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.
be a prime number. Define the set![\[
M_q = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2q^{2025} x} \in \mathbb{Q} \right\}.
\]](http://latex.artofproblemsolving.com/9/8/b/98b266a869cf5686cea86be51152ef234486993e.png)
.
and 
be points on a plane such that 
Let 
be a point on that plane such that 
Prove that ![$$AP \in \left[\frac{5-\sqrt{5}}{10}, \frac{-1+\sqrt{5}}{2}\right] \cup \left[\frac{5+\sqrt{5}}{10}, \frac{1+\sqrt{5}}{2}\right].$$](http://latex.artofproblemsolving.com/7/8/e/78e428cbc99ce4e9448f5c4cec86f50567123812.png)
such that
for all nonnegative real numbers 
, 
, and 
with 
.
be three non-negative real numbers satisfying 
such that any real numbers 
and 
satisfy
Proposed by EmilXM
board has some of its cells coloured red, blue, or green. Each cell is coloured with at most one colour, and some cells may remain uncoloured. Additionally, there is at least one cell of each colour. Two coloured cells are said to be friends if they have different colours and lie in the same row or in the same column. The following conditions are satisfied:
be an acute-angled triangle with 
and let 
be the centre of its circumcircle 
. Let 
be a point on the line segment 
such that 
. Let 
be the second point of intersection of and the line 
. If 
, 
and are the midpoints of the line segments 
, 
and 
, respectively, show that the points , and are collinear.
be two positive real numbers such that 
. Prove that
such that for any 
,![\[f(xf(y)) + xf(x - y) = x^2 + f(2x).\]](http://latex.artofproblemsolving.com/1/d/4/1d47636bc583ed4732cf2ea5723bc8e49c70b80b.png)
Proposed by hyay
be the sequence of all natural numbers which are sum of squares of two natural numbers.
which 
.
, are arranged so that one has one vertex in the center of the other. Determine the area of the gray area.
.
.
.
no matter the rotation of the second square with rotation respect to the center of the first one. Here is my way of proving it...
is bounded.
such that 
?
.
be strictly in 
and 
.
does not depend on the choice of ![$[OPBQ]=\frac{[ABCD]}{4}$](http://latex.artofproblemsolving.com/b/3/5/b35973eea8178d35dad127e662788458160eaf82.png)
.
.![$[OMBN]=\frac{[ABCD]}{4}$](http://latex.artofproblemsolving.com/b/b/f/bbfca51b7b2e517770849054c1d7237ee031d044.png)
,
from it and then adding/subtracting the same area back from 
.
.
since 
since 
and 
,
and 
,