AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
is strictly increasing : 
such that:
satisfying the following conditions:
divides 
,
divides 
,
.
be positive reals such that 
. Show that ![\[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]](http://latex.artofproblemsolving.com/3/b/d/3bd44cede90261ab949beabd32f8c08411ddc051.png)
such that for all 
,![\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]](http://latex.artofproblemsolving.com/d/1/0/d10c01d7d416243614b24f04218d157e8b025ff2.png)
such that for all 
.
such that for all and 
such that 
there exist 
such that 
and 
. Note : 
and 
for all 
.
such that 
. Find the minimum value of:
such that for all . 
.
are colored blue. A set of six cells is called gremista when the cells are the intersection of three rows and two columns, or two rows and three columns, and are painted blue. Determine the greatest value of 
for which it is possible to color chessboard cells blue such that there is not a gremista set.
be non-negative real numbers such that 
Prove that
-excenter of 
is 
and 
. The foot from 
to 
is 
and 
. The intersection of 
and 
is 
. The circumcenter of 
is 
.
is tangent to the -excircle if is on the incircle of .
![[asy]
import olympiad;
import geometry;
size(13cm);
pair A= dir(104); pair B=dir(230); pair C=dir(310);
pair I=incenter(A, B, C); pair O=excenter(B, C, A);
pair D=foot(A, B, C);
pair E=foot(I, B, C);
pair X=intersectionpoint(line(E, O), line(A, D));
pair F=foot(O, B, C);
pair P=circumcenter(X, B, C);
pair J1[]=intersectionpoints(line(X, I), circle(O*2-F, F));
pair J=J1[0];
dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE);
filldraw(A--B--C--cycle, invisible, red);
draw(circle(I*2-E, E), orange); draw(circle(O*2-F, F), orange);
draw(A--D, red); draw(X--O, red);
draw(A--(A+(B-A)*1.8), red); draw(A--(A+(C-A)*1.5), red);
dot("$I$", I, NE); dot("$O$", O, S); dot("$D$", D, SW); dot("$E$", E, SW);
dot("$X$", X, NW); dot("$F$",F, S);
dot("$P$", P, NE); dot("$J$", J1[0], N);
draw(X--J, magenta+dashed); draw(B--X--C, red);
draw(circle(B, J, C), lightred+dashed);
[/asy]](http://latex.artofproblemsolving.com/b/e/0/be0c2d2e3b312ab7ba20d6348e71e9f6ac105614.png)
be the touchpoint of the excircle at , and let 
be the point on the -excircle so that 
is tangent to the excircle. It suffices to prove that 
bisects 
, since this would imply that is the midpoint of arc of .
bisects 
. Since lies on the incircle, we have 
; thus 
and 
are isogonal WRT . Hence 
are collinear. Note that it is well-known that lies on this line as well.
are collinear, and that 
bisects . But this is the excenter version of ISL 2002 G7 (which can be proved analogously), so we're done.
coincides with the midpoints of 
and also lies on line 
.![[asy] size(13cm); pen zzttqq = rgb(0.6,0.2,0); pen fuqqzz = rgb(0.95686,0,0.6); pen qqwuqq = rgb(0,0.39215,0); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw((-2.43454,2.03316)--(-2.6,0)--(-0.2,0)--cycle, linewidth(1) + zzttqq);
draw((-2.43454,2.03316)--(-2.6,0), linewidth(1) + zzttqq); draw((-2.6,0)--(-0.2,0), linewidth(1) + zzttqq); draw((-0.2,0)--(-2.43454,2.03316), linewidth(1) + zzttqq); draw(circle((-1.4,0.92565), 1.51553), linewidth(1)); draw(circle((-1.89059,0.65401), 0.65401), linewidth(1) + fuqqzz); draw((-2.43454,2.03316)--(-2.43454,0), linewidth(1)); draw(circle((-0.90940,-1.83376), 1.83376), linewidth(1) + qqwuqq); draw((-2.43454,0)--(-1.4,-0.58987), linewidth(1)); draw((-2.43454,2.03316)--(-1.4,-0.58987), linewidth(1)); draw(circle((-1.4,0.32644), 1.24360), linewidth(1) + qqwuqq); draw((-2.6,0)--(-2.73712,-1.68502), linewidth(1)); draw((-2.43454,1.01658)--(-0.90940,-1.83376), linewidth(1)); draw((-0.90940,-1.83376)--(-0.90940,0), linewidth(1)); draw((-1.4,-0.91688)--(-0.90940,0), linewidth(1)); dot("$A$", (-2.43454,2.03316), dir((1.347, 3.366))); dot("$B$", (-2.6,0), dir(225)); dot("$C$", (-0.2,0), dir(-45)); dot("$I$", (-1.89059,0.65401), dir((1.479, 2.608))); dot("$D$", (-2.43454,0), dir((1.347, 2.711))); dot("$E$", (-1.89059,0), dir((1.479, 2.711))); dot("$M$", (-1.4,-0.58987), dir((1.224, 2.797))); dot("$O$", (-0.90940,-1.83376), dir((1.306, 2.649))); dot("$X$", (-2.43454,1.01658), dir(135)); dot("$P$", (-1.4,0.32644), dir((1.224, 2.716))); dot("$F$", (-0.90940,0), dir((1.306, 2.711))); dot("$N$", (-1.4,-0.91688), dir(225)); [/asy]](http://latex.artofproblemsolving.com/e/8/d/e8dfcb147bc4ae8be21d4be8b0f8397d890e60b5.png)
is tangent to the incircle and passes through the midpoint 
of 
.
the line 
passes thru .
and the -excircle are orthogonal.
is fixed under inversion around the -excircle. To this end, we prove that ![\[ ON \cdot OX = OF^2. \]](http://latex.artofproblemsolving.com/d/2/9/d291af0f9f22a34f9786cb69021a918dc7486338.png)
Indeed, this follows from the similar isosceles triangles ![\[ \triangle ONF \sim \triangle OFX \sim \triangle EIX. \]](http://latex.artofproblemsolving.com/b/1/8/b181f4241bbfb456ff0155e00114b4cb08c230f6.png)
with radius 
will fix the -excircle. Since is tangent to the -excircle at , the inverse image of is the desired tangency point.
be the -extouch point. Let 
be the antipode of in the incircle. Let 
. Let 
intersect the -excircle again at 
. Let 
intersect the incircle again at 
. Let 
. Let 
intersect 
again at 
. sending to the vertical diameter of the -excircle, we get that must be the midpoint of , therefore 
collinear.
bisects 
.
.
.
concur at 
which must be the harmonic conjugate of w.r.t. 
. This implies the claim.
, bisects and therefore 
(isogonal lines). But since it also bisects 
, must lie on 
.
is a rectangle, 
is the center 
. We also have 
cyclic.
cyclic. Since we have 
, by shooting lemma the circle through and tangent to must be tangent to 
, but since this circle is unique it must be the -excircle, hence we're done.
and 
. Prove that
.
determine the minimum and maximum value of![\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]](http://latex.artofproblemsolving.com/6/f/2/6f2427edf9a5e8e8f2e12276d0a841765e8322f1.png)
