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.
such that for all positive integers 
,![\[
f(mn) = \mathrm{lcm}(m, n) \cdot \gcd(f(m), f(n)),
\]](http://latex.artofproblemsolving.com/6/0/7/6071692e2716901e88cc1cfc19ed60388c10701d.png)
where 
and 
denote the least common multiple and the greatest common divisor of 
and 
, respectively.
has a closed walk of length at most 
passing through each edge of 
at least once.
satisfying the following condition: for any real numbers 
and 
, the number 
is equal to 
or 
contains no term of the form 
and 
.
?
, similarly to 
, by SOS
be a triangle with 
. Let 
, 
, and be the midpoints of the sides 
, 
, and 
respectively. A circle 
passing through 
and tangent to at meets the segments 
and 
at 
and 
, respectively. The points 
and 
are symmetric to and about and , respectively. The line 
meets 
and 
at 
and 
, respectively. The line 
meets again at 
. Prove that 
.
of acute triangle is 
. The ray drawn from through the circumcenter 
intersects at 
. Let the midpoint of 
be 
. Define 
, 
, 
, 
, 
, 
similarly. Prove that 
, 
, 
are concurrent.
![$P(x) \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/5/3/453a624c3b002c1b0e78e0023b24dd22ddd03557.png)
such that:
with 
is an 
th prime and 
is an arbitrary positive integer.
grid with each cell colored with an integer from 
to 
. Two players take turns. On a turn, a player may increase any one cell’s value by a power of 2 (i.e., add 1, 2, 4, or 8 mod 16). The first player wins if, after their move, the sum of each row and the sum of each column is congruent to 0 modulo 16. Prove whether or not Player 1 has a forced win strategy from any starting configuration.
be a parallelogram with 
A point is chosen on the extension of ray past 
The circumcircle of 
meets the segment 
again at 
The circumcircle of triangle 
meets the segment 
at 
Prove that lines 
are concurrent.
. This is achieved when 
.
. Then
Hence shifting any 
to 
is non-increasing, so we may assume 
. Minimising the function 
may be done with standard calculus tricks.
Thus we just minimise the function 
over 
which can be done with standard calculus tricks.
be positive real numbers . Find the minimum value of 
, we have
But 
when 
, hence the minimum value of 
is 
. 
Let be positive real numbers . Then 
![\[a^3+ b^3-kab + \frac{k^3}{27} = \frac{(3b+3a+k)}{27} \cdot \frac{(6a-3b-k)^2+3(3b-k)^2}4.\]](http://latex.artofproblemsolving.com/2/8/5/2859187e2d5806549f3b7008886a518160c84b52.png)
So![\[a^3+ b^3-kab \ge -\frac{k^3}{27}.\]](http://latex.artofproblemsolving.com/e/8/f/e8fa110c024aca972b043a4d2e27e359eff97039.png)
and 
Find the maximum value of 
implies that
where the last inequality holds because 
for all 
,
.
is 
.
Let 
Proof:
in terms of 
,
But 
when 
,
is 
.![$$a^3+ b^3-kab\geq 3\sqrt[3]{a^3\cdot b^3\cdot\frac{k^3}{27}}-kab-\frac{k^3}{27}=-\frac{k^3}{27}.$$](http://latex.artofproblemsolving.com/c/5/f/c5ff0bed94583e9cba00e7118483debb07a75332.png)
The equality occurs when 
.
,
,
be positive reals such that 
Show that
and 
,
thereby 
.
We are done. 
be positive reals such that 
Prove that 
,
be positive reals such that 
Show that
,
which yields 
。
As stated. 
and 
we obtain the quoted result.
be positive reals such that 
Then 
Prove that 
.
.
and the inequality is equivalent to 
which is obvious.
be positive reals such that 
Find the maximum value of 
.
where 
.
.
.
.
It's clear that for positive 
is increasing, so maximum of 
Find the minimum and maximum value of 
and 
Prove that
Old.
Prove that