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.
such that 
are the lengths of a right triangle.
from the domain 
and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!
. Let the perpendicular line from 
to 
meets 
at points 
, respectively, and the foot from 
to 
is 
. 
intersects line 
at 
, 
intersects line at 
, and lines 
intersect at 
.
is tangent to 
.
, 
, 
is the point on such that 
. Let 
be the circle through 
tangent to 
at , and 
the circle through tangent to at . Let 
be the second intersection of and . Prove that 
lies on .
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
, the maximum number of good cells over all possible starting configurations.
, let 
be the excircle opposite to . Let 
and be the points where is tangent to 
, and , respectively. The circle 
intersects line at 
and 
. Let 
be the midpoint of . Prove that the circle 
is tangent to .
positive real numbers such that 
. Prove that :
such that for all 
,![\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]](http://latex.artofproblemsolving.com/d/1/0/d10c01d7d416243614b24f04218d157e8b025ff2.png)
be an acute triangle. Points 
, and lie on a line in this order and satisfy 
. Let and be the midpoints of and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points and lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
the midpoints of and are and , respectively. Prove that the circumcircles of triangles 
and 
intersect all in one point.
be points on and . Then there exists a point lying on the circumcircles of triangles and .
and 
. Notice that 
. But 
. So 
. 
with radius 
. The pairwise radical axes of these circles are the perpendicular bisectors of the sides of . They meet at the radical center, 
, of the three circles, a point where 
. This is obviously the circumcenter. Now, with all the right angles created by the perpendicular bisectors, it is easy to check that 
, 
, and 
are all cyclic meaning that the circumcircles of triangles and intersect all in one point, namely the circumcenter, which is what we wanted.
with 
such that
.
![\[
\frac{(x+y+z)^5 - x^5 - y^5 - z^5}{(x+y)(y+z)(z+x)} = 0
\]](http://latex.artofproblemsolving.com/7/8/f/78f8e6c2570684631dc288144e13568d71e68ec8.png)
![$A_n=\left[n\right]^3{,}f\left(n\right)=\min \left|S\right|{,}S\ satisfy\ S\subseteq \ A_nand\ \forall \ x\in \ A_n{,}\exists \ y{,}z\in \ S\ {,}x{,}y{,}zcollinear$](http://latex.artofproblemsolving.com/a/a/6/aa671ae30c20a584500d6362790124ad6cbb67be.png)
.