[/quote]
,
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
and orthocenter 
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
.
be the point on the circumcircle of triangle 
is perpendicular to 
be the point on line 
such that 
.
and 
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
,
intersects the extension of 
at point 
.
is the midpoint of 
,
intersects the circumcircle of 
.
.
be positive real numbers such that 
.![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
,
.
be reals such that 
P
,
,
is __
satisfying 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
,
is obtuse
is acute
,
is semi perimeter
,![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
satisfy that: 
and 
.
and 
Prove that
Y by anantmudgal09, ATimo, AhirGauss, GeoMetrix, RythBot, A-Thought-Of-God, Adventure10, Mango247, sami1618
In triangle 
let 
be the 
-excenter. Let 
be an arbitrary circle that passes through 
and intersects the extensions of sides 
(extended from 
) at 
respectively. Let 
be points on segments 
respectively such that 
and 
.Lines 
intersect at 
. Lines 
intersect at 
.
Prove that
are collinear.
Proposed by Hooman Fattahi
let 
be the 
-excenter. Let 
be an arbitrary circle that passes through 
and intersects the extensions of sides 
(extended from 
) at 
respectively. Let 
be points on segments 
respectively such that 
and 
.Lines 
intersect at 
. Lines 
intersect at 
.Prove that
are collinear.Proposed by Hooman Fattahi
This post has been edited 4 times. Last edited by bgn, Apr 7, 2017, 6:42 PM
.
.
we get:
.
and 
.
and 
.
.
and Trig Ceva in 
we get:
.
,
and 
.
are cyclic and 
hence if 
then:
note that:
Observe that by ceva_sin theorem:
combining 
we conclude that 
.
.
be the point on the line 
and 
.
and 
are concyclic. Since 
and 
are reflections in 
;
are reflections in 
,
we conclude that 
bisects angle 
.
so 
as desired.
at 
,
cyclic quickly implies that triangles 
and 
are congruent and 
lies on segment 
.
,
.
and 
.
and 
cut segments 
at 
.
,
is cyclic and let 
is cyclic and 
are collinear by well-known properties of the Miquel Point. Let 
,
is perpendicular to 
then 
is perpendicular to 
is cyclic, and hence we just want to show that 
,
.
;
)
;
)
;
)
;
)
)
,
lie on a circle
;
)
;
)
(mod 
;
)
)
;
)
)
;
or 
= 
= 
. 
= 
or 
. 
. 
. 
. 
= 
. 
= 
;
)
)
;
)
+ 
)
is a parallelogram
Let 
,
is parallelogram, 
is the miquel point of 
and since, 
and, 
bisects 
and 
be the reflection of 
.
. Similairly for 
and 
Our goal is to prove 
.
is projective . Now 
is projective. And also since 
we have 
is projective. So it suffices to check for three values of 
so done 
.
is a point such that 
so done again 
are concyclic.
and we are done 
.
and 
,
similarly
Therefore
Hence
which implies 
are concyclic.
hence 
Hence
Now
Hence
Now
similarly
Hence
together with 
be the point on 
.
and 
.
is cyclic.
Also, in the same manner, it is easy to see that 
and 
are cyclic.
is cyclic.
.
On the other hand,
Subtracting those we get that 
is cyclic with 
.
collinearity to 
concyclity. Since 
,
.
.
and 
and Law of Sines on 
and 
gives
Since 
and 
,
,
hence 
,
,
instead of 
throughout the rest of the solution. Let 
and 
.
and 
.
is cyclic. It follows that 
is cyclic. Then 
and 
whence 
which yields 
is cyclic. It's worth noticing that 
is also cyclic since 
Next, we claim that 
and 
are parallel to each other. To see this, let 
and 
then 
which yields 
.
is cyclic. By Reims, it follows that 
as desired.
.
.
are collinear, then via Menelaus we get 
It suffices to show 
are collinear. To finish, combining all previous results with the fact that 
is cyclic, we have
which follows from 
.
as 
,
and 
.
is the external angle bisector of 
,
and 
,
.
,
and 
,
.
is cyclic, so![\[\measuredangle ZES=90^\circ-\measuredangle SI_AT=\measuredangle OTS = \measuredangle OPS=\measuredangle ZPS\]](http://latex.artofproblemsolving.com/1/5/c/15ca3d0bf7b9c927a04475c2479a8b3c5c77898c.png)
and so 
is cyclic too. Similarly, 
is also cyclic. Consequently,![\[\measuredangle PEI_A=\measuredangle PZS=\measuredangle PZT=\measuredangle PFI_A\]](http://latex.artofproblemsolving.com/d/e/b/deb3c2d2f35162bdbc34a054882633da522941b1.png)
and thus 
is also cyclic. Now to finish, since 
is the Steiner line of ![[asy]
defaultpen(fontsize(10pt));
size(10cm);
pen mydash = linetype(new real[] {5,5});
pair A = dir(120);
pair B = dir(220);
pair C = dir(320);
pair I = incenter(A,B,C);
pair I1 = 2*circumcenter(I,B,C)-I;
real s = 0.3;
pair P = s*B+(1-s)*C;
pair X = 2*foot(P,B,I1)-P;
pair Y = 2*foot(P,C,I1)-P;
pair S = 2*foot(circumcenter(C,P,I1),I1,B)-I1;
pair T = 2*foot(circumcenter(B,P,I1),I1,C)-I1;
pair K = extension(B,T,C,S);
pair Z = extension(I1,K,S,T);
draw(A--B--C--cycle, black+1);
draw(B--X);
draw(C--Y);
draw(B--I1);
draw(C--I1);
draw(B--T);
draw(C--S);
draw(S--T);
draw(K--I1);
draw(X--Y, mydash);
draw(P--X, dotted);
draw(P--Y, dotted);
draw(circumcircle(A,X,Y));
draw(circumcircle(B,P,T));
draw(circumcircle(C,P,S));
draw(rightanglemark(P, foot(P,B,I1), B, 2));
draw(rightanglemark(P, foot(P,C,I1), I1, 2));
dot("$A$", A, dir(A));
dot("$B$", B, dir(180));
dot("$C$", C, dir(30));
dot("$X$", X, dir(270));
dot("$Y$", Y, dir(0));
dot("$P$", P, dir(90));
dot("$S$", S, dir(225));
dot("$T$", T, dir(10));
dot("$K$", K, dir(90));
dot("$I_A$", I1, dir(270));
dot("$Z$", Z, dir(315));
[/asy]](http://latex.artofproblemsolving.com/d/4/0/d4075f241228fd0b1b2bb7cdef29cb6d6fd758dd.png)
![[asy]
defaultpen(fontsize(10pt));
size(10cm);
pen mydash = linetype(new real[] {5,5});
pair K = dir(35);
pair S = dir(95);
pair I1 = dir(195);
pair T = dir(345);
pair B = extension(I1,S,T,K);
pair C = extension(S,K,I1,T);
pair Z = extension(S,T,I1,K);
pair O = (0,0);
pair P = extension(O,Z,B,C);
pair E = extension(Z,foot(Z,I1,T),I1,B);
pair F = extension(Z,foot(Z,I1,S),I1,C);
pair X = 2*foot(P,B,I1)-P;
pair Y = 2*foot(P,C,I1)-P;
draw(K--S--I1--T--cycle);
draw(S--B);
draw(K--B);
draw(T--C);
draw(K--C);
draw(S--T);
draw(I1--K);
draw(B--C);
draw(O--P, dotted);
draw(foot(Z,I1,T)--E, mydash);
draw(foot(Z,I1,S)--F, mydash);
draw(circumcircle(K,S,T));
draw(circumcircle(Z,F,T), dotted);
draw(circumcircle(Z,E,S), dotted);
draw(rightanglemark(Z,P,C,2));
draw(rightanglemark(Z,foot(Z,I1,T),I1,2));
draw(rightanglemark(Z,foot(Z,I1,S),S,2));
draw(I1--E--F--cycle, fuchsia+1);
draw(circumcircle(I1,E,F), fuchsia+1);
dot("$I_A$", I1, dir(I1));
dot("$S$", S, dir(S));
dot("$T$", T, dir(315));
dot("$K$", K, dir(45));
dot("$E$", E, dir(90));
dot("$F$", F, dir(315));
dot("$B$", B, dir(90));
dot("$C$", C, dir(0));
dot("$O$", O, dir(270));
dot("$Z$", Z, dir(180));
dot("$P$", P, dir(45));
[/asy]](http://latex.artofproblemsolving.com/3/a/5/3a588214f98fb2383ddac467217942652387bd59.png)
.
meet 
(where 
is a directed angle), 
,
,
,
,
.
,
.
and 
be the intersections of the perpendicular bisector of
by the converse of the Simson line 
and 
is cyclic. We will prove this by phantom points.
to 
and 
to 
.
be the reflection of 
.
,
is cyclic.
.
,
are collinear or 
.
,
,
.
.
,
,
.
is the midpoint of
,
,
and 
hence 
.
.
is a cyclic quadrilateral whose miquel point is 
are the reflections of 
,
.
be the orthocenters of 
.
are steiner lines of 
hence 
be the altitudes from 
to 
and 
be the orthocenter of 
.
.
where 
are collinear as desired
.
and 
meet 
and 
the spiral symmetry also sends 
so 
.
is cyclic as 
.
.
