[/quote]
,
and 
. Prove that
Where 
such that: 
.
be a cyclic quadrilateral, and let 
,
,
,
be the midpoints of 
,
,
,
respectively. Let 
,
,
and 
be the orthocenters of triangles 
,
,
and 
,
have the same area.
be positive real numbers such that : 
. Prove that : 
,![\[
A = \left\{ r \, \middle| \, r = \frac{a}{3}\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) + b\sqrt[3]{xyz}, \quad x, y, z \in \left[1, \frac{a}{b}\right] \right\},
\]](http://latex.artofproblemsolving.com/e/9/2/e92becb6264b8e5a9753f2288436a48523a25f29.png)
and![\[
B = \left[2\sqrt{ab}, a + b\right].
\]](http://latex.artofproblemsolving.com/2/d/2/2d25aa634389b768dc26408480bb71b316c0bf17.png)
.
and 
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
be a set with seven elements, where each element is a positive integer not exceeding 
.
)![\[
x_1 + x_2 + \cdots + x_t = x_{t+1} + x_{t+2} + \cdots + x_m
\]](http://latex.artofproblemsolving.com/7/0/7/707e7e9a0ce45adb307545bbc88c1f0bf90ca0c9.png)
has a solution in the set 
,
are all distinct.
![\[
(AB + CD)^2 + (AD + BC)^2 = (AC + BD)^2,
\]](http://latex.artofproblemsolving.com/7/d/a/7daf2d6ac00609b5252b2b8b05d96a5ca0952a94.png)
then 
touch sides 
lie on the line through 
perpendicular to 
be the intersections of 
with 
be the projections of 
onto line 
be the second intersections of 
with the incircle 
and 
.
bisects segment 
.
can we find three positive integers 
,
and 
such that 
?
satisfy both of the following conditions.
,
,
.
that satisfies the equation 
and 
.
,
,
-
,
.
at 
.
parallel to 
meets 
.
is constant.
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 
.
and 
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 
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 
.
.
,
.
,
.
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 
so that 
be the intersection of 
.
implying that 
,
meaning that 
,
,
.
and we're done. 
,
is isosceles with 
:
.
so 
is cylic, thus 
so 
:
so 
is cylic and 
,
is also cyclic so 
and we're done.
