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.![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut . such that, on the -th move, Jumpy swaps some walnuts 
and 
such that 
.
kawaii if it satisfies 
(
is the number of positive factors of , while 
is the number of integers not more than that are relatively prime with ). Find all that is kawaii.
such that the equation![\[f(xf(x)+y) = f(y) + x^2\]](http://latex.artofproblemsolving.com/2/0/a/20a245460634f0455fc703d7ce41c475c1ae4970.png)
holds for all rational numbers 
and 
.
denotes the set of rational numbers.
with 
and 
. Lines 
and 
intersect at 
, and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
a point in the space and 
the centroid of a tetrahedron . Prove that:
and be positive integers. If 
is divisible by 
for every positive integer , prove that 
.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
, 
, 
be nonzero real numbers such that 
and 
. Find the value of
.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers and . Show that there exist positive integers 
and such that 
for all integers .
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
of positive integers with 
prime, 
, and 
, such that 
is a multiple of 
.
be a logarithmic spiral centered at the origin (ie curve satisfying for any point on it, line 
makes a fixed angle with the tangent to at ). Let 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point.
on the spiral, the polar of wrt. is tangent to the spiral.
. We can easily see that 3 angles BCD, CDA, BAD must equal to angle ABC. Therefore, quadrilateral is a rectangle.
such that all four triangles 
, 
, 
and 
are similar to one-another.
Hence we see
Adding these we get 
.
and 
Adding these we get 
Similarly
a square.
however I have a strong feeling that 
could also be true
. We can easily see that 3 angles BCD, CDA, BAD must equal to angle ABC. Therefore, quadrilateral is a rectangle. can be concave.
https://igo-official.ir/events/
,![\[\square ABCD \text{ is cyclic.}\]](http://latex.artofproblemsolving.com/1/6/5/165e371cea25d9eb70647967ff0afd098495aef0.png)
.![\[\triangle ABO \sim \triangle AOD\]](http://latex.artofproblemsolving.com/5/a/d/5ad40729445de534380a3407948dd22964bac6dc.png)
is a rhombus. On the other hand, it is cyclic too. So, must be a rectangle. similar triangles.
such that all four triangles , , and are similar to one-another. be convex quadrilateral because no concave quadrilateral exists that satisfies the similarity condition. 
is a cyclic quadrilateral. Let diagonals intersect at ,
. 
is a rhombus and cyclic 
is rectangle.
