,
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 
,
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 
