
so that 
is center of 
(the coordinate k can be positive of negative). Equation of the circles 
we get
,
are fixed, 
Likewise, since 
are fixed, 
Slopes of the asymptotes are obtained by factoring the 3 leading terms:
and 
be circumcenters and orthocenters of 
respectively. 
(all perpendicul to AC) and 
(all perpendicular to BD). Thus 
and 
are both parallelograms with equal angles. 
are collinear, because 
are oppositely similar and the circumcenter and orthocenter of a triangle are isogonal conjugates. Likewise, 
are collinear, 
are collinear, and 
are collinear. It follows that 
(because 
etc.). Moreover,
are similar and P is their common corresponding point. Using the opposite similarity of 
and the fact that the circumcenter and orthocenter are isogonal conjugates, we find that the reflection 
of 
by lemma 2, P lies on a rectangular hyperbola 
with asymptotes parallel to the parallelogram angle bisectors, and these are parallel to the angle bisectors of the angles formed by the diagonals AC, BD (because 
)
cut by the parallel line 
The points 
are reflections of 
in a line through P parallel to the asymptotes of the normal hyperbola 
hence by theorem 1, the intersection 
also lies on this hyperbola. But the same can be said about any pair of lines 
Each of these lines intersects 
concur at 
meet at the midpoint of OP, for example, by an angle chase leading to 
O, P are then symmetrical with respect to the hyperbola center. Since 
,
as well.
and 
the above equations yield 
the quadrilaterals 
are congruent and the Euler lines 
are all parallel (their concurrency point 
is at infinity).
are concurrent.
In my drawing 
and radius of the circumcircle of 
taken per two of them, intersect each other at points 
as the
respectively.
the orthogonal projections of 
respectively, on 
and as 
the ones of 
respectively, on 
are the orthocenters of the above triangles.
pass through the point 
the midpoints of 
on the lines 
intersect each other, form segments with the same ratio. Hence, based on the below 
without the use of the Lemma 1. It is enough to prove that 
Really, from the cyclic quadrilaterals 
and 
and from cyclic quadrilateral 
where 
are the midpoints of 
and 
and so, 
etc.
and 
and hence, the we conclude that the lines 
We have now, the configuration of the parallelograms 
correlated each other with their sidelines parallel one per one and also
as the Euler lines of the triangles 
respectively, are concurrent at one point and the proof is completed.
on segment 
which intersect the extensions of 
at points 
and 
and parallel to 
respectively, so be it 
are concurrent at one point.
We denote as 
from the line through 
It is enough to prove that 
and we denote as 
the intersection point of the lines through 
and parallel to 
inwardly to it the point 
and the lines 
and 
Hence, based on the below 
by Thales’s thoerem we have that:
and the proof of the part 
and 
So, we have the point 
and 
Hence, by the 
are collinear.
are coincided, because of their common points 
by Desarques’s theorem, we have that the triangles 
are perspective and hence, we conclude that the lines 
are concurrent at one point, so be it 
is also the as the perspector of the triangles 
and so, these triangles are perspective. Hence, we conclude that the points 
and 
are collinear.
as the intersection point of the lines 
lies also on the segment line 
and 
So, we have the point 
and 
Hence, by the 
and 
are collinear.
are coincided, because of their common points 
by Desarques’s theorem, we have that the triangles 
are perspective and hence, we conclude that the lines 
are concurrent at one point, which is again the point 
and so, these triangles are perspective. Hence, we conclude that the points 
are collinear.
respectively and we denote as 
If 
prove that the points 
at points 
Then P lies on a unique normal hyperbola passing through the parallelogram vertices A, B, C, D, centered at the parallelogram diagonal intersection, and with asymptotes parallel to the parallelogram angle bisectors. 
and the proposition should be "if and only if."
a unique normal hyperbola is inscribed in ABCD. The parallelogram diagonal intersection 
the common midpoint of the diagonals, is the hyperbola center and intersection of its asymptotes. Then L is orthopole of the hyperbola isogonal conjugate line 
WRT the 
passing through its circumcenter 
,
is the perpendicular bisector of BD. Conversely, 
and it immediately follows that for 
This means the circumcircles 
of the adjacent 
with the common side AP are congruent. By similar reasoning, the circumcircles 
of the adjacent triangle pairs 
are all congruent and 
The perpendicular bisector 
of BD cuts the circumcircle 
of the 
their isogonal conjugates 
But since 
their isogonal conjugates are at infinity, therefore identical with directions of the hyperbola asymptotes, and 
of the 
are congruent and circumcircles 
of the 
are also congruent. Translating the 
by 
into 
makes a parallelogram APP'D and a cyclic quadrilateral DPCP', the 
with circumcircles 
are congruent, hence 
The rest of the proof is reversible.
and 
and a point 
inside 
the lines 
through 
and 
,
are concurrent. when 
we get this result.
and 
so we get eight Euler lines concurrent.
