be three non-negative real numbers satisfying 
such that ![\[f(x+f(x)+2f(y))+f(2f(x)-y)=4x+f(y)\]](http://latex.artofproblemsolving.com/4/4/2/4423651bb829b92debd65a23f6cee3d4d4b8530f.png)
holds for all reals 
and 
.
,
,
,
,
.
.
,
For all 
.
is 
,
denotes the 
-
.
as the average of all 
for 
.![\[ a = \lim_{k \to \infty} \sup_{n \geq k} \frac{a_n}{n^4} .\]](http://latex.artofproblemsolving.com/5/2/d/52d93af7aafa6657689246d9a74c8eb05d96efa3.png)
be a 
be a prime number?
be an open interval and 
a twice differentiable function such that 
for any 
Prove that 
for any 
![\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/3/a/2/3a235e8be4c61c32f376999e61d64973db21dc75.png)
has property 
if for any real number 
there exists a 
such that 
for all 
that is not monotone on 
is field. Define addition of elements of 
coordinate wise and for 
and 
define 
.
points?
points?
,![\[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]](http://latex.artofproblemsolving.com/0/2/9/029a935e28ca0a9b36e65702689cca91eeb8f614.png)
Y by tenplusten, Davi-8191, yshk, ImSh95, Adventure10, Mango247, Rounak_iitr
Let 
be a convex quadrilateral and let 
and 
be points in such that 
and 
are cyclic quadrilaterals. Suppose that there exists a point 
on the line segment 
such that 
and 
. Show that the quadrilateral is cyclic.
Proposed by John Cuya, Peru
be a convex quadrilateral and let 
and 
be points in such that 
and 
are cyclic quadrilaterals. Suppose that there exists a point 
on the line segment 
such that 
and 
. Show that the quadrilateral is cyclic.Proposed by John Cuya, Peru
.
be the mirror of 
after the inversion.
and four points 
lie on a circle. These imply that the quadrilateral 
is an isosceles trapezoid. Similarly, we also have the isosceles trapezoid 
.
form an isosceles trapezoid with the perpendicular bisector of 
follows the cyclic quadrilateral 
be points on 
such that 
are cyclic quadrilateral.
are cyclic quadrilateral then 
we get 
thus 
the intersection of 
and 
then 
are collinear, similarly 
are collinear
are collinear.
and 
.
,
(1).
(2).
(3).
the distance from point 
and denote by ![$ [MNK]$](http://latex.artofproblemsolving.com/1/c/c/1ccf243c1704482b184f4d1b66cfb4fb4dfaa250.png)
the area of the triangle 
.![$ \frac {BP \cdot PC} {BQ \cdot QC} = \frac {[BPC]} {[BQC]} = \frac {d(P;BC)} {d(Q;BC)}$](http://latex.artofproblemsolving.com/7/0/9/709a1807afc0b7b44d3c6dca8b528378a987c863.png)
.
.
(*).
.
(from cyclic quadrilateral 
)
(from cyclic quadrilateral APQD), implying 
,
(easily follows from 
and 
and then do a bit of algebra.
(*). But using 
etc, (*) is equivalent to 
which is relation 3 from lasha's solution :
and 
are cyclic, it follows that 
and that 
.![\[\frac{PE}{EQ}=\frac{PC \cdot \sin{\angle{PCE}}}{QC \cdot \sin{\angle{QCE}}}=\frac{PB \cdot \sin{\angle{PBE}}}{QB \cdot \sin{\angle{EBQ}}}\]](http://latex.artofproblemsolving.com/a/5/1/a51d1ab13e191e2a69411789e97f62f6e35037dc.png)
![\[\left( \frac{PE}{EQ} \right)^2 = \frac{PC \cdot \sin{\angle{PCE}}}{QC \cdot \sin{\angle{QCE}}} \cdot \frac{PB \cdot \sin{\angle{PBE}}}{QB \cdot \sin{\angle{EBQ}}} = \frac{PC}{QC} \cdot \frac{PB}{QB}\]](http://latex.artofproblemsolving.com/9/e/5/9e52aafedfb66f61486e72ba130c638047229691.png)
intersect 
at 
.
and 
and hence that![\[\left( \frac{PE}{EQ} \right)^2 = \frac{PC}{QB} \cdot \frac{PB}{QC} = \frac{XP}{XB} \cdot \frac{XB}{XQ}=\frac{XP}{XQ}\]](http://latex.artofproblemsolving.com/d/e/6/de6167969c52bc822b7f5be3db36b2eb62ca4ede.png)
is the intersection of 
and ![\[\frac{XP}{XQ} = \left( \frac{PE}{EQ} \right)^2 = \frac{YP}{YQ}\]](http://latex.artofproblemsolving.com/d/c/5/dc5f23a02779da756346db28a80a903e8db3335e.png)
and 
which implies that 
is cyclic.
and 
.$](http://latex.artofproblemsolving.com/3/5/7/357a4200c0185a90bee7dd3962377bd7d20d9cf7.png)
denote the power of 
.
is cyclic and 
,
.
,
.
and 
,=[ADPQ](T)$](http://latex.artofproblemsolving.com/1/0/8/108a43aeda32d88f98b1d4685c2ead4db7508833.png)
and =[ADE](T)$](http://latex.artofproblemsolving.com/d/d/c/ddc0a9516a7f661271343104deae46ff9ee26350.png)
.=[BCE](T)$](http://latex.artofproblemsolving.com/4/7/9/47927036f51eb7eb4de6ba1ddd0bed3f4da56a02.png)
,=[ADPQ](T)$](http://latex.artofproblemsolving.com/c/f/7/cf7567f3ce21722131fec921943e0e886155aafb.png)
implying 
lies on 
,
is cyclic.
is tangent to 
and 
is cyclic
and 
intersect at 
,
and 
intersect at 
.
bisects 
and 
bisects 
.
and 
be points, and let 
be a point on segment 
be on the circumcircle of 
and let 
be on the circumcircle of 
such that 
is cyclic. Prove that 
passes through the circumcenter 
of 
.
and 
are both isosceles trapezoids by the angle condition, so 
is tangent to 
is cyclic, and see what you can notice.
.
such that 
is cyclic. Then note that 
,
.
,
.
is cyclic, then
,
.
,
,
,
,
.
be the intersection of 
and 
,
be the intersection of 
.
,
.
,
)
,
,
and 
are cyclic, therefore ![\[\angle Q'P'A'=\angle PP'A'=\angle PAA'=\angle PAE=\angle QDE=\angle QDD'=\angle QQ'D'=\angle P'Q'D'\]](http://latex.artofproblemsolving.com/4/0/4/404fbb18392d820ffa36df843d8892b96ecee70f.png)
so 
and 
share a perpendicular bisector. Similarly, 
and 
is an isosceles trapezoid and therefore is cyclic.
cut circle 
again at 
.
,
and 
similarly.
,
and 
both being isosceles trapezoids. In particular, 
itself must be an isosceles trapezoid, and is thus a cyclic quadrilateral.
for 
,
,
and 
.
so 
is an isoceles trapezoid with 
.
is an isosceles trapezoid with 
.
is an isosceles trapezoid and is therefore cyclic, so 
must be sent to cyclic isosceles trapezoids, so 
,
be the second intersections of 
with lines 
and 
respectively. Note that by the angle condition, 
.
is tangent to 
.
,
is tangent to 
.
and 
be the reflection of 
and 
.
,
are concyclic.
is cyclic and 
,
,
and the circumcircle of 
.
and 
and 
and 
so we can solve that 
and 
so 
become isosceles trapezoids from which 
Therefore 
is tangent to 
Similarly 
is tangent to 
If any of 
do not exist then it is easy to show by symmetry that 
and then 
Therefore either 
or 
and 
Hence 
so by symmetry 
are reflections of each other over the perpendicular to 
and similarly for 
and thus 
a contradiction. QED
