[/quote]
,
be a function such that for every regular 
-
we have 
.
for all reals 
.
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
and 
.
passes through the midpoint of 
& 
.
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 
Y by Adventure10
In triangle 
, 
is the centre of the incircle. There is a circle tangent
to
at which passes through 
. This circle intersects 
once more
in
and intersects 
once more in 
. The line 
intersects 
in 
.
Prove that
, 
is the centre of the incircle. There is a circle tangentto
at which passes through 
. This circle intersects 
once morein
and intersects 
once more in 
. The line 
intersects 
in 
.Prove that
This post has been edited 1 time. Last edited by 61plus, Jul 18, 2014, 7:25 AM
is perpendicular to 
,
.
and 
by 
and 
,
is perpendicular to 
.
.
.
,
and 
have equal radii we get:
,
,
and 
.
bisects 
,
not containing 
.
.
and 
,
in 
property. So 
.
,
,
,
.
so 
.
.
.
so 
.
.
so 
.
.
