[/quote]
,
,
units horizontally (left or right); or
and 
,
to 
using at most 
moves.
be a set of 
elements and 
are subsets of 
)
elements.
and 
,
,
and 
is greater or equal to: 
,
are parallel and the angles 
and 
are acute. Show that it is possible to divide the triangle 
and the triangle 
into 4 disjoint triangles 
such that the triangles 
and 
are congruent for all 
such that
has side lengths 
.
inside 
,
,
and denote them as 
,
,
.
?
be non-negative real numbers such that 
Prove that
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of 
(the number of positive divisors of 
with its center 
.
and one moving point 
on 
.
is a fixed point on 
intersect
,
intersects 
at 
.
when 
such that:![\[
a^2 + b^2 \mid a^3 + b^3
\]](http://latex.artofproblemsolving.com/a/1/d/a1ddbd20c854e291da5670de449229d1664fa56c.png)
.
.
.
be the midpoints of segments 
intersects 
.
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 
.
.
be points on a circle, and let 
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.
.
,
centred at 
.
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
