[/quote]
,
and 
be isogonal conjugates with respect to 
and 
be their respective pedal triangles. Let![\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]](http://latex.artofproblemsolving.com/7/f/e/7fe3ec4f69ec074635b2202daad9371e3bb8c72d.png)
Prove that the points 
,
and 
lie on the line 
.
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)
be positive reals strictly smaller than 
,
.
integers are such that 
is prime and 
.
such that 
.
is given. 
is a circle with diameter 
is a circle with diameter 
and 
.
meets 
and 
.
meets 
at 
,
meets 
at 
.
meets 
.
.
, such that 
Prove that :
equality cases : ?
Prove that
![\[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)\]](http://latex.artofproblemsolving.com/e/9/a/e9a57f0700d37fbb28f7d219cc7ac7e52b43acd9.png)
holds for all positive reals 
,
,
.
be a triangle in which 
.
and 
be the incentre and circumcentre of 
be the midpoint of the arc 
.
given that 
.
.
and 
intersect at point 
is symmetric to point 
intersects the circumcircle 
of the triangle 
at the point 
.
is tangent to the circle 
and 
intersect at the point 
and 
.
intersect at point 
.
be the reflection of 
and 
intersect at point 
.
is inscribed into a circle 
.
be an altitude of this triangle. A ray 
meets 
is tangent to 
Y by Adventure10, Mango247, ddami
On sides 
and 
of 
given are 
and 
, respectively. Let 
(
) be a point of intersection of circumcircle of 
and line that is parallel to 
and passing through C. Let 
be a point of intersection of line 
and side , and let 
be on line such that 
and 
. If 
, prove that point of intersection of 
and 
lie on angle bisector of 
.
Proposed by Milos Milosavljevic
and 
of 
given are 
and 
, respectively. Let 
(
) be a point of intersection of circumcircle of 
and line that is parallel to 
and passing through C. Let 
be a point of intersection of line 
and side , and let 
be on line such that 
and 
. If 
, prove that point of intersection of 
and 
lie on angle bisector of 
.Proposed by Milos Milosavljevic
This post has been edited 1 time. Last edited by mihajlon, May 22, 2015, 10:28 PM
and 
.
are concyclic ,
,
are concyclic 
are collinear (Reim theorem (C-D-B and F-E-H)) .
,
,
is the bisector of 
. i.e. the intersection of 
and 
lie on the bisector of 
are cyclic, 
.
.
,
are cyclic.
to 
(the midarc 
say 
concur by radical at say 
.
and since 
is exterior bisector 
must be the interior bisector.
.
.
and 
are cyclic, since
and
Therefore, we have 
and since 
,
.
,
and 
multiplying those together we obtain 
where the first is true, since 
and we used the fact that 
and we conclude by Ceva's theorem, that 
are concurrent, i.e. 
and 
is cyclic.
,
,
.
.
. So this claim is also done. 
and 
,
and 
.
,
b
in 
.![[asy]
import geometry;
size(12cm);
pair foot(pair P, pair A, pair B) {
return foot(triangle(A,B,P).VC);
}
path rect = (2,2)--(2,-1.1)--(-3.7,-1.1)--(-3.7,2)--cycle;
pair C = dir(140);
pair A = dir(220);
pair B = dir(320);
pair X = intersectionpoint(bisector(line(C,A),line(C,B)),line(A,B));
pair P = (2C+7X)/9;
pair D = intersectionpoint(line(A,P),line(B,C));
pair E = intersectionpoint(line(B,P),line(A,C));
pair F = intersectionpoints(parallel(C,line(A,B)),circle(C,D,E))[1];
pair Y = intersectionpoint(line(D,E),line(B,A));
pair G = intersectionpoint(line(F,D),line(A,B));
pair H = intersectionpoint(line(F,E),line(A,B));
pair N = intersectionpoints(line(Y,C),circle(C,D,E))[1];
draw(A--B--C--cycle , red+1.3);
draw(C--X,dotted+blue);
draw(D--Y,red);
draw(A--Y,red);
draw(N--Y,red);
draw(circumcircle(A,E,D),green);
draw(circumcircle(B,E,D),green);
draw(circle(F,G,H),purple+dashed);
draw(circumcircle(A, B, C) , blue);
draw(circumcircle(C, E, D) , orange);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$D$", D, dir(20));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$G$", G, dir(G));
dot("$H$", H, dir(H));
dot("$Y$", Y, dir(Y));
dot("$N$", N, dir(N));
clip(currentpicture,rect);
[/asy]](http://latex.artofproblemsolving.com/7/c/1/7c1d1bb9d2261749774af55fd598bb94924944c5.png)
![\[\measuredangle AGF = \measuredangle CFG = \measuredangle CFD = \measuredangle CED\]](http://latex.artofproblemsolving.com/0/1/1/0113421f818db7c2c5ef8cdaece2817c51aad2a1.png)
which implies that ![\[\measuredangle HDE = \measuredangle ADE + \measuredangle HDA = \measuredangle AGE + \measuredangle GEB = \measuredangle GBE = \measuredangle HBE\]](http://latex.artofproblemsolving.com/7/4/2/7427fd92ba380dab9f1470235edc77888020ffcd.png)
from which it follows that ![\[\measuredangle FED = \measuredangle FDC = \measuredangle FCB = \measuredangle HBD\]](http://latex.artofproblemsolving.com/9/7/f/97fba7d2088746378209be407038191558cf7b0c.png)
which since 
and 
,
and 
.![\[\measuredangle HEN = \measuredangle FEN = \measuredangle FDN = \measuredangle GDN\]](http://latex.artofproblemsolving.com/d/6/7/d674ec5f6541b52a64ccb8fcd569dcc73e0843ed.png)
which implies that 
.![\[\measuredangle NHF = \measuredangle NHE = \measuredangle NGD = \measuredangle NGF\]](http://latex.artofproblemsolving.com/1/1/e/11ed70dbbd42c1f4b89ac92bb17b80841c5ab6e9.png)
from which it follows that 
is a cyclic quadrilateral. Further,![\[\measuredangle AGN = \measuredangle HGN = \measuredangle HFN = \measuredangle EFN = \measuredangle ECN = \measuredangle ACN\]](http://latex.artofproblemsolving.com/2/3/e/23e4bf301dd77c29e829451d3085dd0469ed2a45.png)
and thus 
must also be cyclic. Now, by the Radical Center Theorem on circles 
, 
and 
,
, 
and 
intersect at a point 
,![\[-1=(YZ;ED)\overset{C}{=}(YX;AB)\]](http://latex.artofproblemsolving.com/d/9/f/d9f5d181f0a46e75ea455a2c8cc3a411899084a7.png)
Now, say 
.
,![\[-1=(YX';BC)\]](http://latex.artofproblemsolving.com/4/f/b/4fb351ec875c8904f91fb07d23e11838cc1eb2aa.png)
which implies that 
,
b
are collinear
intersect extended 
at 
.
.
.
Hence, 
Hence 
Hence 
intersect 
.
,
From Ceva's Theorem
