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are collinear, and 
are arbitrary points
be an acute triangle and 
its orthocenter. Let 
be a point on the parallel through 
to 
such that 
. Define 
and 
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
be an acute triangle with circumcenter 
. Points 
and 
are chosen on segments 
and 
such that 
. If 
is the midpoint of the arc 
and 
is the midpoint of the arc 
, prove that 
.
board with a number written in each cell. Every two neighbour rows sum up to at least 
, and every two neighbour columns sum up to at most 
. Find the sum of all numbers on the board.
where ![$a \in [10, 16]$](http://latex.artofproblemsolving.com/2/3/7/2379f0f94d6dcb54024b80fa1bdce0e7baaf7b53.png)
fits this condition which means the sum can be anywhere in ![$ [20, 32]$](http://latex.artofproblemsolving.com/7/4/3/743b32f68a688d26582878dbdf576921cfb3dcd5.png)
. There might be even loser restrictions too but I don't actually know how to solve this mathematically.
where fits this condition which means the sum can be anywhere in . There might be even loser restrictions too but I don't actually know how to solve this mathematically.
we get that total sum is at least 
. By taking pairs of columns without some odd column we get that sums in columns are 
. So the total sum is .
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
board, not a 
board.