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 
.
?
we have 
.
.
be triangle with circumcenter 
and orthocenter 
. 
intersect again at 
respectively. Lines through parallel to 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
satisfying
Prove
and 
be positive integers. Suppose that the sum of integers between and , including and , are equal to 
.
, find the pair such that achieves the minimum.
be an acute-angled triangle with 
. Let 
be the circumcircle of . Let 
be the midpoint of the arc 
of containing 
. The perpendicular from to 
meets 
at and meets again at 
. The line through parallel to meets line 
at 
. Denote the circumcircle of triangle 
by 
. Let meet again at 
. Prove that the line tangent to at 
meets line on the internal angle bisector of 
.
be a triangle with circumcircle . Let 
and 
be two lines through the points 
and 
, respectively, such that 
. The second intersections of and with are and 
, respectively. Assume that and are on the same side of as . Let 
intersect at 
and let 
intersect at 
. If 
, 
and 
are circumcenters of the triangles , 
and 
, respectively, and is the circumcenter of the triangle 
, prove that 
.
is . Two moving points 
moves on sides 
, satisfy that 
. Prove that 
is a definite value.
be real numbers less than 2.Then prove that 
is such that 
for any positive integers . Assume there exists a positive integer 
such that 
for all positive integers 
. Determine all possible values of 
.
is such that for any positive integers . Assume there exists a positive integer such that for all positive integers . Determine all possible values of .
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
.![$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$](http://latex.artofproblemsolving.com/0/5/0/0502ba65f0545b3af84c41b2549d6256550b473b.png)
.
.
Where 
Prove that
Let 
Prove that
satisfies the condition and 
.
.
and 
,
for all 
.
,
for all non-negative integers 
and 
,![\[
g(a, b) + g(c, b) = g(ac, b^2).
\]](http://latex.artofproblemsolving.com/9/1/d/91dd66cfb45616ccfaa1691e942e816060decf3d.png)
denote the assumption on 
in this equation. For 
,
depends only on the value of 
for all prime numbers 
.![\[
h(x) = \sum_{p \text{ prime}} h(p)\nu_p(x). \tag{2}
\]](http://latex.artofproblemsolving.com/6/8/b/68b83a00528564c6a54d5e2ac13d438b4266babf.png)
or 
.![\[
h(45) = g(45, 1) = 2g(3, 1) + g(5, 1) = 0.
\]](http://latex.artofproblemsolving.com/8/e/0/8e02c396ee210441ab72d69160d7f57de8ab2805.png)
and 
follows that 
.
we get that 
.
from 
,
,
,
.
,![\[
g(n, x) = g(n, x^n) + (x^2 - 1)h(x) + (x^4 - 1)h(x) + \ldots.
\]](http://latex.artofproblemsolving.com/4/4/b/44b899e553765ac801138cdc87c7900ea3c368da.png)
for some 
,![\[
g(n, x) + (x^2 - 1)h(x) = g(n, x^2) \le g(n, (x + 1)^2) = g(n, x + 1) + (x^2 - 1)h(x + 1).
\]](http://latex.artofproblemsolving.com/9/d/0/9d07a67f85a3666f1bb39273106e7079e9f29a88.png)
,
are constants. Therefore 
for all 
from the beginning of the problem to case 
.
or 
,
for every prime 
because 
is monotonically increasing. Let 
.
,
(the case 
is analogous). Then we have![\[
h(a^k) = kh(5^k)(7) = kh(5)(7) = h(b^k).
\]](http://latex.artofproblemsolving.com/0/1/7/017694b47f0bd9755bf3e2458aaea70e3f458104.png)
.
,![\[
q + 1 = \frac{a^{q + 1} - 1}{a - 1} > \frac{b^{q + 1} - 1}{b - 1} > b^q + 1.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fdfd3799129cc139d96f18b8d683e47f815745.png)
.
,
and 
.