[/quote]
,
and 
. Prove that
Let 
be reals such that 
. Prove that
be four positive integers in arithmetic progression (that is, 
)
.
ever be a number of the form 
for some 
and some 
,
?
,![\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]](http://latex.artofproblemsolving.com/3/2/b/32b41635c159921e11245369e66185ec69b0d785.png)
and the circle 
of radius 
Let 
be the midpoint of 
The line 
intersects 
Let 
Let 
be the intersection of 
for which there exists an integer 
such that 
is a divisor of 
.
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
are in arithmetic progression and 
then show that the product 
is not a perfect square
and 
be positive integers.
and 
.
.
?
. Prove that
be the side lengths of a triangle. Prove that![\[\frac{ab+bc+ca}{(a+b)^2} + \frac{ab+bc+ca}{(b+c)^2} + \frac{ab+bc+ca}{(c+a)^2} \leqslant \frac{85}{36}.\]](http://latex.artofproblemsolving.com/f/8/0/f80aed170f4f4bd78a2e881ce93b6b7ddfa03c22.png)
such that the reflection of each of these four points about 
![\[
\frac{ab}{b+1} + \frac{bc}{c+1} + \frac{ca}{a+1} \geq \frac{3abc}{1 + abc}
\]](http://latex.artofproblemsolving.com/d/5/8/d582a3269c92041834c02153ee8e946a274e52bd.png)
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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 
.
.
is 
.
,
.
,
,
,
.![\[\angle FBA=\angle FBC-b=(180-\angle FGC)-b=(\angle GYC+\angle GCA)-b=(90-\frac{a}{2})+x-b=\frac{-b+c}{2}+x,\]](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
.
so that 
is a parallelogram, and construct 
.
be the intersection of 
and 
.
and 
.
,
becomes a trapezoid. Therefore 
.
and 
are similar. Hence with the parallelism stated they are necessarily homothetic : the parallelism that was unknown to us is especially 
,
.
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 
,
on both sides this becomes 
,
,
gives 
as the equivalent condition.
is self conjugating, which is equivalent to 
,
,
,
,
,
meet. Let 
be the circumcenter of 
and 
be the midpoints 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, 
.
,
.
