[/quote]
,
& 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
and puts 
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
.
at 
at 
at 
are concurrent.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
,
and 
.
,
are real numbers such that 
is an integral multiple of 
.
,
for all positive integers 
be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
Y by
Let 
be a point on the circle 
centred at 
and 
a circle centred at . For 
, a chord 
of is tangent to at 
and another chord 
of is perpendicular to 
at 
. Let 
be the other tangent from 
to at 
and 
be the intersection of 
with 
. Prove that 
are collinear.
be a point on the circle 
centred at 
and 
a circle centred at . For 
, a chord 
of is tangent to at 
and another chord 
of is perpendicular to 
at 
. Let 
be the other tangent from 
to at 
and 
be the intersection of 
with 
. Prove that 
are collinear.This post has been edited 1 time. Last edited by parmenides51, Mar 18, 2020, 11:32 AM
be the intersection of 
.![\[\angle AQT = \angle AQS = \overarc{AP} = \overarc{AR} = 180 - \angle AQR\]](http://latex.artofproblemsolving.com/e/2/c/e2ca6ba6445d1dfc5efccbc71b47e0f929d8cc2a.png)
This proves that 
are collinear.
are collinear. Observe that 
,
is cyclic. This means![\[\angle MSA = 180 - \angle MPA = \angle RQA = 180 - \angle AQT = 180 - \angle AST\]](http://latex.artofproblemsolving.com/7/5/8/758cbdf7edc41149ad34d34544a393ca0373be36.png)
Therefore, 
are collinear. This means that 
is the inverse of 
with respect to 
are collinear, since 
bisects both 
and 
.
are collinear.
and 
are cyclic with diameters 
and 
respectively, 
and 
,
and hence 
