[/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 
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 
.
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 
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 
,
,
.
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 
.
Y by ineX, Davi-8191, Wizard_32, Amir Hossein, HWenslawski, ImSh95, Adventure10, Rounak_iitr, and 4 other users
Given a cyclic quadrilateral 
, let the diagonals 
and 
meet at 
and the lines 
and 
meet at 
. The midpoints of 
and 
are 
and 
, respectively. Show that 
is tangent at to the circle through the points , and .
Proposed by David Monk, United Kingdom
, let the diagonals 
and 
meet at 
and the lines 
and 
meet at 
. The midpoints of 
and 
are 
and 
, respectively. Show that 
is tangent at to the circle through the points , and .Proposed by David Monk, United Kingdom
be the intersection of 
and 
cuts 
.
then 
and 
.
,
is a cyclic quadrilateral.
and 
we obtain 
.
and a dilation of factor 
(equality holds because 
is the power of 
such that 
is similar to 
.
,
is a parallelogram. Similarly, 
such that 
is a parallelogram. Now, 
are on the symmetric to 
with respect to the internal bisector of 
are collinear. Note also that 
are the respective midpoints of 
,
.
,
and 
intersect at a point 
is tangent to the circumcircle of 
.
be on opposite sides of 
,
,
is increasing, so the above would imply that 
,
.![\[\frac{\sin \angle GHE}{\sin \angle HGE}=\frac{\sin\angle GEJ}{\sin\angle HEK}\]](http://latex.artofproblemsolving.com/6/5/c/65cfc2764135607738197921d45d8a44bb647887.png)
is similar to 
,
is similar to 
,
is similar to 
.![\[\frac{GJ}{JB}=\frac12\left(\frac{AJ}{JB}-1\right)=\frac12\left(\frac{CF\cdot DA}{BC\cdot FD}-1\right)=\frac{CF\cdot DA-BC\dot FD}{2BC\cdot FD}\]](http://latex.artofproblemsolving.com/8/0/9/8099e977101eabab21c1b1a780a782e8226a254a.png)
![\[\frac{HK}{KC}=\frac{AD\cdot BF - FA\cdot CB}{2FA\cdot CB}\]](http://latex.artofproblemsolving.com/1/7/b/17be1a0ca305146c93b260833e0d58fb3df652d5.png)
and 
are the corresponding medians in similar triangles 
and 
.
and get 
,
.
and 
hit 
.
and 
,
and 
,
are collinear.
be 
a
and 
are tangent, since inversion preserve tangency.
be 
intersect 
.
.
be the intersection of 
and 
.
and 
are cyclic because 
are right, and thus,
,
collinear 
collinear.
are the circumcenters of 
,
intersections of tangents in 
and 
on 
intersections of 
intersection 
.
is cyclic. This is enough because then 
.
so 
is cyclic. Similar 
is cyclic. So we have
;
and we have 
which sends 
to 
, 
to 
. 
and 
,
then.
of 
and is the inverse of 
in circumcircle of 
. Clearly, 
the circle 
passing through 
The tangent 
to the circle 
parallel though 
,
.
The line 
of 
.
.
is the radical axis of circles 
so it's enough to show that 
.
.
Lines 
and 
intersect at 
, 
on the circle 
Let 
.
is harmonic and so is the quadrilateral 
.
,
and 
are respectively, the symmedian and tangent to 
is harmonic and so is the bundle 
and so intersecting this bundle with 
and so 
which proves the lemma.
intersect 
in 
. Since the initial spiral symmetry also sends 
to 
,
, 
and 
.
is cyclic, 
.
which implies that the antiparallel to 
in triangle 
is parallel to 
which completes the proof.
.
.
.
,
.
remains true since the angle bisector of 
is parallel/perpendicular to the angle bisector of 
are collinear.
be the pole of 
.
and so 
are concyclic.
,
is tangent to 
,
,
.
we get that JI.JG = JK.JH 
.
.
.
real. By the Brocard argument, we can replace 
with 
and instead show it is imaginary. Furthermore, since 
,
with 
,
.
is imaginary where 
but this is clearly true as this becomes 
but if we conjugate and multiply by 
,
with 
.
intersect 
,
,
is cyclic, which follows from both 
and 
being harmonic and applying power of a point.
and 
so since 
are cyclic.
are collinear.
and 
and we are done.
intersect 
and 
at 
respectively. Let 
,
then by the midpoint lemma 
,
is cyclic by PoP. Now since 
by the midpoint lemma 
which implies the tangency.
intersect 
.
are cyclic.
,
, 
and 
.
are collinear.
and 
are all harmonic due to Ceva-Menelaus and projecting.
due to a well known property of harmonic bundles.
due to a similar property as above.
but since 
due to Power of Point
so 
are concylic
which shows that 
which is 
and 
.
.
.
.
and 
.
so 
is cyclic.
is the polar of 
and projecting from 
.
.
which implies that 
be points such that 
and 
be parallelograms. Let 
,
and, 
.
.
.
.
.
and 
is parallel to 
,
.
.
and 
.
and 
.
.
and 
.
and 
.
.
.
collinear is isogonal to 
and 
can be dissected along diagonal 
and 
into the similar triangles shown below, making the quadrilaterals themselves similar. However, we can then dissect along diagonals 
and 
.
![[asy]
size(270); defaultpen(linewidth(0.4)+fontsize(8));
pair A, B, C, D, E, F, K, L;
A = dir(95);
B = dir(55);
C = dir(350);
D = dir(190);
E = extension(A, C, B, D);
F = extension(A, D, B, C);
K = A+B-E;
L = C+D-E;
draw(A--B--C--D--F--B^^A--E--B--K--cycle^^C--E--D--L--cycle^^K--E--L--F--E^^unitcircle);
draw(E--(E+.98*(K-E))--(E+.98*(B-E))--cycle^^L--(L+.99*(E-L))--(L+.99*(D-L))--cycle, red+linewidth(2));
draw(F--(F+.96*(K-F))--(F+.96*(B-F))--cycle^^F--(F+.99*(E-F))--(F+.99*(D-F))--cycle, blue+linewidth(2));
dot("$A$", A, dir(120));
dot("$B$", B, dir(40));
dot("$C$", C, dir(330));
dot("$D$", D, dir(210));
dot("$E$", E, dir(315));
dot("$F$", F, dir(90));
dot("$K$", K, dir(130));
dot("$L$", L, dir(270));
[/asy]](http://latex.artofproblemsolving.com/4/4/4/444b34800afa9113ee225a9a10ad53bc494290b2.png)
,
.
and 
,![\[JG \cdot JX = JA \cdot JB = JC \cdot JD = JH \cdot JY,\]](http://latex.artofproblemsolving.com/f/e/d/fedf850c6cab4a7d411d8b57c531a0f194a4750e.png)
is cyclic. We finish with the bundle 
,![\[ME^2 = MX \cdot MY = MG \cdot MH. \quad \blacksquare\]](http://latex.artofproblemsolving.com/a/2/8/a28045795b9b714f1e4dd6fd1e1741036d5e0250.png)
![\[e=\frac{ac(b+d)-bd(a+c)}{ac-bd}\]](http://latex.artofproblemsolving.com/4/e/e/4eebfcca36a21000addf7ba15a2d96a57a7e06e3.png)
And![\[f=\frac{ad(b+c)-bc(a+d)}{ad-bc}\]](http://latex.artofproblemsolving.com/8/1/8/81826668146f3e74b4b47ba575f3b4b415229ed8.png)
Further, 
and 
.![\[\frac{\frac{\frac{ac(b+d)-bd(a+c)}{ac-bd}-\frac{ad(b+c)-bc(a+d)}{ad-bc}}{\frac{ac(b+d)-bd(a+c)}{ac-bd}-\frac{a+b}{2}}}{{\frac{\frac{c+d}{2}-\frac{ac(b+d)-bd(a+c)}{ac-bd}}{\frac{c+d-a-b}{2}}}}
\]](http://latex.artofproblemsolving.com/e/a/8/ea8028aa5643297ac570384d9d2ecd28040a2b38.png)
is self conjugate.
Further, we will evaluate:
Hence the numerator of our expression simplifies to![\[\frac{2(c-d)(ab(c+d)-cd(a+b))}{(2cd-ac-bd)(ad-bc)}\]](http://latex.artofproblemsolving.com/6/7/9/6795366d2cf928efd595d57ed111482669221faa.png)
Now by an analogous argument, we have![\[\frac{c+d}{2}-\frac{ac(b+d)-bd(a+c)}{ac-bd}=\frac{-(c-d)(2ab-ac-bd)}{2(ac-bd)}\]](http://latex.artofproblemsolving.com/b/1/f/b1f5a42ae630db59b0286492874774a27051e991.png)
Thus our expression evaluates to![\[\frac{ac-bd}{ad-bc}\times \frac{-2\left(ab(c+d)-cd(a+b)\right)(c+d-a-b)}{(2cd-ac-bd)(2ab-ac-bd)}\]](http://latex.artofproblemsolving.com/4/8/0/480c6ea7586ba9ee5775c854c15fd6567ff296c1.png)
It suffices to show this is self-conjugate.![\[\frac{\left(\frac{1}{abc}+\frac{1}{abd}-\frac{1}{cda}-\frac{1}{bcd}\right)\left(\frac{1}{c}+\frac{1}{d}-\frac{1}{a}-\frac{1}{b}\right)}{\left(\frac{2}{ab}-\frac{1}{ac}-\frac{1}{bd}\right)\left(\frac{2}{cd}-\frac{1}{ac}-\frac{1}{bd}\right)}=\frac{(d+c-a-b)(abd+abc-bcd-acd)}{(2cd-bd-ac)(2ab-bd-ac)}\]](http://latex.artofproblemsolving.com/2/d/f/2df19fe02f4f581c424cd21c3f3954232edec0e1.png)
And![\[\frac{\frac{1}{ac}-\frac{1}{bd}}{\frac{1}{ad}-\frac{1}{bc}}=\frac{bd-ac}{bc-ad}\]](http://latex.artofproblemsolving.com/8/a/3/8a31b7247e06b4c954a42c659305fd9d4a93f73b.png)
Thus our expression is the product of several self conjugate terms hence self conjugate. We are done.
.
if 
is the chord subtended by the Newton Gauss line in 
.
collinear. We shall prove that if 
tangent to 
(which is a parallelogram clearly), or first isogonality lemma, we get 
.
.
,
,
too.
are involutions on a line it must be an inversion, and we check that 
,
