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.
and 
such that divides 
. Show that there exists integers 
, 
, 
satisfying 
.
be positive real numbers such that : 
. Prove that : 
be an acute triangle with 
. Let 
be the centroid of triangle 
and let 
be the foot of the perpendicular from 
to side 
. The median 
intersects the circumcircle 
of triangle at a second point 
. Let 
be the point where intersects . The line 
intersects the circle at a point 
, such that lies between and . Prove that the points 
and lie on a circle.
be non-negative real numbers such that 
Prove that
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and be points on the line segments 
and respectively such that 
is a parallelogram. Prove that 
.
be the set of positive real numbers. Determine all functions 
such that ![\[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]](http://latex.artofproblemsolving.com/e/7/1/e71b6e0d0b858eb46b81149f1e6be8c41e13d301.png)
for every 
.
, and be pairwise distinct nonnegative real numbers. Prove that![\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\]](http://latex.artofproblemsolving.com/2/0/9/209f9807481e7a1e97f023bf65a603cb462c62cb.png)
(Karl Czakler)
that can be written in the form![\[
n = \frac{a^2 - b^2}{b},
\]](http://latex.artofproblemsolving.com/f/9/0/f90a3ce2e1b51d6e1cb53a15cad6f36c2b30492a.png)
where and are positive integers.. Initially, the numbers 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
, or 
as 
, or 
as 
.) is it possible to end with only one number remaining on the board?
st sees the other three, the 
nd sees the rd and 
th, and the rd sees only the th. The devil has 
hats, numbered from to , he puts one hat on each wise man, and hides the extra 
hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
, and are such that 
. Prove that ![\[\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}.\]](http://latex.artofproblemsolving.com/3/e/4/3e4b336137b17e55bf172d45bd18fe6d5561a827.png)
be real numbers such that 
.![\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]](http://latex.artofproblemsolving.com/5/a/8/5a83eaa7470bcca6546599ff4e296e4efec69b16.png)
and determine all the cases when the equality occurs.
, let be the midpoint of side 
. Let the line perpendicular to at point intersect 
at 
. If 
is tangent to at 
, find 
.
such that 
divides 
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 
.
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