[/quote]
,
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)
and 
points on a plane is 
,
such that: in every set of 
points of a plane, no three of which are collinear, consisting of 
points of every color there exists a 
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 
.
be positive reals strictly smaller than 
,
.
integers are such that 
is prime and 
.
such that 
.
is a circle with diameter 
,
is a circle with diameter 
.
meets 
and 
.
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 
,
,
.
.
and 
be the incentre and circumcentre of 
be the midpoint of the arc 
.
given that 
.
and 
.
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 
.
Y by Davi-8191, tenplusten, anantmudgal09, OlympusHero, Adventure10, HWenslawski, Jc426, math_comb01, Mango247, Rounak_iitr, and 3 other users
In a convex quadrilateral 
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](//latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.Attachments:
This post has been edited 2 times. Last edited by djmathman, Aug 1, 2015, 2:53 AM
Reason: formatting
Reason: formatting
,
,
,
be the orthogonal projections of the point P on the lines AB, BC, CD, DA, and let 
,
,
,
be the reflections of P in these lines, or, equivalently, the reflections of P in the points 
is the line BD. Hence, the points 
, we have 
.
and 
,
.
.
.
and 
,
has the segment AP as diameter. Therefore, by the Extended Law of Sines, we have
,
.
.
along line 
has remained the same, so 
is the perpendicular bisector of 
are the symmetrics of 
respectively wrt 
is an isosceles trapezoid which has 
to 
and 
respectively on the circumcircle of 
and 
since they subtend equal arcs. The circle also gives us 
by equal arcs 
.
since they are vertical, and 
by subtraction. The equal angles and two adjacent equal sides force 
and 
at 
and 
.
and triangles 
and 
are similar in ratio 
.
by subtraction, and triangles 
and 
are also similar in ratio 
and 
are similar in ratio 
,
.
is 
.
is cyclic, its reflection 
is also cyclic and the original quadrilateral 
. ]
.
,
and by an analogous argument that 
.
so that by definition, 
.
,
.
is cyclic, so you agree that 
.
,
,
while 
.
.
Let 
and 
Notice 
so 
and 
Similarly, 
Since 
it lies on the perpendicular bisector of 
which is equivalent to the perpendicular bisector of 
Construct 
on 
such that 
Similarly construct 
on 
such that 
Notice 
as 
are isogonal conjugates wrt 
so 
and 
Hence, 
and 
is cyclic isosceles trapezoid. Also, 
and 
so 
Similarly, 
is cyclic isosceles trapezoid and so 
is cyclic. 
intersect the circumcircle of 
Let 
Note that 
so 
This implies 
Similarly, 
is a cyclic trapezoid so 
Thus it must also lie on the perpendicular bisector of 
be reflections of 
Since 
is a reflection of 
over 
so 
is cyclic. Also, 
Thus, 
is cyclic. This implies 
so 
is isosceles trapezoid. Thus, 
so the reflection of 
and 
and 
are both isosceles trapezoids and therefore 
(ii) Assume 
,
which implies 
,
bisects 
.
is cyclic. Since 
,
and 
or 
is cyclic. However, the latter is absurd because 
and 
,
to the corresponding point 
.
is cyclic, then define 
.
such that 
and 
,
so one direction is completed
.
.
both belong to the 
,
.
and thus 
is also cylcic. Hence 
is cyclic.
and 
be 
.
,
,
are cyclic with diameters 
,
respectively.
and 
Now, because 
.
,
.
In addition, we have 
Thus, 
is cyclic if and only if 
.
It follows that 
is cyclic, as this is the harder portion.
and 
and 
.
and 
to be isogonal in this context.
and 
,
and 
excenters, respectively. Now comes the key claim:
and 
are both cyclic.
is cyclic, as the others are analogous. Observe that 
is the angle bisector of 
,
,
!
on 
,
.
bisects 
.
is the external angle bisector of 
,
.
,
.
and 
.
,
is an isoceles trapezoid, so 
.
.
,
is an isoceles trapezoid as well. This means that 
and 
share a perpendicular bisector.
and 
.
to meet the circumcircle of 
respectively.
so it follows that 
and 
which implies that 
is an isosceles trapezoid with legs 
we have 
so 
It follows that 
is an isosceles trapezoid with legs 
so 
is parallel with 
we have 
so 
Since 
so 
as desired. 
such that the fourth point is not outside the circumcircle of those three points.
then 
and let 
be the second intersections of the lines 
with 
(a property we proved in the necessity that it must be isosceles), but since 
intersects that circle at only two points, either 
is an isosceles triangle. We proceed when 
because the other case is analogous.
Since P lies on the perpendicular bisector of 
,
,
which implies that 
It follows that P lies on the perpendicular bisector of 
,
is an isosceles trapezoid (perpendicular bisectors are parallel). Therefore, since 
by SSS congruence, we have 
Hence, 
,
.
share a perpendicular bisector, and the result follows.
.
and 
and 
are congruent as 
,
.
,
is cyclic, and similarly 
is cyclic. It follows that all the points 
are concyclic, as needed.
and 
,
and 
.
and 
are isosceles trapezoids (with 
)
are collinear, as are 
,
which lies on the perpendicular bisector of 
by symmetry.
be points and 
such that 
are isogonal in 
,
etc. The isogonality condition is equivalent to
Upon expansion and simplification this means that any 
)
,
,
and 
.
,
.
is placed at 
,
so 
but 
.
are valid positions for 
, then construct 
is cyclic. Let 
intersect again a
.
be the midpoint of 
and 
the midpoint of 
are collinear.
.
which finishes.
.
means orange, 
means green and 
and 
are like the diagram that is (hopefully) attached.
,![\[ \measuredangle PAC = \measuredangle ACP \iff \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0. \qquad (\spadesuit). \]](http://latex.artofproblemsolving.com/7/e/5/7e57e142c8f1fa0c11c8588dfc18cc3f5af78fa0.png)
A careless reader would be forgiven for thinking that 
implies the problem or at least one direction, but it turns out the situation is more subtle. The second part analyzes the angle conditions more carefully and provides a complete proof.
.
.
,![[asy]
import geometry;
size(9cm); pair A = (-0.50099,1.69422); pair B = (-1.8,-0.6); pair C = (0.09182,-1.79600); pair D = (1.55802,-0.60384); pair P = (-1.04359,-0.19363);
draw(circumcircle(A,B,C), gray+dashed); filldraw(A--B--C--D--cycle, invisible, black); draw(B--D); draw(A--C);
draw(C--P--A, brown); draw(B--P--D, gray);
markangle(radius=15, n=2, P, A, C, red, StickIntervalMarker(1,1, red)); markangle(radius=15, n=2, A, C, P, red, StickIntervalMarker(1,1, red)); markangle(radius=10, n=3, A, P, B, deepgreen, StickIntervalMarker(1,2, deepgreen)); markangle(radius=10, n=3, C, P, D, deepgreen, StickIntervalMarker(1,1, deepgreen)); markangle(radius=12, n=1, B, A, C, blue); markangle(radius=12, n=1, C, B, D, blue); markangle(radius=12, n=1, D, C, A, blue); markangle(radius=12, n=1, A, D, B, blue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$P$", P, dir(45));
[/asy]](http://latex.artofproblemsolving.com/0/e/c/0eca105f1f7fddffcbb253a96d5434767c875c18.png)
and 
in the claim (colored green with three rings) so that their only dependence on 
and 
(colored red with two rings), as follows: 
Since the claim says 
,
in terms of only 
These four latter angles are colored blue with one ring in the figure. This proves 
.![\[ \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0 \]](http://latex.artofproblemsolving.com/b/8/6/b86d4b46e9417c6e66aa186a00663048a68a1057.png)
is equivalent to 
.
meets line 
.
meets line 
,
.
(that is, 
)
if and only if 
is not degenerate) and the second lemma turns our angle condition into 
be the point for which 
is cyclic and harmonic; then 
.
(the corresponding Apollonian circle only meets line 
