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 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
, a function 
is called interesting if the following two conditions hold:
such that 
., we have 
(where 
denotes the 
-th iteration of 
).
.
for which there is no interesting function
, and be positive real numbers such that 
. Prove the inequality
, the number 
is the product of at least 
(not necessarily distinct) primes.
x square composed of 
unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement 
.
be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement than of arrangement . Prove that arrangement can be obtained from 
by performing a number of switches, defined as follows:
have the property that both![\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]](http://latex.artofproblemsolving.com/d/e/c/dec527c4673470e7023d6aed00fac90268bd2fcc.png)
are consecutive perfect squares.
is the incenter and 
is the circumcenter. The line 
intersects the lines 
at points 
respectively. Let 
be the intersection of 
and 
. The points 
and 
are defined similarly. The incircle of 
is tangent to sides at points 
respectively. Let the lines 
and 
intersect at points 
respectively. Prove that the circles with diameters 
and 
have a common point.
be triangle, inscribed in parabola. Tangents in points forms triangle . Prove that 
. (
is area of triangle 
).
be triangle, inscribed in parabola. Tangents in points forms triangle . Prove that . ( is area of triangle ).
be any point on the Steiner circumellipse of 
and let 
be the cevian triangle of WRT 
Then ![$[DEF]=2 \cdot [ABC].$](http://latex.artofproblemsolving.com/3/3/e/33ec9a123e2b3b63c128a4fe7fe4aa12551261a1.png)
be the antimedial triangle of ( against 
). The Steiner circumellipse 
of is obviously the Steiner inellipse of tangent to its sides at 
are collinear on the polar of 
WRT and similarly 
and 
are respectively collinear. Thus by Pappus theorem for the hexagon 
it follows that 
is at infinity, i.e. 
and similarly we get 
![$[DEF]=[DYZ]=[CYZ]=2\cdot [ABC]. \ \blacksquare$](http://latex.artofproblemsolving.com/b/4/0/b4028b012dc90ae29c0bc707cbf447409f6ab058.png)
be the perspector of the parabola. Since the center of any inconic is the isotomcomplement of its perspector (in the parabola case at infinity), then it follows then that the isotomic conjugate of WRT must be on the line at infinity is on the Steiner circumellipse of and the result follows from the lemma.
