,
,
,
,
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 
,![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
:o
so 
and since 
is just 
therefore 
so 
,![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
degrees about one side. The space through which the hexagon travels during the rotation forms a solid. Find the volume of this solid.
and height 
:
Volume of cone frustum with base radii 
,
and 
are numbers that work, but 
and 
don't work.)
-
?
students want to share 
pencils. If every student gets at least one pencil, how many ways are there to distribute the pencils?
Y by Carpemath, Amir Hossein, PcelicaMaja, AbdulelahAltaf, centslordm, megarnie, jhu08, This_deserves_a_like, Adventure10, Mango247
An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from 
to 
.
![\[\begin{array}{
c@{\hspace{4pt}}c@{\hspace{4pt}}
c@{\hspace{4pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{4pt}}c
} \vspace{4pt}
& & & 4 & & & \\\vspace{4pt}
& & 2 & & 6 & & \\\vspace{4pt}
& 5 & & 7 & & 1 & \\\vspace{4pt}
8 & & 3 & & 10 & & 9 \\\vspace{4pt}
\end{array}\]](//latex.artofproblemsolving.com/4/a/4/4a4607074647a93496d7428f5ce2b8489539b704.png)
Does there exist an anti-Pascal triangle with 
rows which contains every integer from to 
?
Proposed by Morteza Saghafian, Iran
to 
.![\[\begin{array}{
c@{\hspace{4pt}}c@{\hspace{4pt}}
c@{\hspace{4pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{4pt}}c
} \vspace{4pt}
& & & 4 & & & \\\vspace{4pt}
& & 2 & & 6 & & \\\vspace{4pt}
& 5 & & 7 & & 1 & \\\vspace{4pt}
8 & & 3 & & 10 & & 9 \\\vspace{4pt}
\end{array}\]](http://latex.artofproblemsolving.com/4/a/4/4a4607074647a93496d7428f5ce2b8489539b704.png)
Does there exist an anti-Pascal triangle with 
rows which contains every integer from to 
?Proposed by Morteza Saghafian, Iran
This post has been edited 3 times. Last edited by djmathman, Jun 16, 2020, 4:04 AM
Reason: problem author
Reason: problem author
,
is possible, just read the paper('s page two) by the four Chinese authors in posting #4 
was not big enough for the solution to work, though with a minor fix (see the last paragraph of the solution) it does (barely) work.
)
be the number of entries in the triangle. Call a number in the triangle big if it is greater than 
and small otherwise (so, about half the numbers are big and half are small). Note that each big number which is not in the bottom row must be above one small number and one big number. Now, the intuition of the proof proceeds as follows: if a number is very big (close to 
)
covers a big number 
and a small number 
,
.
and 
measure the "bignesses" of 
)
be the number of big numbers which are not in the bottom row. So, these ![\[\sum_{\text{big $k$ not in bottom row}} (m-k) = \sum_{\text{$k$ covers big $i$}} (m-i) + \sum_{\text{$k$ covers small $j$}} j.\]](http://latex.artofproblemsolving.com/3/0/f/30f7be081009e4876a5d58ba3a1726e1b3956620.png)
Let 
and 
be the values of the two sums on the RHS above, respectively. Then ![\[A \ge 0 + 0 + 1 + 1 + 2 + \dots + (\lfloor s/2 \rfloor - 1) = n^4/16 + O(n^3),\]](http://latex.artofproblemsolving.com/6/2/9/629f05b851b42aba3ffb8019c49b5dd6a908909e.png)
![\[B \ge 1 + 1 + 2 + 2 + 3 + \dots + \lfloor s/2 \rfloor = n^4/16 + O(n^3).\]](http://latex.artofproblemsolving.com/c/9/c/c9c3f6902fe3011cf6fa3b24b00aac053ff1d832.png)
On the other hand 
is the RHS of the earlier equation, so![\[A+B \le \sum_{\text{big $k$}} (m-k) = 0 + 1 + 2 + \dots + (\lceil m/2 \rceil - 1) = n^4/8 + O(n^3).\]](http://latex.artofproblemsolving.com/9/9/0/990ee76792a624f929e74978d98dd01aced02da5.png)
Note that the upper bound and lower bound for 
apart, while the quantity is 
.
or a big number below 
which is covered. Let a type 2 number be a small number below 
or a big number above 
which is not covered twice (i.e., is covered by zero or one big numbers). There may be at most 
type 1 and type 2 numbers, since otherwise we can increase the lower bound on 
and contradict the upper bound on 
small numbers 
which are covered twice. Moreover, at most ![$[m/2 + s/2 - n^{3/2}, m - s/2 + n^{3/2}]$](http://latex.artofproblemsolving.com/5/7/0/5702f59ea7c2f3efdda0cf32f7255d04b4ab6331.png)
(since this range has size 
small 
or ![$(m-s/2+n^{3/2}, m]$](http://latex.artofproblemsolving.com/b/1/5/b159d46e9766f68f8944e4bbfcd16563a62b0c32.png)
.
may not exist):
are big (since they cover 
must be small. If either 
or 
is in the range 
.
choices for 
type 1 or 2 numbers, so the contradiction arises because 
.
as the minimum term in row 
as the maximum term in row 
and massive if it is 
.
.
which implies that equality must hold and as such our claim must be true.
massive numbers in every row except possibly the bottom row.
which is not massive. Each massive number must therefore be above a tiny number. Since there is only one tiny number per row, there are at most two massive numbers per row.
massive numbers in the bottom row.
massive numbers in the bottom row.
,
must be at most 
.
such that 
and 
is the larger number below 
for 
.
a contradiction.
must be at most 46 rows from the bottom. Combining the above claims, there are at most 1010 massive numbers in the bottom row and 2 massive numbers in the next 45 rows, totalling 
massive numbers, a contradiction. 
with 
We will prove that 
which yields the conclusion.
,
is the number at top of 
are the lower and the most numbers respectively immediately below 
for 
![[asy]
unitsize(1 cm);
real x=1/sqrt(3); pair a1,a2,a3,b2,b3;
a1=(0,2); a2=(-x,1); b2=(x,1); a3=(-2x,0); b3=(0,0);
label("$a_1$",a1,N); label("$a_1+a_2$",a2,W); label("$a_2$",b2,E); label("$a_1+a_2+a_3$",a3,W); label("$a_3$",b3,E); label("$\dots$",(0,-1),S);
draw((a1--a2),EndArrow); draw((a1--b2),EndArrow); draw((a2--a3),EndArrow); draw((a2--b3),EndArrow); draw((a3--(-3x,-1)),EndArrow); draw((a3--(-x,-1)),EndArrow);
[/asy]](http://latex.artofproblemsolving.com/0/9/e/09e608d185c8aea9393f54fa87770fc2e0c86a05.png)
Observe that 
and since the most number in 
it follows that 
forms a permutation of set 
Thus in particular 
![[asy]
unitsize(1 cm);
real h=sqrt(3); pair a,b,c;
a=(-3,0); b=(3,0); c=(0,3*h);
draw (a--b--c--a); draw ((0.5,0)--(-1.25,1.75*h)); draw ((1.5,0)--(2.25,0.75*h));
label("$a_n$",(0.5,0),NE); label("$b_n$",(1.5,0),NW); label("$T_1$",(-1.25,1.75*h/2)); label("$T_2$",(2.25,0.75*h/2));
[/asy]](http://latex.artofproblemsolving.com/f/1/f/f1fe6fa803eacd669e90c5b5bcf494089f80da19.png)
Partition 
as on picture (they includes all numbers of bottom row except 
)
)
rows. But then it contains a sequence of numbers 
(which is arranged analogously to 
But if 
it does, and this is impossible since 
isn't included in 
so 
For even 
and for odd 
so done.
which yields the conclusion.
For even 
so the proof is over.
(strict) when 
when 
or 
rows?
,
.
,
,
,
,
,
(Sum is done modulo 
numbers, we will get even sum whereas 
is odd. Contradiction.
.
.
,![\[y_i = x_i - x_{i-1}\]](http://latex.artofproblemsolving.com/0/0/0/000f9ed403cf859b9e8a6cdd082ac882626a7f4b.png)
are the numbers in distinct cells of the triangular array. Since![\[x_{2018} = x_1 + y_2 + \cdots + y_{2018} \geq 1 + 2 + \cdots + 2018,\]](http://latex.artofproblemsolving.com/8/3/a/83ac01095f0e832adae36c2e79992c959947a95e.png)
.
are not included in either one. Let the left one have 
.
.
.
for 
,![\[a_A = a_1 + b_2 + \cdots + b_A \geq 2019 + 2020 + \cdots + (2018 + A) \geq 2019 + 2020 + \cdots + 3026 > N.\]](http://latex.artofproblemsolving.com/8/2/6/82608f3cc559eeb8c85b548779e487857f3a8cda.png)
Contradiction.
.
.
.
.
,
.
is at the top of the main chain, and the others are at most one element to the left or right of the main chain because they are equal to the differences between consecutive elements in the chain.
must in a black cell themself or directly adjacent to one.
anti pascal triangle that has a bottom row coinciding with the 
anti pascal triangle inside it that has a bottom row coinciding with the 
leftmost cells of the original triangle. Every number in this triangle is only adjacent to numbers in the 
)![\[ (2k+1) + (2k+2) + \cdots + (3k-1) = \frac{5k(k-1)}{2} \ge \frac{k}{2} \cdot (4k+2) =S, \]](http://latex.artofproblemsolving.com/a/2/4/a24c09ac48e7eb5f20d9edc6154b0e0ccbe33e97.png)
which means that the bottom of this chain must also be equal to 
,
as follows:
.
for each 
.
of 
and its two neighbors are included in neither triangle. Note that both triangles do not contain any of the numbers 
.
rows and 
rows. WLOG 
.
.
,![\begin{align*} \ell_{t_1} &= \ell_1+[\Delta \ell_2+\ldots+\Delta \ell_{t_1}] \\ &\ge (n+1)+(n+2)+\ldots+(n+t_1) \\ &\ge (n+1)+(n+2)+\ldots+(n+n/2-1) \\ &> 1+2+\ldots+n = N, \end{align*}](http://latex.artofproblemsolving.com/6/2/a/62a4565ac2bcf73438899e82cce0d8db630fea20.png)
a contradiction.
mohs I think. I claim the answer is no. Suppose otherwise. Note that if we have an 
,
or 
.
,
.
length equilateral triangles positioned on the bottom left and bottom right equilateral triangle. Then one of these triangles has no numbers between 
(the smallest 
,
.
.
then the if there is a number in the 
.
.
,
as desired.
then equality must hold for the relation![\[b_1+\dots+b_m\ge a_1+\dots+a_m\]](http://latex.artofproblemsolving.com/a/a/0/aa07f76abb1506bd491f4c496cc80b91354f7460.png)
That is, each row contains exactly one of the numbers 
.
it forms a ninja path (as described in IMO 2023/5), and any of the numbers ![\[2019+2020+\dots+3026>N\]](http://latex.artofproblemsolving.com/d/e/3/de3662b552292fa8fba80023c4b581c628a8dfad.png)
contradiction.
must be on or adjacent to the path.
rows inside the original triangle such that none of the numbers in the new triangle is adjacent to the path. Then we can similarly draw another path in the new triangle. The maximum element of the triangle is at least 
contradiction. Therefore, the answer is no.
and let 
immediately below, we draw an edge between 
then clearly any top-to-bottom sequence is increasing. Specifically we'll take the "longest" top-to-bottom sequence, which goes from the topmost vertex 
on the bottom row.
directly adjacent to the 
,![\[v_i+a_i=v_{i-1}\]](http://latex.artofproblemsolving.com/d/8/5/d857bfb7047f5a963e6d0146a19edac114726c93.png)
for 
.![\[\{v_1\}\cup \{a_2,\dots,a_n\}\]](http://latex.artofproblemsolving.com/5/4/d/54dc9a5e40a6206cd2c4cac195aa1b3e6aae6189.png)
and![\[\{1,2,\dots,n\}\]](http://latex.artofproblemsolving.com/6/b/5/6b5cfbc5e7968b0e98f4d8a205439f5c47dfbcd5.png)
are equivalent.
is "too big" most of the time and the rest of the cases are well-behaved. Specifically if 
then![\[1+N\le 1+\dots+(n-1)+v_1\le a_2+\dots+a_n+v_1=v_n\le N\]](http://latex.artofproblemsolving.com/2/2/9/2299915141ab40c0863050efc68b564e1f5ca20d.png)
a contradiction. If 
we repeat this process, except now![\[N=1+\dots+n\le a_2+\dots+a_n+v_1=v_n\le N\]](http://latex.artofproblemsolving.com/8/2/c/82cdb88c3d01dc0a4d917d21fe88e49f4ecba2f8.png)
so equality should hold giving the claim.
.
additions.![\[(n+1)+\dots+(n+1007)=2539654>2037171=N\]](http://latex.artofproblemsolving.com/b/4/6/b46ecfe8a9964880f4505d6bf9d5f9e10e982f40.png)
a contradiction.
fail, this is probably just stronger bounding than what is above. Above probably works for something like 
,
since the number at the bottom is the sum of 
t
so we are done.
.
The integers 
We add the purple numbers to the number up top to get the number at the bottom of the grapevine. This sum is at least 
,![\[ 2019 + 2020 + \dots + (2018 + 1008) = 2018 \cdot 1008 + \frac{1008 \cdot 1009}{2} = 1009(2016 + 504) = 1009(2520) > \frac{2018 \cdot 2019}{2}. \]](http://latex.artofproblemsolving.com/c/9/b/c9b5662be7d77485439ffab453b3e55570c9e9ae.png)
This is too big. Thus unfortunately we can make no wine, as desired.
