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.
such that for all real numbers 
and 
, 
satisfying
Prove
be positive reals such that 
. Find the minimum and maximum value of
and 
Prove that
Let and 
Prove that
be a positive integer. We start with piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. Find (in terms of ) the smallest number of nonempty piles that one can obtain by performing a finite sequence of moves of this form.
and 
if 
then find the coefficient of 
in 
in terms of 
and 
Prove that
be an acute triangle and let 
be the midpoint of 
. A circle 
passing through 
and meets the sides 
and 
at points 
and 
respectively. Let 
be the point such that 
is a parallelogram. Suppose that lies on the circumcircle of . Determine all possible values of 
.
satisfying the following property: if 
, 
and 
are integers such that 
, then
such that 
.
touches its sides , 
, at the points 
, 
, 
, respectively. Let 
, 
, 
be the reflections of these points , , in the external angle bisectors of the angles 
, , 
, respectively. Show that 
, 
and 
.
touches its sides , , at the points , , , respectively. Let , , be the reflections of these points , , in the external angle bisectors of the angles , , , respectively. Show that , and .
is a trapezoid it follows that 
.
is a trapezoid it yields that 
.
., , be the reflections of these points , , in the external angle bisectors of the angles , , , respectively.
and 
in IMO 1982 problem 2. My problem, with external angle bisectors, is slightly harder; you can use the internal bisector case as a lemma (this is Peter Scholze's solution) or solve it directly.
etc.
are collinear(since 
is on the mid-line of 
) and analogously 
are collinear.
; using the fact that 
we get:![\[ \frac{Z'K}{KY'}=\frac{Z'C}{Y'B}=\frac{Z'Z_1}{Y'Y_1} \]](http://latex.artofproblemsolving.com/8/2/6/82654bfd6f1e837c5d1d1d54400774881a163dde.png)
and the conclusion follows.
![$$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$](http://latex.artofproblemsolving.com/b/1/d/b1d283f2e7b8e574b6c0ff72236c52a39d3da9e8.png)
such that![\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]](http://latex.artofproblemsolving.com/9/8/a/98ab64d17bc7d6bf7533b0874177fc963405c1d8.png)
.
