.
and 
,
.
,
is composite, then all positive divisors of 
such that 
is an integer for every positive integer 
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
be the reflection of 
.
intersects line 
and 
.
.
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
and 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
to 
to 
.
and 
is a point not on the line and that 
and 
Find the area of 
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 a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
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
.
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.
