[/quote]
,
,
,
cut three common chords concurrent at 
.
,
which are internally tangent to all 
.
.
be a point on 
,
be a line passing through 
.
at 
and the circumcircle of triangle 
at 
)
and 
are tangent at 
be nonnegative real numbers such that 
.
,
satisfying 
.
satisfying:
.
and 
,
.
,
is composite, then all positive divisors of 
such that 
is an integer for every positive integer 
Y by KevinYang2.71, Pengu14, vincentwant, ihatemath123, MathRook7817, ESAOPS, LostDreams
Let 
and 
be positive integers, and let 
be a 
grid of unit squares.
A domino is a
or 
rectangle. A subset 
of grid squares in is domino-tileable if dominoes can be placed to cover every square of exactly once with no domino extending outside of . Note: The empty set is domino tileable.
An up-right path is a path from the lower-left corner of to the upper-right corner of formed by exactly 
edges of the grid squares.
Determine, with proof, in terms of and , the number of up-right paths that divide into two domino-tileable subsets.
and 
be positive integers, and let 
be a 
grid of unit squares.A domino is a
or 
rectangle. A subset 
of grid squares in is domino-tileable if dominoes can be placed to cover every square of exactly once with no domino extending outside of . Note: The empty set is domino tileable.An up-right path is a path from the lower-left corner of
to the upper-right corner of formed by exactly 
edges of the grid squares.Determine, with proof, in terms of
and , the number of up-right paths that divide into two domino-tileable subsets.This post has been edited 1 time. Last edited by rhydon516, Thursday at 12:09 PM
.
splits 
such that 
swaps 
or 
,
and 
,
.
and the upper right is 
.
move with starting point 
a two-letter designation, where the first letter is 
or 
depending on the parity of 
and the second letter depends on the parity of 
.
'
'
or 
.
,
and 
,
'
moves in each set, giving the desired 
paths. 
(how many points would that be? hopefully graders aren't too harsh)
gives an answer of 
(and 
gives an answer of 
)![[asy]
pen col;
for(int x =0; x<6; ++x) {
for(int y = 0; y <=4; ++y) {
if((x+y)%2 == 0) {
col = royalblue;
}
else {
col = orange;
}
draw((x,y)--(x+1,y), linewidth(3)+col);
}
}
for(int x =0; x<=6; ++x) {
for(int y = 0; y <4; ++y) {
if((x+y)%2 == 0) {
col = royalblue;
}
else {
col = orange;
}
draw((x,y)--(x,y+1), linewidth(3)+col);
}
}
for(int x =0; x<=6; ++x) {
for(int y = 0; y <=4; ++y) {
dot((x,y), black+5);
}
}
[/asy]](http://latex.artofproblemsolving.com/6/9/8/698104e1aaafedbf21124d2cd9f9e475fcddff23.png)
on the right side of the up-right path, consider the square 
to its left, and assign ![[asy]
fill((3,2)--(5,2)--(5,3)--(3,3)--cycle, mediumgray);
draw((0,0)--(0,1)--(2,1)--(2,2)--(3,2)--(3,3)--(5,3)--(5,4)--(6,4), black+linewidth(9));
draw((0,0)--(0,1)--(2,1)--(2,2)--(3,2)--(3,3)--(5,3)--(5,4)--(6,4), white+linewidth(6));
pen col;
for(int x =0; x<6; ++x) {
for(int y = 0; y <=4; ++y) {
if((x+y)%2 == 0) {
col = royalblue;
}
else {
col = orange;
}
draw((x,y)--(x+1,y), linewidth(3)+col);
}
}
for(int x =0; x<=6; ++x) {
for(int y = 0; y <4; ++y) {
if((x+y)%2 == 0) {
col = royalblue;
}
else {
col = orange;
}
draw((x,y)--(x,y+1), linewidth(3)+col);
}
}
for(int x =0; x<=6; ++x) {
for(int y = 0; y <=4; ++y) {
dot((x,y), black+5);
}
}
[/asy]](http://latex.artofproblemsolving.com/6/0/a/60aad7202ca56e20f4aa717dfe3979cd298a9594.png)
![[asy]
draw((0,0)--(3,0)--(3,1)--(4,1)--(4,2)--(5,2)--(5,3)--(6,3)--(6,4), black+linewidth(9));
draw((0,0)--(3,0)--(3,1)--(4,1)--(4,2)--(5,2)--(5,3)--(6,3)--(6,4), white+linewidth(6));
pen col;
for(int x =0; x<6; ++x) {
for(int y = 0; y <=4; ++y) {
if((x+y)%2 == 0) {
col = royalblue;
}
else {
col = orange;
}
draw((x,y)--(x+1,y), linewidth(3)+col);
}
}
for(int x =0; x<=6; ++x) {
for(int y = 0; y <4; ++y) {
if((x+y)%2 == 0) {
col = royalblue;
}
else {
col = orange;
}
draw((x,y)--(x,y+1), linewidth(3)+col);
}
}
for(int x =0; x<=6; ++x) {
for(int y = 0; y <=4; ++y) {
dot((x,y), black+5);
}
}
[/asy]](http://latex.artofproblemsolving.com/0/2/5/02516a426bfa513c5397b03b06b2a50a18012490.png)
denote the set 
on the bottom from the up-right path, where 
denotes the number of cells in 
are that it is nonstrictly increasing, every term is an integer in ![$[0,2n]$](http://latex.artofproblemsolving.com/9/f/7/9f71343e694ccc845a33fe5d68418e41284014e7.png)
,
.
are that it is strictly increasing, every term is an integer in ![$[1,2m+2n]$](http://latex.artofproblemsolving.com/0/b/f/0bf8d34f2392a947f9512813664ec86f62830a5a.png)
,
ways to choose the sequence 
gives 36 in the first 20 minutes of solving which probably allowed me to get the formula .
because you basically have to partition it into two equal groups
to 
)
squares then what we know from this sequence is that all values lie between ![$[0,2m]$](http://latex.artofproblemsolving.com/9/8/b/98b492ac8cadd9e248e9bdca1d50281e9e94f595.png)
and its non-decreasing and that the number of odd values on sequence in odd labeled rows is the same as the number of odd values on even labeled rows in order to balance the condition.
and this will be done by simply considering 
instead and the condition is that this is strictly increasing, ranges in ![$[1, 2m+2n]$](http://latex.artofproblemsolving.com/7/2/e/72e805c4fd981581389651a4740ad8fbf28b70c5.png)
,
.
,
was to check my final answer.
