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.
, diagonals 
and 
are equal, and they intersect at 
. Perpendicular bisectors of 
and 
intersect at point 
lying inside triangle 
, and perpendicular bisectors of 
and 
intersect at point 
lying inside triangle 
. Prove that 
.
denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
be a triangle with circumcenter 
. Let be a point on the perpendicular bisector of (see figure) and , 
be the intersections of the perpendicular bisectors of and , respectively, with 
and 
. Prove that the midpoints of 
and 
and the point are collinear.
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
, where 
and 
are integers, 
. A subset 
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to . Find the number of valuable subsets of set .
rows of coins. There is 1 coin in the first row, 2 coins in the second row, 3 coins in the third row, and so on, up to coins in the th row. Initially, all of the coins show heads (H). Carley plays a game in which, on each turn, she chooses three mutually adjacent coins and flips these three coins over. To win the game, all of the coins must be showing tails (T) after a sequence of turns. An example game with 4 rows of coins after a sequence of two turns is shown. for which it is possible to win the game starting with rows of coins.
satisfying 
, prove that![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
such that there exists a unique integer 
such that 
with the following property:![\[ n!\mid a^n + 1
\]](http://latex.artofproblemsolving.com/e/2/3/e23d39aab685888993d355d8c93d4e6475f53487.png)
(use in a problem for junior) to practice. Can anyone share to me? be a square with 
side length and with center 
. A rhombus 
is drawn where point lies on line 
. What is the area of ?
be an odd integer, 
. Consider 
integers 
such that for any 
, 
and 
. Prove that the number of 
for which satisfy 
for some 
, 
is greater than 
.
such that![\[f(x+yf(x))+f(xf(y)-y)=f(x)-f(y)+2xy\]](http://latex.artofproblemsolving.com/e/b/c/ebccae7e765d9fc3c8605e58a7399a70127b39b1.png)
, such that 
Prove that :
equality cases : ?
is a cyclic, 
. Points 
and 
are chosen on sides and respectfully so that 
. Prove that 
.
Let 
be the projections of on 
(
lying on segment and 
lying on ray 
beyond ). Clearly 
are directly congruent 
Since 
and 
bisects 
we deduce that is the C-excenter of 
is C-excircle of touching 
at 
and 
be the point on the ray beyond , such that 
. Then we have 
and 
, so the triangles 
and 
are congruent. In particular, 
. Moreover, we have 
, hence the triangles 
and 
are also congruent.
.
such that is congruent to the triangle 
(
and 
). Since , 
and 
is common, 
is congruent to 
. Then 
, as desired.
it is easy to find that the triangle 
i
so angles 
a
and angles 
so 
is the perpendicular bisector of 
so 
So 