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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inequality => square
Rushil   12
N 7 minutes ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
+1 w
Rushil
Oct 7, 2005
ohiorizzler1434
7 minutes ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 24 minutes ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
24 minutes ago
H not needed
dchenmathcounts   44
N an hour ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
an hour ago
IZHO 2017 Functional equations
user01   51
N an hour ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
an hour ago
No more topics!
Maximum number of terms in the sequence
orl   10
N Jan 6, 2025 by de-Kirschbaum
Source: IMO LongList, Vietnam 1, IMO 1977, Day 1, Problem 2
In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.
10 replies
orl
Nov 12, 2005
de-Kirschbaum
Jan 6, 2025
Maximum number of terms in the sequence
G H J
Source: IMO LongList, Vietnam 1, IMO 1977, Day 1, Problem 2
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orl
3647 posts
#1 • 6 Y
Y by Adventure10, sohere, HWenslawski, megarnie, Mango247, OronSH
In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.
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enescu
741 posts
#2 • 10 Y
Y by Problem_Penetrator, MintTea, Adventure10, Adventure10, JG666, HWenslawski, megarnie, namanrobin08, kiyoras_2001, and 1 other user
This is in my Top 10 of all IMO problems. Not only because I was a competitor then, but also because I was present to the "problem solving session" organized by the hosts and I wached the Czech student Martin Chadek presenting his solution, which made the jury K.O. Here it is:

Let $x_1,x_2,\ldots$ be the given sequence and let $s_n=x_1+x_2+\ldots+x_n$.
The conditions from the hypothesis can be now written as $s_{n+7}<s_n$ and $s_{n+11}>s_n$ for all $n\ge 1$.
We then have:
$0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,$
a contradiction. Therefore, the sequence cannot have $17$ terms.
In order to show that $16$ is the answer, just take 16 real numbers satisfying
$s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6$.
We have $x_1=s_1$ and $x_n=s_n-s_{n-1}$ for $n\ge 2$. Thus we found all sequences with the given properties.

As far as I know, Martin only solved this problem and got a special prize for it :)

Other solutions are possible, including one using linear algebra. I'll post them if anyone is interested.
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Arne
3660 posts
#3 • 3 Y
Y by codyj, Adventure10, Mango247
Suppose a sequence of $17$ terms exists. Consider the matrix
\[\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} \ \ \ x_{11} \\ x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} \ \ \ x_{12} \\ x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} \ \ \ x_{13} \\ x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} \ \ \ x_{14} \\ x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} \ \ \ x_{15} \\ x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} \ \ \ x_{16} \\ x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} \ \ \ x_{17} \end{bmatrix}.\]
The sum of the numbers in each of the rows is positive. The sum of the numbers in each of the columns is negative. That is a contradiction. It is easy to show that $16$ works :)
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enescu
741 posts
#4 • 4 Y
Y by Adventure10, sohere, Mango247, and 1 other user
Arne wrote:
It is easy to show that 16 works :)
This is the jury solution. They provided the example
\[ 5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5. \]
But how do you find such an example?
That's why I loved Martin's solution :)
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pbornsztein
3004 posts
#5 • 1 Y
Y by Adventure10
More generally, it can be shown that :
If $p,q$ are any two distinct positive integers such that none of them divides the other, and if a sequence has the property that the sum of any $p$ consecutive terms is always positive and the sum of any $q$ consecutive terms is negative then the maximal length of this sequence is $p+q-d-1$ where $d = \gcd (p,q).$

The proof I know is quite long, so that I will not write it here. It can be found in :
Quadrature, n°54 (Oct. Dec. 2004), ex. E-218, p.34-36.

Pierre.
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enescu
741 posts
#6 • 2 Y
Y by Adventure10 and 1 other user
Well, this was proved during the competition by GB7, John Rickard, a gold medalist in 1977!
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mavropnevma
15142 posts
#7 • 2 Y
Y by Adventure10 and 1 other user
enescu wrote:
As far as I know, Martin only solved this problem and got a special prize for it :)
No, he also fully solved Problem 4.
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Blast_S1
352 posts
#8
Y by
Does this work or have I deluded myself? Not fully detailed.
We claim that the answer is $16$. One can find a construction without too much difficulty.

Call an $n$-sum the sum of $n$ consecutive terms. If we suppose, for contradiction, that the sequence has $17$ terms, then every $4$-sum can be written as an $11$-sum with a $7$-sum subtracted, and must hence be positive. Similarly, every $3$-sum can be written as a $7$-sum with a $4$-sum subtracted, and must therefore be negative. And, finally, every term can be written as a $4$-sum with a $3$-sum subtracted, and must therefore be positive again. However, if every term is positive, then it's impossible for every $7$-sum to be negative, and we've reached a contradiction.
This post has been edited 1 time. Last edited by Blast_S1, Mar 19, 2022, 2:08 AM
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PROA200
1748 posts
#9
Y by
enescu wrote:
Arne wrote:
It is easy to show that 16 works :)
This is the jury solution. They provided the example
\[ 5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5. \]But how do you find such an example?
That's why I loved Martin's solution :)

>16 years late, but I think you can motivate this solution by noticing that $\frac 27>\frac 3{11}$, meaning that if we can make it so that we get exactly two negative numbers of pretty large magnitude in each block of length $7$ and $3$ in each block of length $11$ we would be done by making the other numbers all equal to some positive number. It's not so obvious to pick the numbers $5$ and $-13$, but maybe this is easier?

\[20, 20, -51, 20, 20, 20, -51, 20, 20, -51, 20, 20, 20, -51, 20, 20.\]
Note that the block of $11$ reaches from the first $-51$ to right before the last $20$.
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dudade
136 posts
#10
Y by
fairly popular technique in competitive programming - prefix sums!

The answer is $16$. For $1 \leq |S| \leq 16$, we can consider any subset of $\{5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5\}$.

For $0 \leq k \leq 17$, then suppose $P_k = S_1 + S_2 + \cdots + S_k$. Then, $P_{k} < P_{k-7}$ and $P_{k} < P_{k + 11}$. Therefore,
\begin{align*}
P_{0} < P_{11} < P_4 < P_{15} < P_1 < P_{12} < P_{5} < P_{16} < P_2 < P_{13} < P_6 < P_{17} < P_3 < P_{14} < P_0,
\end{align*}which is clearly not possible. So, any sequence of $17$ real numbers has seven consecutive terms which sum to a negative value and has eleven consecutive terms which sum to a positive value. Thus, the maximum number of terms in the sequence is $\boxed{16}$.
This post has been edited 4 times. Last edited by dudade, Aug 24, 2024, 4:27 PM
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de-Kirschbaum
184 posts
#11
Y by
If $n \geq 17$ then take the first 17 numbers $a_1, \ldots, a_{17}$ and write them as follows.
$$\begin{bmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7\\
a_2 & \ddots & &  & &  & a_8 \\
\vdots & &\ddots &  & &  & \vdots\\
\vdots & & & \ddots & &\\
a_{11} & & & & & &a_{17}
\end{bmatrix}$$
Summing across the rows we note that the total sum is negative, while summing across the columns we get that the total sum is positive. That is a contradiction. Thus $n \leq 16$, and $n=16$ can be constructed (posted in other posts above).
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