[/quote]
,
and 
Prove that
integers are such that 
is prime and 
.
such that 
.
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
satisfying 
.
is described on a circle 
,
be 
respectively. A circle with a center 
,
and lines 
.
,
and lines 
.
,
and lines 
,
and lines 
are concurrent.
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)
define 
be the product of first 
primes. Determine if there exists positive integers 
with the following property :![\[ m?=n(n+1)(n+2)(n+3) \]](http://latex.artofproblemsolving.com/4/5/c/45cb4a00f9395b8c6edb08a343c3bbd664916a4d.png)
be reals such that 
and 
Prove that
Let 
Prove that
Let 
Prove that
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
and 
points on a plane is 
,
points of a plane, no three of which are collinear, consisting of 
?
be real numbers such that![\[a+b+c+d=-2\]](http://latex.artofproblemsolving.com/c/7/3/c73dff68dcdfbe4f6b6b2f52b378548e47193a16.png)
![\[ab+ac+ad+bc+bd+cd=0\]](http://latex.artofproblemsolving.com/9/f/4/9f450dcb150229ee76e3debdaceadddb4e451458.png)
.
and distinct positive integers 
are given. Does there exist a polynomial ![$p(x)\in\Bbb{Z}[x]$](http://latex.artofproblemsolving.com/5/1/a/51a23efd70e495d2fe516b642f0a6e71e5875ba2.png)
of degree 
that satisfies the following conditions?
of positive integers such that
lying inside the rhombus and satisfying 
.
Y by -[]-, sandeepreddy, siavosh, DIGGER1, jam10307, Davi-8191, mathlearner2357, donotoven, megarnie, Adventure10, Mango247, ItsBesi, buddyram, Rounak_iitr, Sadigly, namanrobin08
Let 
be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
Proposed by Christopher Bradley, United Kingdom
be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
Proposed by Christopher Bradley, United Kingdom
is cyclic so that 
.
are altitudes 
is also cyclic, whence 
.
,
is cyclic.
,
.
is isosceles with 
.
lies inside the triangle. When it is outside you do basically the same. 
and 
,
.
.
.
![[asy]
import olympiad;
defaultpen(linewidth(0.65));
defaultpen(fontsize(10));
size(250);
pair C=origin,
A=(5,12),
B=(14,0),
D=foot(A,B,C),
E=foot(B,A,C),
F=foot(C,A,B),
H=orthocenter(A,B,C),
om=extension(E,F,C,B),
nom=extension(E,F,D,(9,3));
path circ = circumcircle(A,B,C);
pair P1=intersectionpoint(F--om, circ),
P2=intersectionpoint(E--nom, circ),
Q1=intersectionpoint(D--F, B--P1),
Q2=extension(B,P2,D,F),
Ep=reflect(A,B)*E,
Cp=reflect(A,B)*C,
Hp=reflect(A,B)*H,
Dp=reflect(A,B)*D;
draw(A--B--C--A--D--F--C);
draw(B--E);
draw(P1--P2);
draw(F--Q2);
draw(P1--B--Q2, dotted);
dot(A^^B^^C^^D^^E^^F^^Q1^^Q2^^H^^P1^^P2);
draw(circ);
markscalefactor=0.05;
draw(rightanglemark(C,F,A));
draw(rightanglemark(B,E,C));
draw(rightanglemark(A,D,B));
label("$A$", A, N);
label("$B$", B, SE);
label("$C$", C, SW);
label("$D$", D, S);
label("$E$", E, NW);
label("$F$", F, dir(340));
label("$H$", H, E);
label("$P_1$", P1, W);
label("$P_2$", P2, NE);
label("$Q_1$", Q1, dir(290));
label("$Q_2$", Q2, N);
[/asy]](http://latex.artofproblemsolving.com/3/3/4/3343a03103393fd04eb50eb11892247b0540abd1.png)
![[asy]
import olympiad;
defaultpen(linewidth(0.65));
defaultpen(fontsize(10));
size(350);
pair C=origin,
A=(5,12),
B=(14,0),
D=foot(A,B,C),
E=foot(B,A,C),
F=foot(C,A,B),
H=orthocenter(A,B,C),
om=extension(E,F,C,B),
nom=extension(E,F,D,(9,3));
path circ = circumcircle(A,B,C);
pair P1=intersectionpoint(F--om, circ),
P2=intersectionpoint(E--nom, circ),
Q1=intersectionpoint(D--F, B--P1),
Q2=extension(B,P2,D,F),
Ep=reflect(A,B)*E,
Cp=reflect(A,B)*C,
Hp=reflect(A,B)*H,
Dp=reflect(A,B)*D;
draw(A--B--C--A--D--F--C);
draw(B--E);
draw(F--Hp, dashed);
draw(P1--Dp);
draw(F--Q2);
draw(P1--B--Q2, dotted);
draw(Dp--A--Cp--B--Ep);
dot(A^^B^^C^^D^^E^^F^^Q1^^Q2^^H^^P1^^P2^^Hp^^Cp^^Dp^^Ep);
draw(circ);
path circ2=circumcircle(A,B,Cp);
draw(circ2);
markscalefactor=0.05;
draw(rightanglemark(C,F,A));
draw(rightanglemark(B,E,C));
draw(rightanglemark(A,D,B));
draw(rightanglemark(A,Dp,Cp));
draw(rightanglemark(Cp,Ep,B));
label("$A$", A, dir(100));
label("$B$", B, SE);
label("$C$", C, SW);
label("$D$", D, S);
label("$E$", E, NW);
label("$F$", F, dir(340));
label("$H$", H, E);
label("$P_1$", P1, W);
label("$P_2$", P2, NE);
label("$Q_1$", Q1, dir(290));
label("$Q_2$", Q2, N);
label("$D^\prime$", Dp, dir(330));
label("$E^\prime$", Ep, dir(100));
label("$H^\prime$", Hp, S);
label("$C^\prime$", Cp, NE);
[/asy]](http://latex.artofproblemsolving.com/3/c/7/3c76f476737cb03a3b6bcfbd756b1b1feb92c9a1.png)
to be the intersection of line 
and 
to be the intersection with minor arc AB. Then let 
be the intersection of 
and 
the intersection of 
and 
about one of its sides, it lies on the circumcircle of 
.
maps to 
.
and 
,
is also the circumradius of 
which is the circumradius of both 
and 
.
and 
subtend to the same arc, they are equal, so 
,
,
is cyclic.
is cyclic.
.
,
is a cyclic quadrilateral. Similarly, 
be the image of 
after it has been inverted with centre 
and radius 
and 
and on the line 
and 
.
is a cyclic quadrilateral 
and 
,
congruence rule, 
,
and 
,
congruence rule, 
,
,
and 
,
as required.
be the circumcenter of 
,
.
.
,
.
.
.
and 
have congruent circumcircles.
and 
are both intercepted by 
in these circumcircles, 
at 
at 
is cyclic, 
.
,
is cyclic, so 
.
,
by AA similarity. Thus, 
,
is cyclic. From 
,
,
,
.
and 
,
by AA similarity. It follows that 
,
.
since 
is cyclic, so 
is cyclic. Hence, 
,
is cyclic, with circle 
.
where the last follows from anti-parallel 
.
,
is harmonic.
,
.
and intersect it with 
is a harmonic quadrilateral, where 
.
bisects 
,
is the midpoint of arc 
,
.
the second point of intersection of 
the mirror image of 
(as arcs of circles). But this is equivalent to 
(the tangent through 
)
:
,
.
.
,
.
We have that 
and 
so 
so 
From cyclic 
we have that 
so 
which implies 
is isoceles with 
is cyclic.Finally 
.
denote the orthocentre of 
,
all lie on 
.
is cyclic, 
.
,
.
is cyclic. Then 
,
.
,
.
,
,
.
are cyclic.
is cyclic, this is trivial by observing 
.
,
and 
,
which is equilavent to 
as desired.
denote directed angles.
is cyclic.![\[\angle QFA = \angle DFA = \angle DCA = \angle BCA\]](http://latex.artofproblemsolving.com/f/7/5/f756b19f6207270611aaf0090e9c196792a8456b.png)
where the second equality comes from cyclic quadrilateral ![\[\angle QPA = \angle BPA = \angle BCA\]](http://latex.artofproblemsolving.com/8/0/4/80437d81c0d42d62f8cf1fa5935fa6406a76ba89.png)
where the last equality comes from cyclic quadrilateral 
.![\[\angle QPA = \angle QFA\]](http://latex.artofproblemsolving.com/2/1/8/2188be6cb7a152c46c27b078897cdffa42860781.png)
so 
.![\[\angle AQP = \angle AFP = \angle BFE = \angle BCE = \angle BCA\]](http://latex.artofproblemsolving.com/5/4/c/54cbfc1c463d1913f2030d3e446192cbd9655b21.png)
and ![\[\angle QPA = \angle BPA = \angle BCA = \angle AQP\]](http://latex.artofproblemsolving.com/b/9/a/b9a30af6187946008131a0c517d5ca9d00429e0f.png)
Thus, 
as desired.
,
and 
,
,
,
is cyclis because 
.
but 
as well.
so it suffices to show that 
completing the proof. Note that the case when 
is cyclic can be dealt with similarly. QED
so 
so we are done.
![\[ \angle AFQ = 180^\circ - \angle BFD = 180^\circ - \angle C = 180^\circ - \angle APQ \]](http://latex.artofproblemsolving.com/d/b/5/db51167075e051715dfe4c620efcfa05bd6bcc91.png)
so ![\[ \angle AQP = \angle ACE = \angle C = \angle APQ \]](http://latex.artofproblemsolving.com/8/7/1/871b749dd11a06185e82bcfe75ad60db8203f282.png)
which finishes.
be the intersection of 
be the intersection of 
.
implying that 
,
meaning that 
,
,
.
and we're done. 
,
is isosceles with 
