[/quote]
,
chessboard is 
.
satisfying 
,
denotes the set of all real numbers.
are real numbers such that 
prove inequality 
steps, can all the numbers on 6 vertices be equal?
individual tables arranged into 4 rows and 9 columns (all the rows are numbered from 1 to 4, all the columns are numbered from 1 to 9). Number of students in this class is 
.
row, 
column is numbered with an integer number 
.
.
having 
and 
satisfy that: 
,
,
,
be 
integers. If 
is an integer, prove that the only solution to
is is 
.
such that for every 
the following equality holds: 
and 
Prove that
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
,
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
,![$P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$](http://latex.artofproblemsolving.com/0/5/1/0514f85850830af9a9e1a4dd5e7a4e8e68b2dff9.png)
.
polynomial with only one root.
,
,
.
.
: 
-is integer.
,
and 
.
is prime, 
with 
and 
,
inequalities, these given comparisons satisfy the following three constraints:
and 
,
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
Y by Rounak_iitr
Let 
be a triangle, with incenter 
and 
-excenter 
. The lines 
, 
, 
and 
intersect the circumcircle of at 
, 
, 
and 
respectively. Let 
, 
be diameters in the circumcircles of triangles 
and 
respectively.
Prove that
.
Proposed by Ivan Chan Kai Chin
be a triangle, with incenter 
and 
-excenter 
. The lines 
, 
, 
and 
intersect the circumcircle of at 
, 
, 
and 
respectively. Let 
, 
be diameters in the circumcircles of triangles 
and 
respectively.Prove that
.Proposed by Ivan Chan Kai Chin
is cyclic, let 
center of 
and let 
then let reflection of 
over 
then if we redefine 
are reflections of 
over 
respectively but in addition we have 
parallelogram but also by reflection 
and thus more parallelograms are found and thus 
but also by I-E Lemma spam and orthocenter reflection things and the parallelograms which give 
this is enough to imply by size of radiuses that 
be the A-sharkydevilpoint and 
be its "extraverted" image, then from reflection over 
and radax that 
are isogonal in 
we end up getting that 
which implies the desired angle condition (just use directed angles lol), thus done 
.
and let 
be the antipodes, 
.
and 
are collinear.
concur, after reflection over 
concur hence 
are collinear. Similarily since 
concur, after reflection over 
concur thus, 
are collinear. If 
is the midpoint of arc 
,
and 
hence 
so 
.
as desired
