,
,
,
,
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
.
,
be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
.
intersects line 
and 
.
.
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, 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 
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
)
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 
of white cells and black cells under the path differ by 
)
and this is possible when 
for order the 
