[/quote]
,
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
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.
Y by Adventure10, Mango247
Let 
be a triangle . The internal bisector of intersects the side 
at 
and the circumcircle of again at 
Let 
be the perpendicular projection of 
onto 
the circumcircle of 
intersects line 
again at 
Lines 
and 
meet at point 
Prove that 
be a triangle . The internal bisector of intersects the side 
at 
and the circumcircle of again at 
Let 
be the perpendicular projection of 
onto 
the circumcircle of 
intersects line 
again at 
Lines 
and 
meet at point 
Prove that 
This post has been edited 1 time. Last edited by sqing, Sep 3, 2018, 12:02 AM
is cyclic.![\[\angle AKP+\angle PBA=\angle AKC+\frac{\beta}{2}+\angle PBL=\angle AKC+\frac{\beta}{2}+\angle PCL=\angle AKC+\frac{\beta}{2}+\angle KCA=180^\circ+\frac{\beta}{2}-\angle CAK=180^\circ\]](http://latex.artofproblemsolving.com/c/1/8/c1842b26638438c43d80771b440e685b6d360af7.png)
where the last equality is since 
.
is the midpoint of 
.
this is equivalent to showing that 
are on a common circle around 
by Thales. However, we can't really do angle chasing with the angle 
since 
and 
and 
have the radical axis 
we find that it intersects the lines 
and 
and the circle 
so that is a good one. This motivates the following definition.
be the intersection of the circle 
is cyclic.
it suffices to prove that 
.![\[\angle ACD=\gamma+\angle BCD=\gamma+\angle BLD=\gamma+270^\circ-\angle LBA-\angle BAK=\gamma+270^\circ-\alpha-\frac{\beta}{2}-\frac{\beta}{2}=90^\circ\]](http://latex.artofproblemsolving.com/9/7/8/978098d53253525264952153f82cde7770eeb412.png)
and 
are radical axes of three circles, hence have a common point which then must be 
are collinear. But 
,
.
are on a Thales Circle. Since 
by definition of 
meets 
at 
respectively. Let 
be the projection from 
and 
meets 
at 
.
and 
.
is concyclic thus 
.
,
.
yields
hence we are done.
intersect 
at 
gives that 
,
intersects 
at 
which forces 
and in particular 
.
So 
which means 
and hence the result.
meet 
at 
.
,
.
at 
,
is an isosceles trapezoid.
intersect 
,
is an isosceles trapezoid so since 
,
.
,
(since 
,
.
is tangent to 
again at 
, then by Pascal Theorem of the hexagon 
, we have 
, so this implies that from angle chase that 
is diameter of 
,which implies that 
is cyclic.
.
, where at last we uses Fact-5 , so 
and we are done. 
and let 
,
is cyclic.
is right so 
is cyclic, then 
(last quality follows from noticing that 
)
be the point such that 
and let 
.
,
,
is cyclic, we have 
,
,
is cyclic, then 
which completes the solution.
