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 a set consisting of 
positive integers, none of which is a sum of two other distinct members of . Prove that the elements of may be ordered as 
so that 
does not divide 
for all 
.
be an acute triangle with 
and 
. Let 
be the midpoint of the side 
. The circumcircle of triangle 
intersects the side 
again at 
and the circumcircle of triangle 
intersects the side 
again at 
. The point 
lies on the perpendicular bisector of the segment so that the points and 
lie on the same side of 
and 
. Prove that the circumcircles of triangles 
and 
are tangent.
![\[ p = \frac{2ab}{a + b} \]](http://latex.artofproblemsolving.com/0/f/4/0f4297ec59fe97c92fda75af4cc835be6b9a2507.png)
, ![\[ p, a, b \in \mathbb{C} \]](http://latex.artofproblemsolving.com/b/4/4/b4499970ac49d770a810d7345fedef7773cd03f5.png)
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
and 
.
such that
Prove that
has rational sides and equal angles. Show that it is regular.
such that the equation
derivable on 
with 
be a right angled triangle with 
and 
its altitude. We draw parallel lines from 
to the vertical sides of the triangle and we call 
their points of intersection with and respectively. The parallel line from 
to 
intersects the line at the point 
. Let 
be the symmetric of with respect to the line and 
the projections of 
onto and respectively. If 
is the point of intersection of the lines 
and 
, prove that 
.
. Get all the 
such that 
is uniformly continuous on 
.
and 
that satisfy the equation 
be a function, two times derivable on 
for which there exist 
such that![\[ \frac { f(b)-f(a) }{b-a} \neq f'(c) ,\]](http://latex.artofproblemsolving.com/a/3/1/a31861ef909a6e17ed8b495b9b9b16d887ba573f.png)
for all 
.
.
defined for 
being an interval (connected subset of the real line) and 
not being in this image, it follows that we may assume that 
for all . But then 
for all 
and so 
is a minimum point of 
. Of course, this can be written in 11-th grade vocabulary. 
has the same sign for all 
.Can you be give me more detailes please 
?The official solution is based on the fact that the function 
is injective ,hence monotonic.
such that is a connected subset of the plane and the function 
is continuous on this domain, thus its image is a connected subset of the line, thus an interval. I agree however that this is not a solution of an 11-th grade student, but that's how life is. I will not be amazed if in a few years I see complex analysis, Lebesgue integration and other such stuff at RMO. It's quite à pity, it gives huges advantages to some people. 
should be defined on the set of pairs such that 
to ensure that its domain is connected. Of course, all the rest works with this modification, I don't know how I could write such a stupid thing. 
, then 
is strictly concave up or down on some neighbourhood of the point , thus one can draw a close enough parallel to the line tangent at to the function's graph that intersects the graph in two points 
and 
. The slope of that tangent would be 
, equal to the slope of the tangent at , that is,.
is connected, but why is connected ? has clearly two connected components. However with this spirit we can also solve the problem, extend 
to whole plane defining 
and now 
is continuous on whole plane and do similar thing.
has the same sign for all .Can you be give me more detailes please ?The official solution is based on the fact that the function is injective ,hence monotonic.
there exists an 
with the desired property. So you get 
only for values in some set , which has to be dense in the reals, but need not be an interval, so the IVP you use later on in your proof does not give a contradiction. (See the solutions above for proofs that avoid this problem.)
is continuous on 
and that is true because 
is twice differentiable.
. In the first paragraph you show that for all 
has to contain some 
(which means that is a dense subset of the real numbers). This part of your proof is fine. But in order to make the proof of the green claim work you need to show that contains all real numbers. This is missing in your proof (if I interpret you proof correctly).
Prove that
,
Prove that
Let 
Prove that
for all 
and now make 
close to 