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 positive integer not divisible by 3. Prove that there are infinitely many positive integers 
such that 
and 
are divisible by , where 
is the sum of digits of 
.
books are given as prizes to students (each students gets one book). In how many ways can the organisers give these prizes if they have copies of one book and 
other books each in one copy?
. Show that there exists 
such that![\[a_1x_1^2+a_2x_2^2+\ldots +a_nx_n^2\geqslant (a_1x_1+a_2x_2+\ldots +a_nx_n)^2.\]](http://latex.artofproblemsolving.com/4/d/2/4d269e5b682be6778e89237aa91da5e0399042a7.png)
so that the sum of its edges' lengths is minumum.
find, in terms of , the minimum value of 
for which there is a set of 
distinct positive integers that has sum greater than but every subset of size has sum at most 
with 
equilateral. Let 
be the area of the triangle of sides 
, 
and 
. Show that 
where 
is the circumradius and 
is the volume of the tetrahedron. be circumscribed about circle 
, and let 
be the orthocenter of 
. The circle touches line 
at 
. The tangent to the circle 
at meets at . Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
where 
.
such that for every 
we have 
. is irrational then for every 
there exists 
such that![\[u<P(n)-\lfloor P(n)\rfloor <v.\]](http://latex.artofproblemsolving.com/9/4/a/94a8cb1f064e1dea37e8894fb4fd4e34c4c6c845.png)
Generalize the problem!
we consider a variable point 
which moves on line 
towards 
. Halfway there, it stops and starts moving in a straight line line towards 
. Halfway there, it stops and starts moving in a straight line towards 
, and halfway there it stops and starts moving in a straight line towards , and so on. Show that will get as close as we want to the vertices of a fixed triangle with area 
.
satisfying the equation![\[
(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/2/5/e/25eed911f303ed7cc4e03765a2f943ad4641c451.png)
in the plane of a square 
for which ![\[\max (PA, PC)=\frac1{\sqrt2}(PB+PD).\]](http://latex.artofproblemsolving.com/3/9/8/3984b1970837a85af0bac93fd711843630a1fd99.png)
such that 
and 
for any real numbers and 
. be a triangle with incenter and -excenter 
, whose incircle touches 
at 
. The line 
meets at and is the midpoint of 
. Show that 
.
If you just only draw on your paper (and maybe wrong) and think that wrong, you should prove your statement before saying something. If you draw on computer, please give me the figure.
to 
same with , it's will be IMO 2009 problem 2. so, any idea now, I still thinking and cannot find how to apply that problem for this 
be midpoints of 
，
，
，
be midpoints of 
，
be the circumcircle of 
，
be parallelogram，
，
.
，
，the 9-point center 
is the circumcircle of 
，so， is midpoint of 
.
. On another hand，we can easy found 
is a isosceles trapezoid，and 
， is on perpendicular bisector of 
，the same way， is also on perpendicular bisector of 
. is midpoint of 
，therefore，we get 
. Q.E.D
be midpoints of ，，， be midpoints of ， be the circumcircle of ， be parallelogram，， .，，the 9-point center is the circumcircle of ，so， is midpoint of . . On another hand，we can easy found is a isosceles trapezoid，and ， is on perpendicular bisector of ，the same way， is also on perpendicular bisector of . is midpoint of ，therefore，we get . Q.E.D
,
,
,,
is the food of the perpendicular line from to .
is the line pass throught and paralley 
, meet 
at and meet at , we have is the midpoint of 
,
meet at 
, apply "Butterfly Theorem" we have is the midpoint of 
so we have 
so 
![\[ Q.E.D\]](http://latex.artofproblemsolving.com/9/9/4/9944bb829d83b6c8c3731e0b66fafd134f718701.png)
is the midpoint of 
is the mid point of , you should read "Butterfly Theorem" again. is the midpoint of because 
and is midpoint of 
be a triangle and its circumcircle. 
. Let be the midpoint of , be the midpoint of 
, be the midpoint of . the circumcircle pass through 
meets at . Prove: 
