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.
and 
be two sequences of integers such that 
and 
for all 
. Suppose that 
divides 
for an infinity of integers . Prove that there exist an integer 
such that 
for all
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 
,
, the incircle of triangle BCD. Let X and Y be the intersections (closer to D) of DM and DN with . Prove that XY is parallel to BC.
, the incircle of triangle BCD. Let X and Y be the intersections (closer to D) of DM and DN with . Prove that XY is parallel to BC.
to be the incircle, we take it to be the reflection of the circle tangent to 
in the perpendicular bisector of (all we need is for to be a circle tangent to 
), so let's do that. cuts in the points 
, the reflections of 
(respectively) in the midpoint of . Assume it touches in 
respectively, and consider the circle 
tangent to in 
s.t. 
lies between 
, the same for , and 
. After making the computations which say that the power of wrt is equal to the power of wrt and vice-versa, we see that cuts precisely in , so 
are the homothetic images of through the homothety centered at which maps to , and this shows that 
(the last equality means equality of lines).
wrt is equal to the power of wrt and vice-versa, we see that cuts precisely in ... at 
? Perhaps I am missing something obvious....
be the circle tangent to AB, AC at S, T and intersecting BC at M, N. 
Let 
be points such that BS' = BS and CT' = CT. We also require that S, S' are on the opposite sides of the line BC and the same for T, T'. WLOG, assume 
hence 
A circle 
tangent to DB at S' and passing through M passes also through N. Likewise, a circle 
tangent to DC at T' and passing through N also passes through M. Assuming the circles 
intersecting at M, N are not identical, MN is their radical axis. The tangent length to from D is equal to 
while the tangent length from D to is equal to 
Since the sides 
and the diagonals 
of the isosceles trapezoid ABCD are equal, and since tangent lengths 
to the circle from A are also equal, we get 
This means that D is on the radical axis 
of 
which is a contradiction. The only way to avoid this contradiction is to have the intersecting circles 
identical, so that their radical axis does not exist. Call this unique circle 
D is the external similarity center of the circle 
and the incircle of 
the triangles 
are centrally similar with the same similarity center, therefore their corresponding sides are parallel, 
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 
