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.
, and 
be the angles of a triangle, and let 
, and 
. If 
, show that![\[s(s - a)(s - b)(s -c) \geq 0.\]](http://latex.artofproblemsolving.com/1/3/4/1344a0cbbd4e1aa4bcf6a1e777501c62daff32b3.png)
has 
. Point 
is the foot of the perpendicular from 
to 
. Points 
and 
lie on sides 
and 
, respectively, such that lies inside triangle 
and ![\[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \]](http://latex.artofproblemsolving.com/f/8/c/f8c35912c531eb2dae8f62b74666b7757d2b9ba3.png)
Prove that line is tangent to the circumcircle of triangle 
.
of positive integers that satisfy
be an acute triangle. Let 
be the reflection of point over the line 
. Let 
and be the circumcenter and the orthocenter of triangle , respectively, and let 
be the midpoint of segment 
. Let 
and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle , where point lies on the arc not containing . If 
is a point on the line 
such that 
, prove that 
.
be a positive integer and let be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of around the circle such that the product of any two neighbors is of the form 
for some positive integer 
.
be real numbers satisfying ![\[u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.\]](http://latex.artofproblemsolving.com/2/d/0/2d09d5e894fdd13ec466aa664f3f10da5ca8d134.png)
Let 
and 
. Prove that![\[
a b \leqslant-\frac{1}{2019}.
\]](http://latex.artofproblemsolving.com/8/7/0/870973eb68893864ee8f8494efcae4c34c11d2d8.png)
such that 
is divisible by 
and 
. Prove that is a composite number.
be a prime number and 
. Prove that![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
such that 
, 
, and 
are all equal to
for which there exist positive integers 
, 
, 
, 
such that 
and 
divides 
.
cannot cover entirely a square surface of side 
, but they can cover more than 
of it.
such that the sequence 
defined by ![\[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \]](http://latex.artofproblemsolving.com/2/1/b/21b2a1f93b11a94b84e7b55f4b4f679aa20e36c6.png)
contains at least one integer term.
is chosen inside a convex quadrilateral . Could it happen that
, because last four digits would have to be divisible by 
, but in each of this cases numbers 
are not. Similarly, we can see that 
leads to contradiction. is coprime to 
. We know that 
is coprime with 
and 
for all 
, so function 
is bijective on integers relatively prime to 
(we count modulo ). To see that this function is bijective, we can also see that 
iff 
divides 
, but 
cannot divide neither 
, nor 
if 
is not divisible by , so we would have to have 
. The same argument works modulo 
. Therefore, we can have 
. Obviously, we can also have 
, so our answer is 
.
such that for any positive integer 
there exists a positive integer 
are equal to 
