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 an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
with incircle 
. Let 
be the point on such that the circumcircle of 
is tangent to and let the -excircle be tangent to at 
. Prove that the tangent from to and the tangent from to (distinct from ) meet on the line parallel to and passing through .
by 
, 
for every positive integer 
. Prove that 
leave distinct remainders when divided by 
.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
are coprime. Consider the sequence 
given by 
. Suppose that there exist positive integers 
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be distinct subsets of 
. Show that there exists 
such that the n subsets 
are also disjoint.
satisfies 
. If 
, find the maximum possible value of 
.
such that for all irrational numbers 
and 
,
be a triangle and be a point on the segment such that 
. Let be the mid-point of . Let and 
intersect in . Determine the ratios 
and 
.
Hence, 
and 
Hence, and
Hence, and 
be the midpoint of 
,
are points of trisection of .
, is the mid-point of
and is that of 
. Hence 
. Thus 
. Hence we find
. Hence 
.
. We obtain 
.
, since we get 
. Thus 
.
. Hence 
. Therefore 
.
- 
.
.
midpoint of 
.
so 
.
.
by midpoint theorem, 
.
.
.
.
.. 
just memorize a bunch of commands 
midpoint of . so .. by midpoint theorem, ...... just memorize a bunch of commands 
. Set 
. Since and are cevians, 
. Normalizing everything, we seek ![$\frac{BP}{PE}=\frac{[ABP]}{[APE]}$](http://latex.artofproblemsolving.com/5/d/e/5de99b1eadea6f7b5c14c34434a36c720bcf44ec.png)
and ![$\frac{AP}{PD}=\frac{[ABP]}{[PBD]}$](http://latex.artofproblemsolving.com/8/5/7/857bdba401c4e484870c78e2ef48b9db13e598d6.png)
which are immediate by the area formula.
![$[APE]=\delta$](http://latex.artofproblemsolving.com/0/2/c/02cc358796d269ec4937f6d856b660cacc59bb0f.png)
as in triangle APC : PE is the median then ![$[APE]=[EPC]=\delta$](http://latex.artofproblemsolving.com/d/8/a/d8a87926b268a46883ad1dbbbb3eed2a56a57748.png)
call ![$[BPD]=\beta$](http://latex.artofproblemsolving.com/b/7/9/b79c71fe181678cc21fc86b95abfee90fb493521.png)
then by BDT in triangle ![$BPC : [PDC]=2\beta$](http://latex.artofproblemsolving.com/f/0/6/f062cd0c4d86f4b0101e4491201084670af25af7.png)
now call [ABP]=X then by BDT in triangle 
which gives 
now again by BE as median in trinagle ABC we get 
which means 
which means and 
by BDT in triangle ABD and ABE respectively
, and 
, 
is the midpoint of , which gives 
.
and 
, we have that 
.
and 
. Thus, 
. Again, using BPT in 
and 
, we get that 
. We are thus done.
and 
.
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
s
where 
satisfy that 
.![$$\sum_{cyc} \sqrt[4]{\frac{a^3}{b+c}}$$](http://latex.artofproblemsolving.com/e/f/0/ef089c4f2915bbf75b0f60b044bbe9273b301a81.png)
.
