[/quote]
,
such that
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.
,
,
are points on segments 
and 
respectively with 
.
be a point such that 
and 
.
intersect 
and 
respectively. Let the line passing through 
intersect 
and 
respectively. Prove that points 
,
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 
.
meets the circumcircle of 
and 
such that 
and 
.
and 
meet at 
.
and 
and the line 
concur.
aand 
Prove that 
![$P(x) \in \mathbb{Q}[x]$](http://latex.artofproblemsolving.com/9/1/0/9104130500a15ab4bffb68bc493faa4ea2891610.png)
be a polynomial with rational coefficients and degree 
.
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 
Let 
aand 
Prove that 
at 
respectively. The line 
meets the circumcircle of 
.
meets the line 
and the line 
meets the circumcircle of 
.
similarly. Prove that the lines 
are concurrent.
intersect at point 
.
lies between the parallel lines 
and 
such that 
,
separates points 
.
Y by GeoKing
Let 
be a triangle with circumcenter 
, angle bisector 
with 
and altitude 
with 
. The lines 
and 
meet at 
. The circumcircle of 
meets 
at 
and 
meets at 
. The circumcircles of triangles 
and meet at 
. Show that 
is a symmedian in 
be a triangle with circumcenter 
, angle bisector 
with 
and altitude 
with 
. The lines 
and 
meet at 
. The circumcircle of 
meets 
at 
and 
meets at 
. The circumcircles of triangles 
and meet at 
. Show that 
is a symmedian in 
be 
s
.
and 
be the midpoint of 
.
are cyclic.
with radius 
and reflect inverted points according to 
And 
is the intersection of 
.
are collinear and 
.
is the miquel point of 
.
are collinear.
and 
are midpoints, we have 
.![\[90-\angle XCH^*=90-\angle ACH^*=90-\angle AMH^*=\angle H^*MF^*=\angle H^*NC=90-\angle XNH^*\implies \angle XCH^*=\angle XNH^*\]](http://latex.artofproblemsolving.com/0/d/e/0de3160759e698d605f3dd8bb0d13a7477271e96.png)
Thus, 
are cyclic.![\[\angle CH^*N=\angle CXN=90-\angle C=\angle CAI^*=\angle CH^*I^*\]](http://latex.artofproblemsolving.com/7/2/5/725d567ab7691f450e46a24c429bc4e2869e7047.png)
Which gives that 
are collinear as desired
be point on 
is symmedian. we will prove 
be midpoint of 
.
are collinear.
Let 
meet 
.
.
is the external angle bisector of 
so 
.
and 
so 
so 
which implies 
is cyclic.
is cyclic so 
so 
is cyclic.
so 
is cyclic which implies 
= 
=
=
at 
is collinear ( followed by angle chasing) and now we have 
=
the reflection of the 
denote the 
and 
be the intersections of line 
with 
and 
.![\[\frac{BX}{\sin(90-\angle C)} = \frac{BN}{\sin BXN)} \text{ and } \frac{CY}{\sin(90-\angle B)} = \frac{CN}{\sin \angle NYC}\]](http://latex.artofproblemsolving.com/5/2/d/52dda2c34c706e869ff5247a2ef78c633651fd15.png)
Obtaining the ratio of these two results we have,![\[\frac{BX}{CY} = \frac{\sin(90-\angle C)}{ \sin (90- \angle B)}\]](http://latex.artofproblemsolving.com/6/f/8/6f87156bfc6deb37189d2071104d6be7a733054f.png)
However, we further note that,![\[\frac{Q'B}{Q'C} = \frac{QC}{QB} = \frac{P_bC}{P_cB} = \frac{\sin (90-\angle C) \cdot BC}{\sin (90 - \angle B) \cdot BC} = \frac{\sin(90-\angle C)}{ \sin (90- \angle B)}\]](http://latex.artofproblemsolving.com/4/1/6/416922d6f5e684e60da72e5d052ba5273483706e.png)
where 
and 
are the feet of the altitudes from 
and 
respectively. Thus,![\[\frac{BX}{CY} = \frac{\sin(90-\angle C)}{ \sin (90- \angle B)} = \frac{Q'B}{Q'C}\]](http://latex.artofproblemsolving.com/7/2/7/72711a1eb27b42127ae3b71c637c23ec41f124ef.png)
which in addition with![\[\measuredangle Q'CY = \measuredangle Q'CA = \measuredangle Q'BA = \measuredangle Q'BX\]](http://latex.artofproblemsolving.com/c/0/6/c065af2429ea81e5c38640daa998f81b659c874e.png)
implies that 
which in turn implies the claim.
followed by a reflection across the internal 
and 
passes through the midpoint of segment 
passes through the midpoint of segment 
.
.
,
,
concur. Now see that ![\[FB=BC \cos B=\frac{ac}{b+c} \cos B \text{ and similarly } CG=\frac{ab}{b+c} \cos C\]](http://latex.artofproblemsolving.com/4/9/d/49d7dbec6ff4d6d2aca3a2c73bf0c725e66874f5.png)
Hence see that 
And by Ceva's theorem (bar directed lengths because bleh) we are done. 
invert at 
replaces with 
,
and we want to prove it is the midpoint of 
,![\[-1=(H^*,A';B,C) \overset {H_A}= (J^*,\infty;B,C)\]](http://latex.artofproblemsolving.com/9/3/6/936d041da1dc0b71222554c79d89a1ee68d6c46a.png)
And we are done.
