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.
converge?
be a parallelogram with 
A point 
is chosen on the extension of ray 
past 
The circumcircle of 
meets the segment 
again at 
The circumcircle of triangle 
meets the segment 
at 
Prove that lines 
are concurrent.
be a twice differentiable function such that![\[
f(0) = 2, \quad f'(0) = -2, \quad \text{and} \quad f(1) = 1.
\]](http://latex.artofproblemsolving.com/1/e/1/1e178debf8ae0b6b76d2e60ffca95f5a4c4ad4e5.png)
Prove that there exists a point 
such that![\[
f(\xi) \cdot f'(\xi) + f''(\xi) = 0.
\]](http://latex.artofproblemsolving.com/4/6/b/46b8d103981b838ae0aa5afdee104edfae4e4b8f.png)
be real numbers. For any set 
let 
. Assume that the function 
takes on at least 
values where 
runs over all 
subsets of 
. Prove that the number of sets 
for which 
does not exceed 
.
numbers 
are written on the blackboard . An allowed operation consists of choosing two different numbers 
and 
of the blackboard such that is a multiple of and delete exactly one of the two. María performs a sequence of permitted operations until she observes that it is no longer possible to perform any more. Determine the minimum number of numbers that can remain on the board at that moment.
, Elbek writes 
. Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
- Greatest common divisor of numbers 
and 
such that there exist two permutations 
and 
of the set 
, satisfying the condition
for all 
.
and 
be positive real numbers such that 
. Prove that
satisfying the following two conditions for all 
:![\[ f(x+y)=f(x)+f(y), \;\;\; f \left( \max \{x, y \} \right) = \max \left\{ f(x),f(y) \right\}. \]](http://latex.artofproblemsolving.com/3/9/1/391164166941b57fe54bde43b1f3ef3b49494180.png)
with their incenter . Point 
lies on 
on the same side as 
wrt 
. Point 
lies on the shorter arc of the circumcircle . It is given that 
Prove that 
is the angle bisector of 
.
be the set of all sequences 
of non-negative integers such that
; and
such that 
for all 
.
to be the sequence 
, where, for 
,
is the number of 
's which are greater than or equal to 
. belongs to . is the original sequence itself. are distinct.
, where 
and 
.
satisfies the following property:![\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]](http://latex.artofproblemsolving.com/3/b/d/3bd9fbf82b0b1d98a3c68745bb3f159cd9e2529a.png)
and for all 
.
be a subset of 
. Suppose there is a positive integer 
such that for any integer 
, one can find positive integers 
so that 
and all the digits in the decimal representations of (expressed without leading zeros) are in . Find the smallest possible value of 
.
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
Let 
be such that any positive integer 
where the non-negative integers 
have all their digits in 
Find the smallest possible number of elements in 
for all 
and now make 
close to 