[/quote]
,
be an acute-angled scalene triangle with circumcenter 
.
,
,
the midpoints of sides 
,
,
,
be the circle passing through 
and tangent to 
at 
at points 
and 
,
be the midpoint of segment 
,
be the intersection of lines 
.
and that triangle 
is isosceles.
chessboard has a real number written in it in such a way that each number is equal to the geometric mean of all the numbers a knight's move away from it.
is inscribed in circle 
.
intersects side 
and 
and 
(other than 
be the midpoint of side 
intersects sides 
and 
again at 
and 
(other than 
,
be the foot of the perpendicular from 
.
is tangent to the circumcircle of triangle 
.
Prove that
Where 
and 
Prove that
and 
be different points on a circle 
is not a diameter. Let 
be the tangent line to 
is such that 
.
is chosen on the shorter arc 
of triangle 
intersects 
meets 
.
is tangent to 
satisfying
Prove
.
.
over 
and 
.
lies on 
.
be a trapezoid with parallel sides 
.
and 
lie on the line segments 
and 
,
.
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points 
and 
are concyclic.
be a positive integer. A board with a format 
is divided in 
(or 
)
positive integers 
such that 
for any 
,
such that 
?
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
.
,
is the smallest number different from 
which doesn't divide 
.
Y by jhu08, Sprites, Adventure10, Mango247
Let 
and 
be three positive integers. Prove that there exist a positive integer 
and a set of 
positive integers 
, such that, for every 
, the 
-ary expansion of is a 
-digit palindrome, and the 
-ary expansion is exactly 
.
proposed by Bojan Basic, Serbia
and 
be three positive integers. Prove that there exist a positive integer 
and a set of 
positive integers 
, such that, for every 
, the 
-ary expansion of is a 
-digit palindrome, and the 
-ary expansion is exactly 
.proposed by Bojan Basic, Serbia
Notice that 
since obviously we can select 
hence 
and at the 
t
given the new 
which satisfies 
and there is at least one 
with 
.
satisfying the following conditions: (i) each of them is coprime with 
;
for all 
is coprime with 
.
,
are distinct large primes. Another option is to choose some large odd 
coprime with 
.
,
represent all residue classes modulo 
such that 
.
.![\[ b_0=B, ~ c_i =\frac{B}{d_i + 1}, ~ b_i=d_ic_i=B-c_i, ~ \text{and}~ N = \overline{xyx}_B = x(B^2 + 1) + yB. \]](http://latex.artofproblemsolving.com/4/e/b/4ebecec757b02cc60687accb7c9e6c636787adbb.png)
We claim that these values satisfy all the requirements; moreover, for every 
the 
we have![\[ N = x(d_i + 1)^2 c_i^2+y(d_i + 1)c_i + x = (x \cdot (d_ic_i)^2 + x)+ (2c_ix + y) \cdot (d_ic_i) + (xc_i^2+yc_i). \]](http://latex.artofproblemsolving.com/a/d/7/ad716d5662296e27857890a4e93e7eb99cd24088.png)
Clearly, 
.
and 
.
is divisible by 
,
.![\[ xc_i + y = \frac{xB + y(d_i + 1)}{d_i + 1}=\frac{(xB + y) + d_iy}{d_i + 1}. \]](http://latex.artofproblemsolving.com/9/5/2/9525dbc187f28b5f2a442b204c60b3b5bf0416fc.png)
The numerator of the last fraction is divisible by 
(due to the choice of 
)
.
;
.
.
,
.
.
