AoPS Academy
with 
. Prove that:![\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]](http://latex.artofproblemsolving.com/c/b/5/cb551e6d6ce94902b7bcb78f6da210d6c27b96d5.png)
with equality if and only if 
.
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
. Prove that
Where 
and 
are real numbers such that
with 
, inscribed in circle 
.
be the orthocenter of triangle 
, and 
be the midpoint of 
., parallel to 
, intersects lines 
and 
at points 
and 
, respectively.
be the circumcenter of triangle 
. Prove that: Points 
are collinear.
and radius 
.
. and radius ..
with the following property: for each integer 
it is possible to partition a cube into 
smaller cubes.
smaller cubes or 
smaller cubes easily.(
)
smaller cubes we can also split it into 
and 
smaller cubes. Since we can "split" it into 
cube we can split it into 
for any 
. 
. Let 
be such that 
. Let 
, then 
and we can split it into smaller cubes whenever 
. QED
works; 
is the largest positive integer that can't be written as a sum of sevens and 26s. Obviously, we can do better.
could be 
; rather than replacing a 
cube with 
cubes, we replace it with one 
and cubes. Since adding is possible by doing the above and then splitting the large cube into eight pieces, we don't need to include it separately. From sevens and nineteens, we get a new estimate of 
. into a 
and 
cubes; this allows us to add . This is very useful, since 
and 
. The multiples 
form a complete system of residues mod 
, and 
is the last number we can't get as a sum of copies of 
. The new estimate is 
.
, 
, or 
smaller cubes, which effectively replace my dissections into 
and 
cubes respectively. You can look them up at mszew's link, or try to find them yourself if you're feeling clever. That gets down to 
.
.
Let 
.
[/hide]