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.
are pairwise distinct 
numbers and 
, find the minimum value of![\[
\left( \frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b} \right)^2 + a^2 + b^2 + c^2.
\]](http://latex.artofproblemsolving.com/3/6/9/3694d92e77e5049af2ec32f1dc1879dcc74f5936.png)
?
be a triangle and call 
the intersection of the internal angle bisector from 
with the side 
. The perpendicular bisector of 
intersects the circumcircle of in 
and 
, with and 
on two opposite sides of the line . Let 
be the intersection of lines 
and 
, and let 
be the intersection of lines 
and 
.
and are parallel.
is called a good prime if there exists a positive integer 
such that![\[
n^2 + 1 \text{ is divisible by } p^{2025}.
\]](http://latex.artofproblemsolving.com/3/3/4/334cd7b5e4a4bfb6b0dd3a4607cdd793ecadd66c.png)
Prove that there are infinitely many good primes.
, 
, 
be real numbers such that 
and 
. Prove that 
Let , , be real numbers such that and 
. Prove that 
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 orthocenter 
and let 
be the midpoint of 
Suppose that 
and 
are distinct points on the circle with diameter 
different from 
such that lies on line 
Prove that the orthocenter of 
lies on the circumcircle of 
, , be real numbers, such that and 
. Determine the value of 
.
, determine the smallest integer 
, such that it is possible to place numbers 
around a circle so that the sum of every consecutive numbers takes one of at most values.
. 
amount of people stand on coordinates 
where 
. Every person got a water cup and two people are considered to be neighbour if the distance between them is 
. At the first minute, the person standing on coordinates 
got litres of water, and the other 
people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups. will not have more than 
litres of water.
, 
, 
. Show that: ![\[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]](http://latex.artofproblemsolving.com/2/6/4/264d110e2a40c7f787c650aa8b8dc30af5d28263.png)
. By one question, we may take three different indices 
and find out the set of numbers 
(unordered, of course). Find the minimal number of questions, which are necessary to find out all numbers 
.
. By one question, we may take three different indices and find out the set of numbers (unordered, of course). Find the minimal number of questions, which are necessary to find out all numbers .
, actually, and, for general instead of 
, I think it's 
.
questions, where 
is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other 
indices, the ones we haven't touched so far, and the procedure is repeated. questions suffice, assume first that is even (but 
; for 
we need 
questions, 
don't suffice). We ask questions 
for odd 
between and 
, and after that we ask the question 
. For odd we do something very similar.
numbers after few questions concerning next group? For example: 123, then 234, then 567, then 514?
is the third one.
If there are more questions using numbers already used, just choose one of them. We do this until we no longer can, i.e. until there are no questions using the numbers previously used.
questions, where is the number of indices which have been included in our questions so far. After we have to stop, it means that we can start asking questions about the other indices, the ones we haven't touched so far, and the procedure is repeated.
. Ask 
. Let 
of be asked once. If 
are asked once, then they cannot be asked in the same set hence there are at least sets asked. Consider the bipartite graph whose vertices are 
and the asked sets. If one could find in 
questions, then by looking at degrees, we get 
or 
however, there must be at least sets which contradicts as desired.
.
.
Where 
?
questions 
where the 
'
'
'![$[1, 2005].$](http://latex.artofproblemsolving.com/b/a/0/ba07258b82d281889375fb4235b2b902ca265f90.png)
Call a positive integer 
sketchy if it only appears one time among the 
,
is. Now, let's define the function 
for 
,
is exactly equal to 
for all 
Indeed, if it was greater than 
are 
However, this means that the values of 
and 
are indistinguishable, impossible. Therefore, as 
contradiction. Hence, our assumption that we finished in 