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.
have the same incenter 
.Prove that 
is concyclic iff 
is concurrent
, 
, 
be real positive numbers. Show that ![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\]](http://latex.artofproblemsolving.com/0/9/5/0959464bbb9b821a11df292f9e6addded7326371.png)
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 real numbers such that 
and
be integers such that, for each 
, there exists a subset 
with![\[
\sum_{j \in S} a_j \equiv i \pmod{2048}.
\]](http://latex.artofproblemsolving.com/8/b/e/8be39aaf83795f611b1c3fa0d5971905bd6ebdfe.png)
Show that for each 
, there is exactly one 
such that 
is divisible by 
but not by 
.
is the summation notation, for instance, 
, while for the empty set 
, one defines 
.
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 an acute-angled triangle with 
. The circle with diameter intersects the sides 
and 
at 
and 
respectively. Denote by 
the midpoint of the side . The bisectors of the angles and 
intersect at 
. Prove that the circumcircles of the triangles 
and 
have a common point lying on the side .
. Given an infinite sequence of positive real numbers 
, Turbo successively crawls distances around the circle, each time choosing to crawl either clockwise or counterclockwise.
with the following property: for every sequence of positive real numbers with 
for all 
, Turbo can (after studying the sequence) ensure that there is some point on the circle that it will never visit or crawl across.
is inscribed in circle . Let denote its incenter and 
its -excenter. Let denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides and at 
and 
respectively. The circumcircle of 
meets 
a second time at , and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
, prove that![\[
2+ \frac{B+C-A}{\sin A} +\frac{C+A-B}{\sin B} +\frac{A+B-C}{\sin C}
> \frac{\pi}{2}\left( \frac{\sin^2 A+\sin^2 B+\sin^2 C}{\sin A \sin B \sin C } \right)
\]](http://latex.artofproblemsolving.com/2/0/1/2012bbef4475a5c433a4e54c53195c729c5bab99.png)
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 .
.![$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$](http://latex.artofproblemsolving.com/0/5/0/0502ba65f0545b3af84c41b2549d6256550b473b.png)
.
, such that for all 
:![\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]](http://latex.artofproblemsolving.com/1/4/1/1418fad9fd00b31c38bfbc5657e2a322d66ef450.png)
satisfying
Prove
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 
.