[/quote]
,
of acute triangle 
is 
.
intersects 
.
be 
.
,
,
,
,
,
similarly. Prove that 
,
,
are concurrent.
![$P(x) \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/5/3/453a624c3b002c1b0e78e0023b24dd22ddd03557.png)
such that:
with 
is an 
th prime and 
is an arbitrary positive integer.
grid with each cell colored with an integer from 
to 
.
A point 
The circumcircle of 
meets the segment 
again at 
The circumcircle of triangle 
meets the segment 
at 
Prove that lines 
are concurrent.
,
and 
,
intersects at point 
,
intersects 
.
. Prove that points 
,
is incenter of triangle 
and 
is a line tangent to the incircle. Let 
be another line such that intersects 
respectively at 
.
to the incircle other than 
,
.
are similarly defined. Prove that 
are concurrent.
and 
.
and 
is right.
is called 
if there exists a permutation 
of the numbers 
,
have different parities for every 
;
the sum 
is a quadratic residue modulo 
for every 
.
is tiled in two ways - only with dominoes and only with squares 
.
,
,
points 
and 
and 
is a parallelogram. The circle with center the midpoint 
intersect for the second time at the point 
.
and 
intersect in a point.
and 
Show that
Where 
Show that
Where 
Y by tenplusten, anantmudgal09, xhenisa, Problem_Penetrator, AlirezaSh, megarnie, Adventure10, Mango247, ItsBesi, Rounak_iitr, PikaPika999
Let 
be a convex quadrilateral with 
and 
. Let 
and 
be points on segments 
and 
, respectively, such that line 
intersects lines 
and 
at points 
and 
, respectively. It is given that 
.Let the midpoint of 
be 
and the midpoint of be 
.Prove that the points 
and 
lie on a circle.
be a convex quadrilateral with 
and 
. Let 
and 
be points on segments 
and 
, respectively, such that line 
intersects lines 
and 
at points 
and 
, respectively. It is given that 
.Let the midpoint of 
be 
and the midpoint of be 
.Prove that the points 
and 
lie on a circle.
implies 
,
and 
are right angled it follows that
too, done.
,
,
is an exterior angle of 
.
,
is a cyclic quadrilateral.
,
,
.
.
and easy angle-chasing yields that 
.
or 
is cyclic.
and 
simultaneously. (Thus triangles 
and 
play morally dual roles.)
,
,
.
because 
from right angle 
.
and 
.
,
.
and we are done. 
(probably should have 
)
.
and 
be the feet of the altitudes from 
to 
are right triangles, 
is the reflection of 
over the angle bisector of 
.
the reflection of 
over the angle bisector of 
.
and 
are pairs of parallel lines, 
,
as desired.
and 
have common circumcenter at 
as desired.
concur. (And similarly, 
concur.)
Because 
we deduce that the point 
and 
Therefore,
and this proves that the points 
to the center of circumcircle 
(they become collinear in the new configuration) map to a circle in the original configuration?
are concyclic which isn't something projectivity saves .(for example a perspectivity).
and
are concyclic
and 
; 
, 
and 
and 
and 
is cyclic.
and 
,
, as desired.
and 
.
and 
.
and 
.
,
.
and 
,
.
and 
,
.
and 
is cyclic.
,
,
,
.
,
.
thus we have that 
is right angled and 
,
.
.
so 
while 
Notice 
so 
Similarly, we have 
.
,
by utilizing that 
.![\[ 2 \angle D = \angle CMA \]](http://latex.artofproblemsolving.com/3/a/8/3a8b36ccbda8fceedc7dd154d76a5ff32c5ae0ff.png)
and ![\[ \angle CNA= \angle CNQ +\angle ANQ = (180-2 \angle PQB )+(180 - 2 \angle NAP) = 360 - 2(\angle NPB+\angle PQB) = 2\angle D\]](http://latex.artofproblemsolving.com/b/8/d/b8d73ef5491777dee10d8704ffcec52ebdec59bd.png)
and 
THen, 
However, since 
Then, 
so 
hence done.
.
.
is cyclic qed.
,
and 
are medians in right angled triangles
and 
and 
.
and 
and 
.
.
,
.
are concyclic.
.
first. We can see that ![\[\angle AMC = \angle AMB + \angle BMC \Rightarrow \angle MAB + \angle MCB = 180 - \frac{\angle AMC}{2} \Rightarrow \angle ADC = \angle DAM + \angle DCM = \frac{\angle AMC}{2} \]](http://latex.artofproblemsolving.com/8/1/c/81cf2d0379146053a72a4344390ea566a6132168.png)
For 
we have ![\[ \angle ANC = \angle ANP + \angle NQC \Rightarrow \angle NAP + \angle NCQ = 180 - \frac{\angle ANC}{2} \Rightarrow \angle DAN + \angle ANC + \angle NCD + \angle ADC = 360\]](http://latex.artofproblemsolving.com/9/f/a/9fa8b658acf10bcdadc56a3da61dd31443fdc762.png)
(note that this is not convex). Thus we can see that, 
.
since 
as well. Thus, we have![\[\angle ANC = \angle ANP + \angle QNC = 2(\angle ASP + \angle QRC) = 2(\angle DSR+\angle DRS) = 2 \angle ADC\]](http://latex.artofproblemsolving.com/6/9/3/693a0fa7cfa075bff940970e133c24706e1c604c.png)
.
,![[asy]
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is the midpoint of 
is cyclic. 
is cyclic with center 
and 
.
.
and 
So![\[\angle ANC=\angle CNQ+\angle ANP=2(\angle CRQ+\angle ASP)=2(\angle DRS+\angle DSR)=2\angle ADC=\angle AMC.\]](http://latex.artofproblemsolving.com/4/e/2/4e2dff70690e60d36a29b7eb197cea9c39d0eff1.png)
while 
Therefore, 
and the desired result follows. QED
and 
obviously. Also, 
which finishes.
thus, we are done.
