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 circumscribed about circle 
, and let 
be the orthocenter of 
. The circle touches line 
at 
. The tangent to the circle 
at meets at 
. Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
be positive real numbers satisfying 
. Prove that![\[
3(x^{\frac{2}{3}} + y^{\frac{2}{3}}) \geq 4 + 2x^{\frac{1}{3}}y^{\frac{1}{3}}.
\]](http://latex.artofproblemsolving.com/f/3/c/f3cd114f909f388eacee142d3f90d07f6b149131.png)
be the maximal number of the vertices of a convex lattice polygon with exactly 
lattice points in the interior. Show that:
for 
for some constant 
.
and a set consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
, for 
in at least 
different ways.
be real numbers such that 
. Prove that
Let be real numbers such that 
. Prove that
and , we define 
as follows:
where 
are all the positive divisors of . For example,
.
Find all positive integers such that 
.
Find all positive integers such that 
.
be a convex quadrilateral. Suppose that the circles with diameters 
and 
intersect at points and 
. Let 
and 
. Prove that the points 
, 
, and are concyclic. and are not the diagnols)
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
. Prove that there exists a positive integer satisfying 
.
?
be a triangle with 
and 
, 
-incenter, 
-circumcenter. Prove that perpendicular bisector of 
, line 
and line have a common point.
and 
intersect circumcircle of at 
and 
.
that means that point lay on the perpendicular bisector of . Analogously we get that lay on the perpendicular bisector of .
is perpendicular bisector of .
are cyclic since 
are cyclic since 
and are three radical axis of three circles (circumcircle, 
, 
).
at 
and that 
Since 
therefore 
and 
are equilateral. Also 
therefore 
and the incircle of bisects 
is the perpendicular bisector of 
be the reflection of in 
Then note that we have 
therefore lies on 
Now we already have 
which give 
leading to the fact that bisects 
This readily implies that bisects 
as the perpendicular bisector of 
so passes through the pole of 
wrt 
Since pole of is 
and because 
is the perpendicular bisector of 
we are done. 
cut the circumcircle of again at the point 
.
, is on a same circle with its center at , so 
.
be a line passing through , parallel to . Denote by and 
its intersections with the circumcircle that 
is a quadrilateral with vertices arranged in this order. is a parallelogram and that 
is an isosceles trapezoid.
, 
, and 
.
, clearly as 
is cyclic, so we find 
where 
is the mid point of the arc 
, also note that 
is a parallogram, hence bisects 
, now if intersects 
at 
then clearly as 
we get 
,but it is well known that 
, hence 
is cyclic.. Also note that as 
,so we have 
, so, 
is isoscles with 
hence the perpendicular from tn 
must bisects Which proves our claim.
and meet circumcircle of at S and 
. It's well known 
is perpendicular bisector of .
so 
is cyclic.
so 
is cyclic., and are concurrent.
projectively along 
. Then 
moves with degree 
, so 
moves with degree 
and since the midpoint of moves with degree 
as well, the perpendicular bisector of moves with degree 
. moves with degree as lies on 
, so we check 
cases.
lies on perp and isn't lies such that is parallel to (two cases). is the antipode of 
wrt .
![[asy]
size(400);
defaultpen(linewidth(0.7)+fontsize(10pt));
pair A = dir(135), B = dir(210), C = dir(330), I = incenter(A,B,C), O = circumcenter(A,B,C), P = extension(O,I,B,C), M = dir(270);
draw(P--B--C--A--B,rgb(0.1,0.1,0.7));
draw(circumcircle(A,B,C), rgb(0.1,0.6,0.1));
draw(A--P--O,rgb(0.5,0.1,0.5));
draw(A--M--O,rgb(0.9,0.5,0.1));
draw(circumcircle(A,P,O),linetype("3 3")+rgb(0.8,0.1,0.1));
draw(circumcircle(B,O,C),rgb(0.9,0.3,0.1));
clip((-5,-1.1)--(1.5,-1.1)--(1.5,1.5)--(-5,1.5)--cycle);
dot("$A$",A,N,linewidth(3.3));
dot("$B$",B,SW,linewidth(3.3));
dot("$C$",C,SE,linewidth(3.3));
dot("$M$",M,S,linewidth(3.3));
dot("$P$",P,SW,linewidth(3.3));
dot("$O$",O,NE,linewidth(3.3));
dot("$I$",I,NE,linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/a/9/e/a9e3c5af6393dda089cafd931605227d4609fde1.png)
. Let be the intersection point of and ; it suffices to prove 
. be the midpoint of minor arc 
. Because 
, points 
, , , and 
lie on a circle centered at (Fact 5). Compute
It follows that quadrilateral 
is cyclic. Finally, 
, so 
implies .
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
be reals such that 
and 
.
and 
.
through the line 
is the in-radius and 
are internal angles of 
then 
apply cosine law to 
thus 
intersects 
,
are cyclic,we can prove that 
