,
is the foot of the altitude from 
,
is a point on the segment 
.
and 
meet 
at 
and 
,
and 
lie on the circles 
and 
,
and 
.
and 
are concyclic.
and 
be its orthocentre and circumcentre, respectively,
and 
be the feet of the altitudes from 
to 
to 
be the midpoint of the segment 
,
be the midpoint of the segment 
.
,
again at 
,
meets the circle 
again at 
.
and 
are similar.
.
is extended beyond 
.
,
be a convex quadrilateral such that diagonals 
and 
intersect at right angles, and let 
be their intersection. Prove that the reflections of 
are concyclic.
Y by
Let 
be a simple polygon and let 
and 
be positive integers with 
. Suppose that it is possible to partition into rectangles and shade of them gray such that no two shaded rectangles share a positive amount of perimeter. (It is permissible for any two of them to share a vertex.) In terms of and , what is the fewest number of sides that could have?
Benny Wang
be a simple polygon and let 
and 
be positive integers with 
. Suppose that it is possible to partition into rectangles and shade of them gray such that no two shaded rectangles share a positive amount of perimeter. (It is permissible for any two of them to share a vertex.) In terms of and , what is the fewest number of sides that could have?Benny Wang
can be seen to be the union of rectangles aligned to the same axis. Let 
.![\[
4 \cdot \min (k - w, 1).
\]](http://latex.artofproblemsolving.com/d/f/c/dfcf2b1739d4b1c4c130b64a6e92bb3c1206659a.png)
,
rectangles and shade non-adjacent ones. If 
,
rectangles and shade every other piece such that we shade the ending. We then cut one of the ending pieces into 
distinct shaded rectangles, which creates an additional 
corners as desired.
:![[asy]
path rect = (0,0)--(0,3)--(1,3)--(1,0)--cycle;
pen shade = lightgrey;
filldraw(rect, shade);
filldraw(shift(1,0)*rect, white);
filldraw(shift(2,0)*rect, shade);
filldraw(shift(3,0)*rect, white);
filldraw(shift(4,0)*((0,0)--(0,1)--(1,1)--(1,0)--cycle), shade);
filldraw(shift(4,0)*((0,2)--(0,3)--(1,3)--(1,2)--cycle), shade);
[/asy]](http://latex.artofproblemsolving.com/8/e/2/8e284f971c5b2ce8bb52bc2cc1394602bc664e0d.png)
![[asy]
path rect = (0,0)--(0,3)--(1,3)--(1,0)--cycle;
pen shade = lightgrey;
filldraw((0,0)--(0,3)--(1,3)--(1,0)--cycle, shade);
filldraw((0,4)--(3,4)--(3,5)--(0,5)--cycle, shade);
filldraw(shift(4,2)*((0,0)--(0,3)--(1,3)--(1,0)--cycle), shade);
filldraw(shift(2,-4)*((0,4)--(3,4)--(3,5)--(0,5)--cycle), shade);
filldraw(shift(1,0)*((0,0)--(0,3)--(1,3)--(1,0)--cycle), white);
filldraw((shift(0,-1)*((0,4)--(3,4)--(3,5)--(0,5)--cycle)), white);
filldraw(shift(3,2)*((0,0)--(0,3)--(1,3)--(1,0)--cycle), white);
filldraw(shift(2,-3)*((0,4)--(3,4)--(3,5)--(0,5)--cycle), white);
filldraw(shift(2,2)*unitsquare, shade);
[/asy]](http://latex.artofproblemsolving.com/0/2/c/02c27fd2ff17af895db4a7391a24a9c4009e7db9.png)
edges. Since 
