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.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
, where 
is the binomial coefficient 
choose 
.
satisfy that: 
and 
. Find the maximum: 
be a convex quadrilateral. Suppose that the circles with diameters and 
intersect at points 
and . Let 
and 
. Prove that the points , , and are concyclic. and are not the diagnols)
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
be a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be positive integers, where 
. It is given that there exist a positive integer 
such that
. Prove that the above equation is true for all positive integers 
. (Here 
is the sum of digits of 
taken in base 
).
and on the side of a triangle 
are such that 
. Prove that the midpoint of the altitude drawn from , the midpoint of 
and the circumcentre of are collinear.
be the circumcentre of , 
be the circumcentre of 
, be 
. Let 
. Note, 
so 
and similarly 
so a homothety wrt takes 
. So it is internal centre of similitude i.e. 
. If is the midpoint of then 
gives the conclusion. is the external centre of similitude as 
have the same angle bisectors at , so the isogonal of the common altitude means 
are collinear.
be the diameter of 
, 
is the feet of altitude from , reflect over the midpoint of 
to obtain 
, hence we just have to prove 
are collinear. It's clear that 
are homothetic. Since 
, therefore is the center of homothetic and the collinearity follows.
and i think 
is paralllel to 
is not correct
, 
is wrong and where is ? Could you explain?
. At first look like a mud , but in the end i understand your value and this solution.
let 
on 
and 
on such that quadrilateral 
is rectangle let be the center of this rectangle then the locus of point is the line 
where 
are midpoints of and altitude ( lies on ). is rectangle and is its center we prove that 
are collinear. Let 
because 
so 
similarly bisects 
at 
midpoint of clearly 
note that 
so 
are collinear by thales theoream and the lemma proved. foot of on and 
because 
points 
and 
are collinear so 
is a rectangle with center using the lemma we deduce that 
are collinear.( midpoints of 
)
is the projection of on , 
are the midpoints of 
and the circumcenter of 
be the antipode of 
WRT 
and 
The midpoints of 
are collinear (ERIQ)
be the antipode of 
in 
, respectively .
be the circumcenter of 
and 
be the midpoint of 
.
and 
.
(A-median of 
) passes through the midpoint 
of 
,
(M-median of 
) passes through the midpoint of 
(
) .
and 
where is the orthocenter of and so is the antipode of in 
.
and 
, we have that the constructed parallelograms are homothetic, and the homothetic center is 
which is precisely .
, we get our conclusion.
the midpoint of the two intersections 
of the -circumcircle and 
say 
. Let 
the respective midpoints of 
and the altitude. By simple angle chase we deduce that 
and 
are isogonal WRT 
so 
are tangent at hence 
besides 
implies that is their symmetry center i.e .
.Since 
then 
cuts the altitude at its midpoint .
be the midpoint of 
and 
be the and -antipodes in 
; obviously 
and 
are collinear. If is the reflection of over the circumcenter of , then it is also the midpoint of segment 
and so line 
bisects 
, implying the conclusion.
satisfying 
such that 
