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.
for which the following statement holds: there exists at least one triple 
of integers such that
and all triples of real numbers, satisfying the equations, are such that 
are integers.
+1 w
is called 
if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
;
the sum 
is a quadratic residue modulo 
for every 
.
th power.
let 
denote the intersection of diagonals 
and 
, and let 
denote the intersection of diagonals 
and 
. Prove that if 
and 
, then 
bisects angle 
.
are 
and 
, respectively. The excenter and exradius relative to vertex 
is 
and 
, respectively. Let 
denote the circumradius. Prove that if 
, then 
.
-gon, and at each vertex there is a black or a white cap. (Caps of the same colour are indistinguishable from each other.) Under one of the caps 
dollars are hidden, and there is nothing under the other caps. The host rotates the table, and then Ádám chooses a cap, and take what is underneath. Ádám's accomplice, Béla is working at the company behind this game. Béla is responsible for the placement of the dollars under the caps, however, the colors of the caps are chosen by a different collegaue. After placing the money under a cap, Béla
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.
and 
are points on the segments 
and 
, respectively, such that 
and 
. If the area of triangle 
is 
, then what is the area of triangle ?
consists of the union of disjoint pieces. Compute the distance between the endpoints of any one piece
such that there exists 
not necessarily distinct positive divisors of where adding those divisors of gives another divisor of . (Source:Illinois middle school math Olympiad 2024 problem 2)
by AM-GM used thrice.
, expanding the objective expression and applying AM-GM to pairs of reciprocal terms gives 
with equality along the ray 
.
satisfies the constraint, this makes the constraint redundant., prove 
. But this isn't challenging either as it follows immediately by AM-GM.
about the plane 
is given by 
.
,
and
Prove that 
and 
we have 
and 
thus, 
multiplying, the lhs of the inequality is equal to 
the term under the square root is equal to 
or 
we simplify to claiming 
moving non-radicals to the lhs, the inequality claims that 
which is false, as 
is a bit more than 
(*note that 
and 
)