.
,
.
be nonnegative integers such that 
Prove that![\[
\sum_{i=0}^n i\binom{a_i}{2}\le\frac{1}{2}\binom{a_0+a_1+\dots+a_n}{2}.
\]](http://latex.artofproblemsolving.com/c/4/9/c49d0ccfef69402d05dc44d721172498c605cf7e.png)
Note: 
for all nonnegative integers 
.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
,
such that 
.
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
is an integer for every positive integer 
.
and 
,
.
,
is composite, then all positive divisors of 
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
.
intersects line 
and 
.
.
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, Mar 20, 2025, 12:09 PM
.
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 ![$[0,2n]$](http://latex.artofproblemsolving.com/9/f/7/9f71343e694ccc845a33fe5d68418e41284014e7.png)
,
.![$[1,2m+2n]$](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
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.
consecutive moves of the same tipe the set under the path will be a triangle and so cannot have the same number of white and black.
and color all the cells in the quadrant 
and 
.
under the path and under the line from the point where we are parallel to 
(so if we have done 
)
of white area and black area in this region in every moment. After a right moves this quantity don't change. Now suppose that at 
otherwise it decreases by 
.
of up moves in even position and odd position. And a value of 
)
and this is possible when 
because we can consider the path of all up and then all right.
for order the 
