[/quote]
,
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
be a triangle with incentre 
.
tangent to 
at 
again at 
.
tangent to 
again at 
.
is tangent to the incircle of 
be a positive integer, and let 
be the set 
.
and 
be difined as:
Which is larger, 
or 
?
be a square with diagonals 
,
touch sides 
and 
at 
and 
respectively. Let 
and 
denote the circumcircles of triangle 
and 
respectively.
and 
intersect at 
and 
and 
intersect at 
and 
and 
and 
Prove that 
and 
Show that 
and 
are concurrent.
, such that for all 
:![\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]](http://latex.artofproblemsolving.com/1/4/1/1418fad9fd00b31c38bfbc5657e2a322d66ef450.png)
such that, for all positive integers 
and 
,
,
.
,
.
,
intersects 
.
is midpoint of 
is orthocenter of triangle 
so that 
.
coincide
,
.
lies on segment 
such that 
.
and 
be the circumcenter and incenter of triangle 
,
and 
be the circumcenter and incenter of triangle 
,
and 
intersect at point 
.
.
the number of positive integer 
satisfying 
is less than or equal to 
where 
is a constant independent of 
,
satisfies 
and 
.
to line 
meets the angle bisector of 
.
holds.
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
be a positive integer divisible by 3, and let 
be non-negative real numbers satisfying 
.![\[
M_n = \min_{1 \le i \le n} \{x_i x_{3i}\},
\]](http://latex.artofproblemsolving.com/d/b/b/dbbf0f6bf20f506d6cb63357b06ec6c52bebed10.png)
where the indices are taken modulo 
.
with integer coefficients which satisfy the following conditions:
is a positive integer for any positive integer 
for all positive integers 
Y by ImSh95
A unit square is removed from the corner of an 
grid, where 
. Prove that the remainder can be covered by copies of the figures of 
or 
unit squares depicted in the drawing below.
![[asy]
import geometry;
draw((-1.5,0)--(-3.5,0)--(-3.5,2)--(-2.5,2)--(-2.5,1)--(-1.5,1)--cycle);
draw((-3.5,1)--(-2.5,1)--(-2.5,0));
draw((0.5,0)--(0.5,3)--(1.5,3)--(1.5,1)--(3.5,1)--(3.5,0)--cycle);
draw((1.5,0)--(1.5,1));
draw((2.5,0)--(2.5,1));
draw((0.5,1)--(1.5,1));
draw((0.5,2)--(1.5,2));
[/asy]](//latex.artofproblemsolving.com/5/7/8/578baca60367441fe9b771689f1a2e91c0ccf984.png)
Note: Every square must be covered once and figures must not go over the bounds of the grid.
grid, where 
. Prove that the remainder can be covered by copies of the figures of 
or 
unit squares depicted in the drawing below.![[asy]
import geometry;
draw((-1.5,0)--(-3.5,0)--(-3.5,2)--(-2.5,2)--(-2.5,1)--(-1.5,1)--cycle);
draw((-3.5,1)--(-2.5,1)--(-2.5,0));
draw((0.5,0)--(0.5,3)--(1.5,3)--(1.5,1)--(3.5,1)--(3.5,0)--cycle);
draw((1.5,0)--(1.5,1));
draw((2.5,0)--(2.5,1));
draw((0.5,1)--(1.5,1));
draw((0.5,2)--(1.5,2));
[/asy]](http://latex.artofproblemsolving.com/5/7/8/578baca60367441fe9b771689f1a2e91c0ccf984.png)
Note: Every square must be covered once and figures must not go over the bounds of the grid.
This post has been edited 2 times. Last edited by Rukevwe, May 6, 2022, 2:54 PM
rectangular, so 
3-unit figure can make 
rectangular.
have been proved. Then when 
,
.
exists, 
with a unit removed from a corner can be made as below.
or 
,
does exist. So we only need to prove whe
.
is trivial.
:
:
:
and 
.
only needs one case: the top-right 
can be reduced to 
we split into three cases, also note that one can build a 
then build a big L shape on the bottom-left and use 
it's the same method but start with a middle L shape
then it's the same method but start with the 
