attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
satisfying 
.
Y by HWenslawski, centslordm, Rounak_iitr
Let 
be a cyclic convex quadrilateral and 
be its circumcircle. Let 
be the intersection of the diagonals of 
and 
. Let 
be the center of the circle tangent to sides 
, 
, and 
, and let 
be the midpoint of the arc of not containing 
and 
. Prove that the excenter of triangle 
opposite lies on the line 
.
be a cyclic convex quadrilateral and 
be its circumcircle. Let 
be the intersection of the diagonals of 
and 
. Let 
be the center of the circle tangent to sides 
, 
, and 
, and let 
be the midpoint of the arc of not containing 
and 
. Prove that the excenter of triangle 
opposite lies on the line 
.This post has been edited 2 times. Last edited by Miku3D, Jun 9, 2021, 6:40 AM
Reason: Changed title from "Collinear Stuff" to current title
Reason: Changed title from "Collinear Stuff" to current title
as the in, A-excenters of 
,
as the in, D-excenters of 
,
all lie on a circle and the problem's done by Pascal's on 
.
and ignore 
.
with 
the excenter of 
,
.
.
and so 
which is equal to 
,
,
and similarly, 
,
,
or 
.
must be equal to 
(half of what 
should be) and 
.
are actually equal.
equal, there exists a point 
conditions plus the fact that the 
.
,
,
,
collinear.
Basically, let 
be the intersection of the perpendicular bisector of 
(if those lines are equal then 
is a kite and there's nothing to prove).
,
and 
,![\[ \dfrac{LB}{\sin(LI_EB)} = 2R_{\triangle LBI_E} = 2R_{\triangle LCI_E} = \dfrac{LC}{\sin(LI_EC)} \]](http://latex.artofproblemsolving.com/6/b/9/6b94a96031cf86a1d592659bbb69f8615d6801be.png)
then 
and 
.
by construction, let 
and similarly, 
.
yields![\[ 1 = \dfrac{BM}{MI_E} \dfrac{MI_E}{MC} = \dfrac{\sin(BI_EM)}{\sin(MBI_E)}\dfrac{\sin(MCI_E)}{\sin(MI_EC)} = \dfrac{\sim \sin(\gamma)}{\sin(C-\beta)} \dfrac{\sin(C-\gamma)}{\sim \sin(\beta)} \]](http://latex.artofproblemsolving.com/6/e/4/6e44108e9b0facc2e1cc2d572763151d53c78f55.png)
Since our goal is to presumably prove that 
,
(as this would also imply that 
.![\[ \sin(\gamma) \sin(C-\gamma) = \sin(\beta) \sin(C-\beta) \Rightarrow \dfrac{\cos(C)-\cos(2\gamma-C)}{-2} = \dfrac{\cos(C)-\cos(2\beta-C)}{-2} \]](http://latex.artofproblemsolving.com/1/d/a/1daf1816edf5e2fb92eb7242e5cae8710eabe6fc.png)
implies 
which basically means 
or 
.
Now construct a line 
and 
so that 
and 
;
cut 
at 
,
,
,
in it.
,
has something cut out of segment 
while 
has something attached to the segment 
,![\[ |BM_B| < |BM'| < |CM_C| \]](http://latex.artofproblemsolving.com/8/6/0/860703e03bb62d601097d477fd8a0a058a4ef34b.png)
and similarly, 
.
I used earlier), 
can be obtained to be equal to 
and 
is equal to 
multiplied by 
;
is equal to 
;
.![\[ \dfrac{M_BI_E}{BM_B} = \dfrac{\sin(LI_EB)}{\sin(\gamma)} = \dfrac{\sin(LI_EC)}{\sin(\beta)} = \dfrac{M_CI_E}{CM_C} \]](http://latex.artofproblemsolving.com/2/9/7/2975ae680a583da38b684bd397acfda00348ba7b.png)
However, this is blatantly wrong by simple inequalities: a fraction (specifically, on the LHS) with a bigger numerator and smaller denominator must have a bigger value.
the thought processes I got during the contest since all papers were to be destroyed (to prevent leaking, presumably) but what I vividly remember is 
'
Firstly, it seemed that the only thing which hindered the circle with center 
and 
back; to be able to define 
the reference point for forever (in my mind, at least) and examined the circle 
concur at one point, and since the line 
passes through 
,
)
in a lot of ways: the angles 
,
(excircle opposite 
,
![\[\begin{tabular}{p{12cm}}
We don't have $\color{red} \overline{\text{lengths}}$, so what better way can we include $\color{blue} \textbf{all points}$ (if say we have captured $L,M$ but not $I_E$, how can 3 floating points must be collinear?) in perspective with $\color{magenta} \textit{one camera snapshot}$ aside from $\color{cyan} \underline{\text{the richness of angles available}}$ in an $\color{green} \boxed{\text{in-excenter-cyclic-mid arc}}$ config?
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/2/c/f/2cf9809fdaa9c302ebb8346dfcaf85858144b9c9.png)
;
is clearly 
.
is actually just equal to 
;
and 
are supposed to be equal.
)![\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]](http://latex.artofproblemsolving.com/8/4/2/842413e566043c96b7a8f2300401b08af9d8ba48.png)
upwards towards 
;
and 
.
,
was chosen as I imagined the config to be most friendly when 
;
first, descending gradually with a movement of 
and 
be the incircle of 
and 
)
the excircle of 
,
,
.
and 
insterect at 
and 
are both external bisectors of 
and 
,
which means that 
is cylic with 
because 
and 
are both diameters of 
.
and 
.
is cyclic
.
and 
are perspective
and 
.
.
,
,
(on line 
)
are collinear. So, 
,
,
are concurrent at 
)
,
be the excenter of 
,
be the isogonal conjugate of 
.
so 
is an isosceles trapezoid. By symmetry, this means that 
is isogonal with respect to 
,
,
denote the incenters of 
,
,
,
denote the excenters of 
,
.
.
is the incentre of 
be the incentres of 
and let 
be the excentres of 
and 
be the excentre of 
is cyclic then by incentre excentre lemma 
lies on the same circle .
and 
are collinear .
and 
is the external angle bisector of 
which implies 
are collinear .similarly 
are collinear .therefore by Pascal's theorem on hexago
is the midpoint of the major arc 
,
is the midpoint of arc 
are collinear.
be the incenter of 
be the incenter of 
.
be the miquel point of the self intersecting quadrilateral 
and so 
and 
(because 
.
,
are also collinear. Since 
and so 
.
,
are collinear or that 
.
since its the excenter and 
and so we are done. 
be the incenters of triangles 
,
and 
are collinear.
,
and 
.
and 
are also collinear.
and 
are collinear as well, since 
i.e., 
is cyclic with its center being point 
.
.
e
,
.
and 
.
,
.
and 
therefore 
.
,
.
,
,
and 
are acute.
where 
is strictly increasing, hence injective.
implies that 
,
hence 
.
by simple angle-chasing: 
therefore we infer that 
and 
form equal angles, and two lines, one from each pair, are perpendicular.
be the midpoint of 
,
be the midpoint of 
and let 
Note that 
Therefore, 
We can similarly show that 
Now construct 
and 
Remark that 
Hence, 
Moreover, by the second isogonality lemma, 
is isogonal to 
Thus, 
collinear, as desired.
and 
and 
respectively. From the incenter-excenter lemma 
.
be the excenter of 
and 
intersect 
and 
respectively. Pascal Theorem on 
finishes the proof.
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[/asy]](http://latex.artofproblemsolving.com/2/3/9/239e4733468d1ead799204bd7bab598121567a1c.png)
,
,
be the excenter of triangle 
be the arc midpoint of 
and 
.
,
,
are the midpoints of the arcs not including 
.
and 
-- this is because we have two of the same midpoints and one of the opposites.
and so on. First choose the sign of 
arbitrarily, then choose the sign of 
so that 
.
such that 
,
.
such that 
,
.
,
,
,
,
,
,
,
,
,![\[\frac{\ell - m}{n - m} = \frac{\ell - bc}{n - bc} \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/7/0/1/701e88e383b436201bbcc3b926cd713001006d94.png)
![\[\ell = \frac{b^2ac(c^2 + bd) - c^2bd(b^2 + ac)}{b^2ac - c^2bd} = \frac{ab(c^2 + bd) - cd(b^2 + ac)}{ab - cd},\]](http://latex.artofproblemsolving.com/7/5/c/75c7362f0504db3bd8df30a03aca64748222d0a0.png)
so then we can factor ![\[\ell - bc = \frac{(b - c)(ad(b + c) - bc(a + d)}{ab - cd}.\]](http://latex.artofproblemsolving.com/b/3/0/b302f9a3fdac22ca73377c5c6dfc6df0db1af125.png)
Similarly we have ![\[n = \frac{b^2cd(c^2 + ab) - c^2ab(b^2 + cd)}{b^2cd - c^2ab} = \frac{bd(c^2 + ab) - ac(b^2 + cd)}{bd - ac},\]](http://latex.artofproblemsolving.com/6/3/a/63a959fc2f214d74b25d71ca05a6e91cb004797c.png)
so then ![\[n - bc = \frac{(b - c)(ad(b + c) - bc(a + d)}{bd - ac}.\]](http://latex.artofproblemsolving.com/1/5/a/15a64f4b4cdea8651a58cd8c64849747b8d2d11d.png)
This means ![\[\frac{\ell - bc}{n - bc} = \frac{bd - ac}{ab - cd},\]](http://latex.artofproblemsolving.com/7/6/6/766a8d8c817277a0c73acd20ebf175994249ce82.png)
which is clearly self-conjugate so is real.
be 
.
lie on 
.
through 
through 
are concurrent.
But, we have 
,
and 
,
but this is true by sine law on 
and 
.
,
and 
,
,
,
,
are collinear. The problem is immediate by complex at this stage, but we choose to goof around instead.
.
.
and 
.
and 
convex. We need to check the maps 
and 
are the same noting 
is a projective map. At 
,
but also 
,
,
and 
so collinearity certainly holds. At 
,
and 
so collinearity certainly holds. gg.
and 
be the incenters of 
and 
,
.
and 
.
and 
,
is a cyclic quadrilateral.
and 
(i.e. the incenter of 
,
.
.
is a cyclic quadrilateral with diameter 
,
.
,
,
and 
are tangent to the circumcircles of 
and 
,
as required. 
,
that's tangent to both 
and 
,
and 
,![\[
\angle LBI = \angle LBC + \angle CBI = \frac 12 \angle ABC + 90^{\circ}-\frac 12 \angle EBC = 90^{\circ}+\frac 12 \angle ABD
\]](http://latex.artofproblemsolving.com/4/6/d/46d0ce9f7e1fc82255c378230fecadd0173ed731.png)
and similarly 
.![\[
\angle BLC=180^{\circ}-\angle LBC-\angle LCB = 180^{\circ} - \frac 12 \angle ABC-\angle DCB
\]](http://latex.artofproblemsolving.com/8/5/1/851de5ed21be73fc0b373f713244ec8f9ad90033.png)
So![\[
\angle LBI+\angle BLC=270^{\circ} - \frac 12 (\angle ABC+\angle DCB-\angle DCA)=270^{\circ} - \frac 12(\angle ABC+ACB)=180^{\circ}+\frac 12 BAC
\]](http://latex.artofproblemsolving.com/d/9/2/d928321ea4e2810d2ee14070afd6f7785ab59e02.png)
which means that 
and similarly 
,
,
,
is cyclic with circumcircle 
,
be the incenters of 
,
,
is cyclic as 
Now, let 
at points 
meaning that 
and 
and consequently, 
is the 
we get that 
are collinear and consequently, line 
and 
be the corresponding angles as shown in the figure for the point 
.
is constant, it is equivalent to show 
,
.
in terms of 
and 
as denoted in the diagram. As in the diagram, let 
and let 
.![\[ \frac{CX}{\sin \theta}=\frac{LX}{\sin (\beta+\beta_2)}, \quad \frac{BX}{\sin \theta'}=\frac{LX}{\sin(\alpha+\alpha_2)} \implies f(X) = \frac{BX}{CX}\cdot \frac{\sin(\alpha+\alpha_2)}{\sin(\beta+\beta_2)}.\]](http://latex.artofproblemsolving.com/4/1/f/41fe76b882a0073bec32bd0e6b5390dd3076b871.png)
We now verify 
,
:
.
:
and 
bisect the angles external to 
and 
respectively. So 
and 
.
since 
.![\[ \frac{BI_E}{CI_E} = \frac{\sin \angle BCI_E}{\sin \angle CBI_E} = \frac{\sin(90^\circ-\tfrac12\angle ACB)}{\sin(90^\circ-\tfrac12\angle DBC)} = \frac{\cos(\tfrac12 \angle ACB)}{\cos(\tfrac12 \angle DBC)}. \]](http://latex.artofproblemsolving.com/8/f/2/8f2db70df8394d0cfdbbafb9c8a5dc02acef3830.png)
angles of to locate them. Then the above tricks are used to make the calculation not an insane trig bash.
and 
intersect the circumcircle of 
again at 
and 
are homothetic, so their center of homothety is 
be the 
and 
.
.
,
on 
is the polar of 
,
is concurrent.
,
and 
,
,![\[ \frac{MP}{MN} = \frac{PL}{NI_E} = \frac{PB}{NB}, \]](http://latex.artofproblemsolving.com/9/2/7/9272b102dc3ec074455d310353a285ac6ee2f9f0.png)
where 
and 
by the incenter-excenter lemma. Now note that![\[ \angle NBM = \angle NBE - \angle MBA + \angle ABE = \angle EBC + \frac{1}{2} \angle BEC - \angle ABC - \frac{1}{2} \angle A = \frac{1}{2}\angle ABD. \]](http://latex.artofproblemsolving.com/4/4/0/440a3cdb75799c941a1684b0583337fa022a3183.png)
Similarly, we can show that 
,
.
.
=
=
=
,
/
/
,
to find 
/
,
and 
to find 
be the excenter of 
be the intersection of lines 
,
is the exterior bisector of 
,
,
\
.
,
and 
meet 
and 
.
an 
.
,
,
and 
.
and 
are homothetic. Hence 
.
yields 
,
,
,
are all collinear triples. By the incenter-excenter lemma, 
intersect 
intersect 
at 
.
;
is similar. We have 
proving the claim.
gives 
collinear, as desired.
)
and 
respectively. By the incenter excenter lemma on 
.
is cyclic with center 
be the 
respectively. Hence, 
and 
are collinear.
and 
are collinear. Analogously, 
are collinear.
and 
are the external and internal bisector of 
.
,
collinear. Analogously, 
are collinear.
,
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, where 
and 
is a cyclic quad. By pascal's on 
we get 
are collinear.
