[/quote]
,
be a positive integer. Find, in terms of 
with integer coefficients such that for every integer 
,
such that 
.
.
,
.
lies on 
such that 
.
passing trough 
at 
.
.
such that
,
.
and 
the midpoints of 
and 
.
and 
meet 
and 
at 
and 
,
,
bisects both angles 
,
,
.
be the circumcircle of triangle 
,
be the tangent line of 
passing 
.
be the points each on side 
such that 
.
meets 
.
meets 
passing 
.
and tangent to line 
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of 
are pairwise different; and
denotes the integer part of real number 
,
Y by Davi-8191, nguyendangkhoa17112003, itslumi, Adventure10, son7, Miku_, ImSh95, OronSH, Mango247, Rounak_iitr, erringbubble, Funcshun840, and 1 other user
The diagonals of a trapezoid 
intersect at point 
. Point 
lies between the parallel lines 
and 
such that 
, and line 
separates points and . Prove that 
.
Author: Vyacheslav Yasinskiy, Ukraine
intersect at point 
. Point 
lies between the parallel lines 
and 
such that 
, and line 
separates points and . Prove that 
.Author: Vyacheslav Yasinskiy, Ukraine
(constructing earlier new point) and solution from now is obvious. But unfortunately this lemma isn't so obvious. I'm not aware of its proof without using trigonometric Ceva's theorem and law of sines :/.
and 
.
.
to 
takes 
Then we have 
and 
so 
is cyclic. Then, we have 
and we are done.
be the second intersection of line 
with 
.
.
in 
in 
.
at a point 
.
so we have 
from which we finish as 
.
to 
to 
,
to 
.![\[\angle DTA=\angle BQC=\angle AQD,\]](http://latex.artofproblemsolving.com/d/7/f/d7fec76fe7b55193969cf830e4f0045d6f56b91f.png)
so 
is cyclic.![\[\angle BQP=\angle DTQ=\angle DAQ,\]](http://latex.artofproblemsolving.com/a/1/c/a1caae8eb6de190aa72c48280be3bd9f468e87e0.png)
as desired 
involution swaps 
so we're done by angle chase.
will also send 
and 
.![\[\angle DRA = \angle BQC = \angle DQA \implies ADQR \text{ cyclic} \implies \angle BQP = \angle QRD = \angle QAD. \quad \blacksquare\]](http://latex.artofproblemsolving.com/7/a/4/7a47554522655b9ed610ff09e7380a1508350b83.png)
and let 
be the image of 
centered at 
Then 
and 
so the given condition is equivalent to 
being cyclic. This means that 
Therefore, we'd like to show that 
or 
This is evident since 
and 
be the homothety sending 
,
.
,
,
is cyclic. From our homothety, we also know that 
,
then the angle condition implies 
cyclic and 
.
is convex instead of 
gives you a cyclic quadrilateral. In fact, when 
,
.
the polar of 
the pole of 
,
are collinear. Also, 
.
are concyclic.
.
and 
are isogonal in 
.
,
.
.
and 
that 
and similarly 
However note that 
because 
,
by the isogonality of 
so we also have 
,
we have 
and 
,
;
;
.
,![\[\angle BQP=\angle TQA=\angle DAQ\]](http://latex.artofproblemsolving.com/d/e/b/deb78d698fa44f4c858342cdc339e3795615dfaa.png)
and we are done.
are isogonal in triangle 
,
,
intersect 
passes through the center of ![\[\angle BQC = \angle AQD = \angle ARD,\]](http://latex.artofproblemsolving.com/9/6/0/9608c1223abecd4f0337391ea1749c20437b1066.png)
,![\[\angle PQB = \angle PRD = \angle QRD = \angle QAD,\]](http://latex.artofproblemsolving.com/f/f/4/ff472df788eb48c081489ceb47a070d160902ee8.png)
