[/quote]
,
,
and 
.
is prime, 
with 
and 
be a positive integer. Egor has 
cards with the number 
”
”
that has 8097 distinct points with area of a triangle that has 3 points belong to 
.
that has the area 
contains at least 2025 points that belong to 
is even it's contradiction. I mean 
.
and 
,
and 
.
.
.
but 
.
we have 2 option. Firstly 
but 
is contradiction. Secondly 
but 
.
.
but 
.
.
of positive integers, a partition of 
and 
is 
if the least common multiple of the elements in 
such that there exists a set of 
good partitions.
are lengths of sides of some triangle. Prove the inequality
where 
.
is the circumcenter of triangle 
is the projection of point 
onto line 
.
,
are 
,
,
,
intersects 
and 
at points 
and 
,
.
,
,
according to the following rule: If 
,
,
if 
(taking 
)
is the symbol other than 
if 
.
for which there exists some integer 
such that 
.
and 
be the feet of the altitudes of triangle 
and 
be the midpoints of 
and 
,
be the intersection of lines 
and 
.
.
and 
Prove that
and 
be two prime numbers, with 
.
such that the polynomial 
is divisible by the polynomial 
with 
is a multiple of 
.
be the circumcircle of acute triangle 
and 
are on segments 
.
and 
intersect minor arcs 
and 
and 
are either parallel or they are the same line.
and circumcircle 
.
is the midpoint of small arc 
is the projection of 
with center 
meets 
,
bisects 
.
Y by Rounak_iitr, GeoKing
Let 
be a triangle, and let 
be centered at 
, 
and tangent to line 
at 
, 
respectively. Let line 
intersect 
again at 
and let line 
intersect 
again at 
. If 
is the other intersection of the circumcircles of triangles and 
, then prove that lines 
, , and 
either concur or are all parallel.
Advaith Avadhanam
be a triangle, and let 
be centered at 
, 
and tangent to line 
at 
, 
respectively. Let line 
intersect 
again at 
and let line 
intersect 
again at 
. If 
is the other intersection of the circumcircles of triangles and 
, then prove that lines 
, , and 
either concur or are all parallel.Advaith Avadhanam
![[asy]
size(10cm);
import olympiad;
import geometry;
pair A = (0.7,3);
pair B = (0,0);
pair C = (4,0);
pair O_b = (0,0.8);
pair O_c = (4,1.8);
pair X = 2*foot(O_b,A,B) - B;
pair Y = 2*foot(O_c,A,C) - C;
pair X_1 = reflect(O_c, circumcenter(A,X,Y)) * Y;
pair T = extension(O_b,O_c,B,C);
pair P = 2*foot(O_b,T,X) - X;
draw(A--B--C--cycle, linewidth(1));
draw(circle(O_b,abs(O_b-B)));
draw(circle(O_c,abs(O_c-C)));
draw(T--X_1);
draw(T--B, dashed);
draw(Y--X_1);
draw(anglemark(P,X,B));
draw(anglemark(X_1,X,A));
draw(anglemark(X_1,Y,A));
dot("$A$", A, dir(107));
dot("$B$", B, dir(-90));
dot("$C$", C, dir(-88));
dot("$X$", X, 1.5*dir(121));
dot("$Y$", Y, dir(-151));
dot("$X'$", X_1, dir(-15));
dot("$T$", T, dir(-169));
dot("$P$", P, dir(131));
[/asy]](http://latex.artofproblemsolving.com/0/3/a/03af4676f637abc43a5efc80cfccc64cea549d68.png)
be the exsimilicenter of 
meet 
.
that sends 
,
Hence, 
,
and 
,
,
,
.
,
,
,
.
,
.
.
.
and 
,
;
and solving gives 
.
.
and 
intersect at 
.![\[\frac{CX}{XD}=\frac{[ABC]}{[ABD]}=\frac{\frac{1}{2} CA \cdot CB \cdot \sin \angle ACB}{\frac{1}{2} DA \cdot DB \cdot \sin \angle ADB}=\frac{CA}{AD} \cdot \frac{CB}{BD},\]](http://latex.artofproblemsolving.com/7/8/7/7872bbed8ae4e3eaac187e2cca11c1caee752edc.png)
as desired. 
and 
intersect at 
.
,
.![\[\frac{ZB}{ZC}=\frac{BA}{AC} \cdot \frac{BQ}{QC}=\frac{AB}{AC} \cdot \frac{BX}{CY}=\frac{AB}{AC} \cdot \frac{2BO_1 \cdot \cos \angle O_1BA}{2CO_2 \cdot \cos \angle O_2CA}=\frac{AB}{AC} \cdot \frac{BO_1}{CO_2} \cdot \frac{\sin \angle ABC}{\sin \angle ACB}=\frac{BO_1}{CO_2}.\]](http://latex.artofproblemsolving.com/d/f/8/df822e915c6acbbb0c5190afd50bc38a860cba8b.png)
Thus, we have 
,
denote two lines intersecting in point 
is a linearly moving point on 
and 
,
is constant.
.
,
,
.
.
is a linear map (spiral similarity centered at 
.
and 
.![\[\frac{O_1B}{O_2C}=\frac{BX}{CY}\cdot\frac{\sin(\measuredangle CBA)}{\sin(\measuredangle ABC)}=\frac{QB}{QC}\cdot\frac{AC}{AB}\]](http://latex.artofproblemsolving.com/3/1/a/31a7489cbe6e5d4a462d8989dd2a52702f7220ef.png)
Let 
.
to be collinear, it suffices to show that ![\[\frac{TB}{TC}=\frac{O_1B}{O_2C}=\frac{QB}{QC}\cdot\frac{AC}{AB},\]](http://latex.artofproblemsolving.com/1/3/6/1362c7df9af48f4bc1eb93a1392d64620220ae6a.png)
which is a well-known identity (sketch: use sine rule in 
)
Define O as the centre of 
, now observe that 
and 
so we can find 
and rest is same as ABOVE
my solution is wayyy overcomplicated oops.
.
,
.
and 
under this inversion. Note that 
and 
.
and 
for reasons to be disclosed later on. This is not hard; notice that 
under 
.
and 
to satisfy these properties; it’s quite clear now that under 
also passes through 
and then note that 
, we have two lines such that one of them intersect 
( It follows from similarity)
Denote the radius of circle tangent to 
the problem will solve by homothety and that three lines are concur.
We know that 
So 
. So the problem solved
