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.
and integers 
, prove that for any 
positive numbers 
satisfying 
for 
(with the convention 
), the following inequality holds:![\[
\sum_{j=1}^n a_j^{2m-1} \leq \frac{mt}{2} \left( \sum_{j=1}^n a_j^{m-1} \right)^2.
\]](http://latex.artofproblemsolving.com/5/c/d/5cd52e9ce307563bde96f697a75079229d4bfcc2.png)
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
.
prove that
be an acute angled triangle with 
and incircle 
. Let touch the sides 
and at 
and 
respectively. Let 
and 
be points outside 
satisfying ![\[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\]](http://latex.artofproblemsolving.com/4/7/0/470b0486fa1342c4b6444ec1e3731c23bbb842fb.png)
Prove that the circumcircles of 
and meet at a point 
.
be a fixed integer. The cells of an 
table are filled with the integers from 
to 
with each number appearing exactly once. Let be the number of unordered quadruples of cells on this board which form an axis-aligned rectangle, with the two smaller integers being on opposite vertices of this rectangle. Find the largest possible value of .
of 
touch at , 
, let 
denote the line tangent to through 
. Define 
such that 
. Prove that the circumcenter 
of lies on 
.
and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching without passing through 
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
be the incenter of triangle . Let 
, 
, and 
be the orthocenters of triangles 
, 
, and 
, respectively. Prove that the lines through , , and , parallel to 
, 
, and 
, respectively, are concurrent.
and 
have a common tangent that meets in and in . The two circles intersect at and where is closer to than . Line 
meets circle a second time in 
. Prove that 
bisects angle 
.
, 
in 
.
and 
are homothetic.
.
are cyclic.
.
touches .
be the point such that 
and lies on circle. Then, 
must be collinear. Applying Pascal's Theorem on 
, we get that 
are collinear. We know that 
, and 
. So, lines 
are concurrent. But we know that 
concur at . Hence, lies on tangent to 
at circle. 
exists. Let 
meet the circumcircle at . Define 
as the intersection of the tangent from and the line parallel to 
thru 
(which is just 
). We then just want to prove 
, which is equivalent to proving that 
are concurrent. By Desargues theorem, this is equivalent to proving that 
are collinear. We can simplify these as 
. The second is just the point at infinity along , so we just want to show that the line formed by 
and 
is parallel to . We just use complex numbers, clearly 
, then 
. Then 
. Thus we need to verify 
is self conjugating. This is obvious.
. Note that since 
, we have
On the other hand, let point be on the segment such that 
. Thus 
implying 
is a cyclic quadrilateral. And since 
, we get that 
From which the result follows that is tangent to 
at .
.
by parallel lines, and ![\[\frac{CM}{BC} = \frac{CM}{DL} \cdot \frac{DL}{BC} =\frac{KC}{KD} \cdot \frac{AD}{AC} = \frac{AC}{BD} \cdot \frac{BC}{AD} \cdot \frac{AD}{AC} = \frac{BC}{BD}. \ \blacksquare\]](http://latex.artofproblemsolving.com/1/e/f/1efb37b2ed7d419bfe2747fda35568683deaab46.png)
, implying desired tangency.
and see that 
and so 
, , collinear. Redefine as 
. Now just use Pascal on 
.
meet 
at .
we have that 
Thus, 
so is on 
proves that 
are collinear which finishes.
such that 
and 
.![\[\sum_{i=1}^n (a_i-a_{i+1})^3\leq \frac{1}{2\sqrt{3}} (\sum_{i=1}^n a_i )^3,\]](http://latex.artofproblemsolving.com/8/1/0/810f07e3394370ce06aec1e7025bcf5cc5433d60.png)
where 
.
is inscribed in circle 
R
intersect at 
is chosen on the diagonal 
is chosen on the segment 
so that 
Prove that line 
and Pascal for 
.
.
,
,
are collinear. Now let the tangent of 
at 
are collinear, which means 
,
.
,
and you are done.
.
is an isosceles trapezoid so 
gives that the intersection of the tangent at