[/quote]
,
(
well). i want to study so if you guys have some easy or normal problems please send me
x
to 
row and 
column get symbolized by square 
. Subset 
of squares on the board is called "good" subset if two random squares 
belong to 
with 
or 
are not belong to 
has a right angle at 
.
such that 
and 
lies between 
is chosen so that 
and 
is the bisector of 
.
is chosen so that 
and 
is the bisector of 
.
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，
is such that 
for any positive integers 
.
for all positive integers 
.
.
,
,
,
,
in some order.
.
be the point such that the quadrilateral 
is a parallelogram. Let 
be the point on the line 
such that 
.
at 
.
.
, 
and 
be points on circle 
divided into three equal parts. Construct three equal circles 
,
,
tangent to 
be any point on arc 
,
,
,
to circles 
.
,
,
,
be points on circle 
equal parts. Construct 
tangent to 
,
,
to circles 
of 
(where 
)
and 
with sum ![\[\sum_{i=1}^n \min\{{x_i^2, x_{i+1}}\}\leq \frac{1}{2}.\]](http://latex.artofproblemsolving.com/7/f/0/7f01a8d065c8e03233fe2254946522a5762f80e2.png)
where 
.
of the form 
with 
,
Determine all ordered pairs of polynomials 
with integer coefficients such that 
.
be the reflections of 
,
be the reflections of 
over 
,
be the orthocenter of triangle 
.
.
is parallel to 
.
,
,
be positive real numbers. Prove the inequality![\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \]](http://latex.artofproblemsolving.com/d/a/5/da5096e4c732244baf08effc43b5d086444e2c62.png)
.
is a root of the cubic equation 
(real or complex), then 
satisfying the equation for all real numbers 
:![\[
f(x^2 + y + f(y)) = f(x)^2
\]](http://latex.artofproblemsolving.com/0/6/f/06f2c3b16aecf900a28c033c45993564afd80055.png)
Y by Batman007, Davi-8191, tenplusten, FlakeLCR, test20, AlastorMoody, Purple_Planet, vsamc, ilovepizza2020, megarnie, pog, EthanTAoPS, jhu08, THEfmigm, Jupiter_is_BIG, sleepypuppy, EpicBird08, Adventure10, Mango247, Aopamy, ItsBesi, cubres, Rounak_iitr, and 1 other user
Let 
be a point interior to triangle 
(with 
). The lines 
, 
and 
meet again its circumcircle 
at 
, 
, respectively 
. The tangent line at 
to meets the line 
at 
. Show that from 
follows 
.
Proposed by Marcin E. Kuczma, Poland
be a point interior to triangle 
(with 
). The lines 
, 
and 
meet again its circumcircle 
at 
, 
, respectively 
. The tangent line at 
to meets the line 
at 
. Show that from 
follows 
.Proposed by Marcin E. Kuczma, Poland
.
is tangent to the circuncircle of 
.
.
be the intersections of 
,
is in the 
)
.
.
are orthogonal so the point of intersection of 
lies on the circle 
and call this point 
.
belong to ![$[S,B]$](http://latex.artofproblemsolving.com/5/1/3/51355ee5df3a30573fcf9db351ea2efbc476df82.png)
.
since 
.
is the external symmedian of the vertex 
,
,
forming a harmonic range and this implies that 
.
of the 
. Define the point 
on the line segment 
is the bisector of the 
.
in 
and 
so the traingle 
and 
are homothetic so the points 
and 
are collinear and we have 
.
and 
are similar, we have that 
Analogously we have that 
So it's enough to prove that 
.
so 
.
is common for triangles 
and 
,
But triangles 
and 
are similar too. So 
Therefore 
and we have done!
which means that the triangles
so 
(1). But 
and 
(2)
and we are done.
we conclude that 
is tangent to the circumcircle of triangle 
at 
.
.
.
is harmonic, because projecting from 
meet 
.
,
.
(if we move 
.
,
,
.
is the circumcenter of 
,
.
.
and 
are both isosceles, 
.
.![\[SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}\]](http://latex.artofproblemsolving.com/0/1/c/01c2b22c06bde3f2954f728fe145ef4476f2bbe6.png)
.
.
.
.
,
so if 
is the reflection of 
then 
,
,
)
so 
is parallel to 
.
.
and permutes 
,
meet 
at 
.
is harmonic. Now it suffices to show 
is a diameter. We will now use phantom points. Let 
be the 
.
.
,
,
and 
are tangents to 
where 
.
tangent to 
.
from harmonic 
,
.![\[\angle SPB = \angle PAS = \angle KLB,\]](http://latex.artofproblemsolving.com/2/3/d/23d86a7c84f326c9e5982824a8cf05538b14bf69.png)
or 
.
.
and 
,
,
and 
at 
and 
are similar, and that 
and 
are similar
to meet 
,
is similar to 
,
,
.
,
Further, we also have,
so the claim follows. 
from the diagram. Now, we note that
which leads to 
,
.
Therefore, it is left to show that 
.
as desired. 
and also 
from which we get that 
.
and 
cyclic implies 
.
to get that 
lies on 
to get that 
lies on 
is parallel to 
.
.
,
.
.
is harmonic, by symmetry we must have that 
,
we see that 
is cyclic.
.
lies on the 
.
.
and 
we have 
and from LOS we have 
.
Now just plug in either condition in this equation to get the other one. 
![[asy]
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.
which in turn implies that 
.
,
.
,
.
and hence 
,
Works though 
gives 
collinear where T = 
now i claim that the quadrilateral 
is cyclic , that ends the problem because, it means 
which means arc 
= arc 
so it finishes.
. notice 
where the last one follows from the fact that 
is tangent to the circumcircle of 
since 
from 
we obtain 
which indicates 
giving 
gives 
.
gives 
.
where 
.
because 
.
.
.
which finishes the proof.
bisects 
,
,
.![\[\angle PAB+\angle PCB=\angle PDB-\angle APD+\angle PCB=\angle CPD-\frac12 \angle APB-\angle ABC.\]](http://latex.artofproblemsolving.com/9/b/b/9bbe5aab0ed3d1409b6d91470419f3a246a46822.png)
But since![\[\angle CPD=180^{\circ}-\frac12 \angle CSD=90^{\circ}+\angle CDS=90^{\circ}+\frac12 \angle C+\angle ABC,\]](http://latex.artofproblemsolving.com/2/f/8/2f8c0397e55f0c9066a5c62d44337efce6cad961.png)
we just get that![\[\angle PAB+\angle PCB=90^{\circ}+\frac12\angle C-\frac12 \angle APB=\angle PBA+\angle PCA,\]](http://latex.artofproblemsolving.com/4/7/2/47261bf80522811f65c9c84673070340bfa42e9c.png)
which implies the result. 
and 
Hence we have that 
.
Hence 
so 
.
