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.
table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
+1 w
, prove that 
is the circumcenter of 
, and 
are points on segments 
and 
respectively with 
. Let 
be a point such that 
and 
.
intersect and at points 
and 
respectively. Let the line passing through and perpendicular to 
intersect and at points 
and 
respectively. Prove that points 
, and are concyclic.
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
and 
be altitudes in an acute triangle 
which meet at 
. Suppose that 
meets the circumcircle of at 
and 
such that lies on the shorter arc of 
and lies on the shorter arc of 
. Let 
and 
meet at 
. Show that the circumcircles of 
and 
and the line 
concur.
![$P(x) \in \mathbb{Q}[x]$](http://latex.artofproblemsolving.com/9/1/0/9104130500a15ab4bffb68bc493faa4ea2891610.png)
be a polynomial with rational coefficients and degree 
. Prove there is no infinite sequence 
of rational numbers such that 
for all 
.
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
aand 
Prove that
Let 
aand Prove that
Let 
aand 
Prove that 
with the center 
touches the sides 
at 
respectively. The line 
meets the circumcircle of at 
. The line 
meets the line 
at 
and the line 
meets the circumcircle of at 
. Define 
similarly. Prove that the lines 
are concurrent.
intersect at point 
. Point 
lies between the parallel lines 
and 
such that 
, and line 
separates points and . Prove that 
.
and 
Prove that 
be an even integer and let 
be real numbers satisfying 
.
-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.
aand 
Prove that 
aand 
Prove that
Let 
aand 
be an integer greater than or equal to 
.
satisfy 
,![\[\sum_{1 \le i < j \le n} \frac{1}{n- a_i a_j} \le \frac{n}{2} \]](http://latex.artofproblemsolving.com/2/b/b/2bbca5c72fe6bf9f3fa50cc34b8bff27f2ec5d4d.png)
must hold.
,
are 
real numbers satisfying 
,![\[ \sum_{1\leqslant i < j\leqslant n}\frac {1}{1 - x_ix_j}\leqslant\frac {n^2}{2}\mbox{.}\]](http://latex.artofproblemsolving.com/7/9/e/79e8a348888aed07bbd757ffbf01e85f33d5ada0.png)
