[/quote]
,
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
such that 
for all 
.
,
,
be positive real numbers such that 
.![$$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$](http://latex.artofproblemsolving.com/2/0/7/207934bca53047858a3820f37705171ff0abfd45.png)
and 
and 
Prove that![\[\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \geq \frac{3}{(abc)^2}.\]](http://latex.artofproblemsolving.com/0/a/b/0ab4b4e6c28a1ae67788525642661dbcd04d0bd8.png)
such that ![\[ f(x^2 + yf(z)) = xf(x) + zf(y) , \]](http://latex.artofproblemsolving.com/2/b/a/2baee1d8464992fc38a16561b48b18f4b35807bd.png)
for all 
.
and in 
is valid but 
is not valid.
be an acute triangle. Points 
,
lie on a line in this order and satisfy 
.
be the midpoints of 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
such for any 
.
satisfy,
Where 
}
be an integer. We are given a balance and 
weights of weight 
.
to satisfy that the equation 
has at least 1 roots 
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
.
of the equation![\[n^3=p^2-p-1\]](http://latex.artofproblemsolving.com/f/f/8/ff88093bb0aa93afcfdff3ae25149ec5fbacb61f.png)
is a prime number and 
?
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 in the bottom row (so 
)
be the number of big numbers which are not in the bottom row. So, these 
big numbers cover one big and one small number each.![\[\sum_{\text{big $k$ not in bottom row}} (m-k) = \sum_{\text{$k$ covers big $i$}} (m-i) + \sum_{\text{$k$ covers small $j$}} j.\]](http://latex.artofproblemsolving.com/3/0/f/30f7be081009e4876a5d58ba3a1726e1b3956620.png)
Let 
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 ![$[m/2 + s/2 - n^{3/2}, m - s/2 + n^{3/2}]$](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)
.
are big (since they cover 
must be small. If either 
.
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 
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 
.
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 
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.
