[/quote]
,
are on sides 
such that 
is a cyclic quadrilateral. Let 
be the intersection of 
and 
.
is a point on 
.
be the midpoints of 
.
.
of positive integers is called central if for every positive integer 
,
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
,
,
.
,
lie on a line in this order and satisfy 
.
and 
be the midpoints of 
and 
,
is acute, and let 
lie on lines 
and 
,
and 
and 
and 
be all the positive integers smaller than 
such that
for all 
.
be non-negative real variables with 
Prove that
Equality holds iff 
or 
where 
?
be a convex quadrilateral such that diagonals 
and 
intersect at right angles, and let 
be their intersection. Prove that the reflections of 
are concyclic.
and 
Prove that
Let 
Prove that
we are given :
and one side length of the triangle equals 
,
be a trapezoid with parallel sides 
.
again at 
respectively. Prove that 
are concurrent.
and 
Prove that
in it. Prove that there are an infinite number of perfect cubes in this progression. (A perfect cube is an integer of the form 
,
is an integer. For example, 
,
and 
there are elements 
such that 
.
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Let 
be the circumcircle of acute triangle 
. Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
be the circumcircle of acute triangle 
. Points 
and 
are on segments 
and 
respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
This post has been edited 2 times. Last edited by djmathman, Jun 16, 2020, 4:02 AM
Reason: problem author
Reason: problem author
and 
is perpendicular to internal bisector of 
be points on smaller arc 
chosen so that 
.
onto two equal parts. That means that 
(Archimed Lemma). Simialrly 
.
to cut the circumcircle at 
, also 
to cut the circumcircle at 
are cyclic (of course with center 
)
are also cyclics)
Given 
.
be the point on 
and let 
be the point such that 
.
.
intersects 
at 
.
so 
and 
.
.
.
so 
as desired.
on arc 
.
and 
congruent, so 
.
are parallel, 
and 
.
be the incenter of the triangle 
.
.
is cyclic, so 
,
.
,
,
,
,
and the circumcircle of 
we have![\[ \frac{AD}{\sin \angle AFD} = \frac{AF}{\sin \angle ADF} = \frac{AF}{\sin \angle BDF} = \frac{AF}{\sin \angle ABF} = 2R \]](http://latex.artofproblemsolving.com/4/7/c/47cba532afa31887d889d490fb02d9560c53f665.png)
![\[ \frac{AE}{\sin \angle AGE} = \frac{AG}{\sin \angle AFG} = \frac{AG}{\sin \angle GFC} = \frac{AG}{\sin \angle ACG} = 2R\]](http://latex.artofproblemsolving.com/a/d/9/ad991c91fe9220922d511ae068e2d94ad206c9d7.png)
and since 
.
,
.
.
and 
.
from the perpendicular bisector, we compute![\[ \angle APQ = \angle AFG + \angle BAF = \angle ACG + \angle BAF = \angle AFG + \angle BAF \]](http://latex.artofproblemsolving.com/9/7/8/9787520ab38141721fa6bcf46c6a981ae0660588.png)
![\[ = \angle GAF + \angle AGF + \angle BAF = x + y + z. \]](http://latex.artofproblemsolving.com/e/0/8/e08c5b4eb40141a421c0502338f2c814bd94b5ac.png)
Analogously 
.
,
and the result follows.
and 
,
.
,
to be the point on segment 
)
and define 
.
.
.
,
.
,
,
,
.![\[\angle FBA=\angle FBC-b=(180-\angle FGC)-b=(\angle GYC+\angle GCA)-b=(90-\frac{a}{2})+x-b=\frac{-b+c}{2}+x,\]](http://latex.artofproblemsolving.com/f/b/f/fbf504da22d09bcc507be5184e8932100e114052.png)
which then gives that![\[\angle FAB= \frac{b+c}{2}-x,\]](http://latex.artofproblemsolving.com/a/c/1/ac1fd4f3d32e07d988b3efe4352ea3299c661265.png)
and![\[\angle CAG=b-x.\]](http://latex.artofproblemsolving.com/b/5/0/b5089f036c12227f1f9bf17ddf73ff3844c61f8d.png)
![\[CG=\sin \angle GAC=\sin (b-x),\]](http://latex.artofproblemsolving.com/3/6/7/3678d628180a4255f122b9b3f72a5636e1b51aab.png)
and since 
,![\[CE'=2\sin (b-x)\cos x,\]](http://latex.artofproblemsolving.com/4/9/c/49cd17c94f4ddb08643968895defe02ee8d88114.png)
which is equal to 
by Product-to-Sum formulas. This means that![\[AE'=AC-(\sin b+\sin (b-2x))=\sin b-(\sin b+\sin (b-2x))=-\sin(b-2x).\]](http://latex.artofproblemsolving.com/5/c/0/5c0ce6bf63dfc34df576c7fbc552869f6f4014bb.png)
Similarly, we get that![\[BD'=2\sin \left(\frac{b+c}{2}-x\right)\cos x,\]](http://latex.artofproblemsolving.com/1/6/d/16dcad67ed3711819daef3446e63c83e2a8dd6c9.png)
which is equal to 
by Product-to-Sum formulas. This means that![\[AD'=AB-(\sin c+\sin (b-2x))=\sin c-(\sin c+\sin (b-2x))=-\sin(b-2x),\]](http://latex.artofproblemsolving.com/1/6/6/166cf3aed36fcda007cd3c44de982c10a0b58365.png)
which is indeed equal to 
,
,
and 
)
or 
and 
also move linearlly. Hence its enough to prove that 
for three positions of the points 
,
and 
.
and 
be the tangents to 
at 
and 
be the angles that 
and 
,
will denote the derivative of the point 
has magnitude 
along 
as the direction of 
,
by the chain rule. Repeating the same argument with 
.
so 
as well.
as functions of 
as when 
are collinear and both angles can be chased to be equal to 
.
with the initial condition 
.
is "nice" enough, the solution to this differential equation is unique, so indeed we have that 
and 
are parallel, so 
are points on 
then we get that:
so 
so 
are collinear.
so 
is cyclic. We are motivated that 
is isosceles. We make the following claim:
and 
are both isosceles.![\[\angle GXA = \angle GCE = \angle GEC = \angle AEX. \phantom{c} \square\]](http://latex.artofproblemsolving.com/1/c/d/1cdb9de727f3b7b8c20d61c76ccf8cea05611564.png)
,
là một hình diều. Tương tự, 
l
.
l
t
l
v
.
is a parallelogram, and construct 
.
be the intersection of 
and 
.
and 
.
,
becomes a trapezoid. Therefore 
.
and 
,
.
and 
are isosceles trapezoids. Then, since 
over the line through 
and 
are both parallelograms. Thus, 
,
is an isosceles trapezoid. Additionally, letting 
and 
.
,
,
,
are all equal, which now gives that 
is an isosceles trapezoid. Thus, 
,
.![\[\angle(DE,BC) = B - \angle{ADE} = B - \left(90 - \frac{A}{2}\right) = \frac{B-C}{2}.\]](http://latex.artofproblemsolving.com/f/b/8/fb85bbfb4236f4b8866f4de8fc0f5a522697cbf0.png)
We will show that this value is equal to 
.![\[\angle(MN,BC) = \angle{NBC} - \angle{MNB} = \frac{B}{2}-\frac{C}{2},\]](http://latex.artofproblemsolving.com/7/7/a/77a75920c1490898b54fee71bf109c61162fa6d2.png)
as desired.
,
.
,
,
gives the condition as 
,
,
,
gives 
as the equivalent condition.
is self conjugating, which is equivalent to 
,
,
,
,
,
meet. Let 
be the circumcenter of 
is a rhombus.
,
.
therefore by sliding principal we have 
.
.
,
and 
are symmetric about 
is a cyclic isosceles trapezoid with 
It is well known that the arcmidpoints of ![[asy]
size(7cm);
pair A = dir(110);
pair B = dir(205);
pair C = dir(335);
pair D = A+0.4dir(degrees(B-A));
pair E = A+0.4dir(degrees(C-A));
pair X = 0.5(B+D);
pair F = intersectionpoints(X+10*dir(degrees(A-B)-90)--X-10*dir(degrees(A-B)-90), circumcircle(A, B, C))[1];
pair Y = 0.5(C+E);
pair G = intersectionpoints(Y+10*dir(degrees(A-C)-90)--Y-10*dir(degrees(A-C)-90), circumcircle(A, B, C))[0];
draw(extension(A, dir(-90), F, G)+dir(degrees(G-extension(A, dir(-90), F, G)))*0.06 -- extension(A, dir(-90), F, G)+dir(degrees(G-extension(A, dir(-90), F, G)))*0.06+dir(degrees(A-dir(-90)))*0.06 -- extension(A, dir(-90), F, G)+dir(degrees(A-dir(-90)))*0.06, red+linewidth(0.5));
draw(A--B--C--cycle);
draw(D--E^^C--G--F--B);
draw(circumcircle(A, B, C));
draw(A--dir(-90), dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(160));
dot("$E'$", E, dir(40));
dot("$F$", F, dir(F));
dot("$G'$", G, dir(G));
dot("$P$", extension(A, B, F, G), dir(125));
dot("$Q$", extension(A, C, F, G), dir(60));
[/asy]](http://latex.artofproblemsolving.com/0/c/d/0cd32b83b651cd2f4a007840d7f75bc2a6aed50f.png)
where 
is the angle bisector of 
such that 
.
and 
.
.
:
which is equal to 
.
.
Doing the same thing on the other side, 
.
,
.
