AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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 
. Prove that
Equality holds when 
or 
and 
, Ivica and Marica play the following game: Given two piles of cookies, on each turn a player takes 
cookies from one of the piles, of which he eats 
and puts of is arbitrary in every move. Players take turns alternatively, with Ivica going
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
. Show that the tangent of 
at 
,the tangent of 
at 
, the tangent of 
at 
are concurrent.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly positive integer factors 
. Assume that no prime larger than 
divides any of the 
's. Show that there must be some perfect cube among the 's.
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
, 
and 
.
, where 
are real numbers such that 
is an integral multiple of 
. Using (a) or otherwise show that if 
, then 
for all positive integers .
be the orthocenter of an acute-angled triangle 
. The circle 
centered at the midpoint of 
and passing through intersects the sideline at points 
and 
. Similarly, define the points 
, 
, 
and 
., , , , and are concyclic.
with real coefficient. It is known that 
and 
. Prove that 
can't have 
distinct real roots.
be an acute triangle. Let 
be a point on side 
and 
be a point on side 
such that lines 
and 
are parallel. Let 
be an interior point of 
. Suppose rays 
and 
meet side at points 
and 
, respectively, such that both and lie between and . Suppose that the circumcircles of triangles 
and 
intersect at a point 
. Prove that the points 
, and 
are collinear.
. Prove that:
be a triangle with 
such that there is a point 
on side with 
and BH perpendicular to . Let and be the midpoints of and respectively. A line through parallel to cuts at 
. Prove that 
.
be a triangle with such that there is a point on side with and BH perpendicular to . Let and be the midpoints of and respectively. A line through parallel to cuts at . Prove that .
. on the perpendicular bisector of which passes through we get by angle chasing that 
is cyclic. Furthermore we get:
Combining this with 
shows 
is tangent to 
giving:
is on the -midline we get 
so:
Which is enough to show lies on the -symmedian in as desired.
is a parallelogram, 
and thus 
is a rectangle and 
, that is, and 
are antiparallel; 
being median for 
, it is symmedian for , done!
and are parallel, 
, Now draw a line parallel to passing through and extend to hit the line parallel to at 
, This gives 
, connecting 
, we are left with proving 
is cyclic, now we have to prove 
, is parallel to as is the midpoint of , so 
, we have to show that circumcenter of 
lies on 
so 
is a tangent, as 
is isosceles, 
, if intersects 
at 
,
, now we have to show 
for to be the circumcenter, by angle chasing we get 
and 
, so for triangle 
, is the circumcenter, giving 
.
is a parallelogram, and thus is a rectangle and , that is, and are antiparallel; being median for , it is symmedian for , done!
is a parallelogram by midline and given conditions. Now 
so 
parallelogram. finishes.
,
.
is a rectangle.
is a parallelogram, we have 
.
and 
,
.
is a rectangle by the congruent right angled triangles 
and 
.
and 
.
and 
.
.
,
,
,
Which is what we wanted to prove.
.
are isogonal, so 