[/quote]
,
be a real or complex inner product space. Suppose that 
and that 
Then
,
such that 
,![\[
f(x^2+f(y))=f(xf(x))+y
\]](http://latex.artofproblemsolving.com/d/e/d/dedd2338779fcc6723077ce586713fb4a68b9931.png)
be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at 
.
and 
and 
.
and 
meet 
and 
at 
and 
,
,
bisects both angles 
,
,
.
Prove that
and 
Prove that
be positive real numbers such that 
Find the minimum value of 
lies on the bisector of an acute angle with vertex 
,
be the projections of 
meets the sidelines at points 
and 
such that 
.
and the circle with center 
are tangent.
are nonzero real numbers such that![\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]](http://latex.artofproblemsolving.com/6/d/9/6d9aabd8130b99d7128de9a3f063eb4c7124febf.png)
let 
be the maximum value of the expression![\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]](http://latex.artofproblemsolving.com/a/b/2/ab28769126365574d4c0108500c370826082de6b.png)
Determine the sum of the numerator and denominator of the simplified fraction representing 
Prove that
Where 
such that for all 
,![\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]](http://latex.artofproblemsolving.com/c/0/c/c0cdf1bcb8053f5dc26e9df6355207b5e479f861.png)
be a set of 
elements. Let 
be subsets of 
of them has more than 
elements. Show that there's a member of 
different subsets 
.
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 
such that 
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 
.
;
.
.
,
.
.
