be a positive integer. Consider a 
array of square unit cells. Two corner cells that lie on the same longest diagonal are colored black, and the rest of the array is white. A move consists of choosing a row or a column and changing the color of every cell in the chosen row or column.
satisfying
Prove
such that for any 
:
is inscribed in circle 
R
and 
intersect at 
is chosen on the diagonal 
so that 
is chosen on the segment 
so that 
Prove that line 
touches 
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Ok so there's been no geo marathon here for more than 2 years,so lets start one,rules remain same.
1st problem.
Let
be a cyclic quadrilateral with 
and let 
and 
be the feet of altitudes from 
to the lines 
and 
,.Prove 
bisects 
.
P.s._eeezy ,try without ss line.
1st problem.
Let
be a cyclic quadrilateral with 
and let 
and 
be the feet of altitudes from 
to the lines 
and 
,.Prove 
bisects 
.P.s._eeezy ,try without ss line.
. Hence SMK is isosceles & since SKQ=90° KMQ is also isosceles. Hence SM=KM=MQ.
,
be the incenter and 
the excenter opposite 
.
meets 
and the circumcircle of triangle 
and 
,
of the circumcircle of triangle 
and 
intersect the circumcircle of triangle 
and 
,
inscribed in circle 
with orthocenter 
,
,
.
take 
: 
perpendicular 
.
cuts 
cuts 
.
cuts 
and 
are tangent to each other.
)
and 
intersecting at points 
and 
,
be a line through the center of 
and 
and let 
be a line through the center of 
and 
.
and 
.
denote the feurbach hyperbola of 
,
be the isogonal conjugate of point at infinity along 
, which is also 
,
;
is cyclic, and now reims finishes.
be given and it's orthocenter 
and 
e
and 
,
and 
.
is cyclic and that it's circumcircle passes through Feuerbach point of 
be the intouch triangle, let 
let 
be the medial triangle, let 
be the orthic triangle, and let 
are concyclic.
to 
is projective and the map from 
is projective so it remains to check three cases.
then 
are concyclic. If 
of 
then 
and 
and 
are concyclic. If 
then there is a common parabola tangent to the intouch triangle 
,
,
is the foci of this parabola and by Miguel's theorem the result follows.
and 
.
is cyclic.
where 
is the extouch point.![\[
(G, C; T, \infty_{CH}) \overset{I}= (B, B'; P, F) \overset{X} = (B, C; \infty{BC}, D)
\]](http://latex.artofproblemsolving.com/a/0/b/a0b46a0db50e6d8fba8fb855da0a13651fbcd943.png)
where 
and 
follows by Iran Lemma.
,
.
by angle chasing we get that 
using Lemma 2.
and 
lies on 
.
is cyclic which follows by angle chasing 
,
.
is known to be cyclic by Lemma 1. We get that 
also lies on it by length. Then since 
,
follows.
as the angle bisector of 
is perp / parallel to 
.
and thus 
is cyclic.
are cyclic. Note that 
don't lie on a line, so 
must be cyclic.
also lies on this point by earlier so we are done.
where 
are the fermat points. Prove the Euler line of the triangle with vertices as those three points goes through the centroid 
of 
passes through the reflection of 
is enough to finish. 
,
as desired.
and 
.
by cos theorem and computations (for three sides), it is evident that 
on ellipse 
,
centered on the longer axis of 
,![\[\frac{\text{radius of $\omega_A$}}{\text{radius of $\omega_B$}}=\frac{CA}{CB}\]](http://latex.artofproblemsolving.com/8/f/5/8f55637d277268f63cf152e06316acf9d696519f.png)
satisfy that (1) 
and 
,
are tangent at 
intersect at 
,
,![\[B_1B_2X\text{ is collinear, and }CX\text{ is tangent to }\omega_{1,2}\]](http://latex.artofproblemsolving.com/b/a/a/baab6dbe19f53e98bcad2ddccba5aa4d55f7b519.png)
is not collinear and 
is not the tangent.
be the tangent to 
at 
.
,
is collinear. It is easy to show ![$$[B_iA_{i1},B_iA_{i2};B_iX,B_iC'_i]=-1$$](http://latex.artofproblemsolving.com/6/5/a/65ae8996646341194d8a63d27766647b44ac2d5a.png)
as 
is the polar line of 
wrt 
and 
,
we get ![$$[C_1,C_2;X_i,Y_i]=-1$$](http://latex.artofproblemsolving.com/b/3/5/b359d5dc749ce20c4d592c32b962ea72ac71e966.png)
Let 
.![$[C_1,C_2;D,C]=-1$](http://latex.artofproblemsolving.com/f/9/8/f98dfb95946a1a93d2131104685be551a3b3244d.png)
;
is on the same "side" of 
are necessarily on different "sides" of 
is inside the segment 
for one 
,
inside it and Menelaus) and Menelaus). It is better to draw your own graph and see why it happens.
are all tangent to 
is 
at 
(index mod 5). Denote 
and 
.
.
![\[\prod_{i}\frac{C_iX_{i+2}}{C_iX_{i+3}}=1\]](http://latex.artofproblemsolving.com/5/5/d/55d8c171347e34d412a6d6b90178f5222862c174.png)
But by Menelaus we have ![\[\prod_i \frac{C_iX_{i+2}}{X_{i+2}B_{i+2}}\frac{B_{i+2}Y_i}{Y_iB_{i+3}}\frac{B_{i+3}X_{i+3}}{X_{i+3}C_i}=\prod_i1=1\]](http://latex.artofproblemsolving.com/1/d/1/1d1ff082231db477e8b9e5d92a09d77cdb1e7d03.png)
The 
term cancels, so we shall only prove ![\[\prod_i \frac{B_{i+2}Y_i}{Y_iB_{i+3}}=1\]](http://latex.artofproblemsolving.com/3/a/a/3aaa95d10d450d51f83c84aa19a2037e17307142.png)
This led to the following lemmas:
and ellipse 
as follows: construct tangents 
and 
with tangent points 
to 
.
,
.
intersects 
.
.
.
such that 
.
.
,
and setting 
does the trick. But we can again stretch ![\[\prod_i \frac{B_{i+2}Y_i}{Y_iB_{i+3}}=\prod_i F_{E_i}(B_{i+2},B_{i+3})=\prod_i F_E(B_{i+2},B_{i+3})=\prod_i\frac{G_E(B_{i+2})}{G_E(B_{i+3})}=1\]](http://latex.artofproblemsolving.com/3/f/1/3f1b5502a3b7afe446fbf32ea86182aedf6cea3e.png)
.
.
respectively such that 
.
,
,
.
.
,
is tangent to 
.
,
.
meet 
for the second time at 
is tangent to 
at 
,
,
Next, we claim that ![\[ \measuredangle AEL = \measuredangle AEC = \measuredangle ABF = \measuredangle AMF = \measuredangle AML. \]](http://latex.artofproblemsolving.com/2/9/7/2979cb20256b16d9d696f945239339863a62b91c.png)
Let 
,
.![\[ ZG^2 = ZM \times ZA = ZL \times ZE, \]](http://latex.artofproblemsolving.com/a/1/d/a1d14133c05fe6bd62484295682368c68dbfaf3a.png)
which implies 
is also tangent to 
.
,
and 
with 
form a line parallel to 
,
,
is another tangent to that circle. 
cuts 
are concyclic.
and 
.
.
and 
.
gives 
are collinear hence 
are concurrent. Pascal at 
implies 
are collinear.
hence 
are concurrent. By using this, the line passing through 
is 
are concurrent. Thus, there is an involution 
so 
.
is 
.
swap with 
which are collinear as desired
be the incenter and the circumcenter of triangle 
at 
and 
cuts 
at 
cuts 
at 
are collinear, then we bash them!
are the midpoints of the minor arc 
of the circumcircle, respectively. Line 
intersects 
Let line 
Let line 
at point 
Prove: 
,
.
and 
.
(the real axis) and ![\[
e = \frac{a^2 + c^2}{a^2c^2 + 1}.
\]](http://latex.artofproblemsolving.com/8/8/f/88f180f12c6063978576f6028b65132b042c8c16.png)
Let 
.![\[
k = \frac{a^2(c-1)^2 + c}{a^2c + 1}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5f58929f1d165a9d7513c06489136ab093d6c52.png)
Let 
and ![\[
l = \frac{2a^2 - a^2c + c}{a^2 + 1}.
\]](http://latex.artofproblemsolving.com/7/9/f/79f08ae94056d21a6b4a8157892a5a1c52b9c6ba.png)
Let ![\[
t = a^2 - a^2c + c.
\]](http://latex.artofproblemsolving.com/1/9/f/19f3021892fca42ceed1bd026cba678f906fad7e.png)
We need to verify that 
are collinear. Indeed, consider the cross-ratio:![\[
\frac{l-t}{k-t} = \frac{a^4c - a^2c + a^2 - 1}{a^4c + a^2c - a^2 - 1}.
\]](http://latex.artofproblemsolving.com/d/d/4/dd474b1de358efae4debe32012d6be1be1ec77e7.png)
Since 
,
is a real number, which confirms that 
and 
.
and 
be the isotomic conjugates of 
and 
passes through the circumcenter of 
