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.
is mundane if there exist positive integers 
and 
less than 
such that 
is a positive integer. Find, with proof, all prime numbers that are not mundane.
be a fixed positive integer. Find the greatest real constant 
that has the following property: Any 
real numbers, not necessarily distinct, that lie in the interval ![$[100,101]$](http://latex.artofproblemsolving.com/8/e/3/8e3fe24e3b899bbf2128eebe08756e1ff963aaa0.png)
, can be partitioned into two groups with sums 
and 
such that
.
, 
and 
.
.
-tromino is a figure made up of three squares, obtained by removing one square from a 
board.
board consisting of 
unit segments. A configuration of several -trominoes is optimal if the -trominoes do not overlap, each one covers exactly three squares of the board, and moreover, no unit segment of the board belongs to two -trominoes. Below is an optimal configuration of 5 -trominoes: for which there exists an optimal configuration of L-trominoes on the board.
such that 
is a divisor of 
.
, 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)
are three points on the sides 
of triangle 
and 
Then the transversal 
goes through 
if and only if the following relation holds: 
passes through the centroid of 
intersect at and median 
at 
. Applying Menelaus to triangles 
and 
with the transversal 
and 
respectively we get 
and 
; adding side by side the two equalities we get 
, but from the problem we know 
and, from 
being midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
we get 
and, substituting in 
we get 
, i.e. 
, done.
, for the centroid 
,
, which gives the collinearity of 
with points 
and 
are in the sides 
and 
respectively.
, then prove that the centroid of 
.
.
,
,
,