Find all integers $n\geq 2$ for which there exist the real numbers $a_k, 1\leq k \leq n$, which are satisfying the following conditions: \[\sum_{k=1}^n a_k=0, \sum_{k=1}^n a_k^2=1 \text{ and } \sqrt{n}\cdot \Bigr(\sum_{k=1}^n a_k^3\Bigr)=2(b\sqrt{n}-1), \text{ where } b=\max_{1\leq k\leq n} \{a_k\}.\]
Problem
Source: Romania
Tags: inequalities, number theory
mihaig
04.05.2019 07:54
This is indeed a premiere on AOPS. I never posted it here before and I'm sure no one else did. It's very difficult.
MathIsNicee
04.05.2019 10:48
mihaig wrote: Find all integers $n\geq 2$ for which there exist the real numbers $a_k, 1\leq k \leq n$, which are satisfying the following conditions: \[\sum_{k=1}^n a_k=0, \sum_{k=1}^n a_k^2=1 \text{ and } \sqrt{n}\cdot \Bigr(\sum_{k=1}^n a_k^3\Bigr)=2(b\sqrt{n}-1), \text{ where } b=\max_{1\leq k\leq n} \{a_k\}.\] Who is the author?
mihaig
04.05.2019 11:46
MathIsNicee wrote: mihaig wrote: Find all integers $n\geq 2$ for which there exist the real numbers $a_k, 1\leq k \leq n$, which are satisfying the following conditions: \[\sum_{k=1}^n a_k=0, \sum_{k=1}^n a_k^2=1 \text{ and } \sqrt{n}\cdot \Bigr(\sum_{k=1}^n a_k^3\Bigr)=2(b\sqrt{n}-1), \text{ where } b=\max_{1\leq k\leq n} \{a_k\}.\] Who is the author? Authors,actually. But this is not important. Important thing that this is a real problem,and it represented Romania. Not anyone represents Romania,even if it has the citizenship.
MathIsNicee
04.05.2019 16:09
Much talking, yet no solution. mihaig, do you have a solution to the proposed problem?
mihaig
04.05.2019 21:41
Hea! It's not polite,but I forgive you.Let me tell you a story. I once had a cute polar little birdie,not a genius though.But very greedy. It once saw a tuna fish. But tuna fish was greedy too. Conclusion: I don't have a birdie anymore. So please,do not spam my post. Come up with a proof. Hea!
MathIsNicee
05.05.2019 12:17
What is this hea? And what are all these anecdotes? Anyway, I think I have a solution to your problem. Not so difficult after all.
mihaig
06.05.2019 00:40
You don't deserve an hea! That is a greedyngs only for the few. Show your solution!
mihaig
08.05.2019 13:34
MathIsNicee wrote: What is this hea? And what are all these anecdotes? Anyway, I think I have a solution to your problem. Not so difficult after all. So where is the solution? Not Hea!
BlazingMuddy
09.05.2019 05:05
MathIsNicee wrote: What is this hea? And what are all these anecdotes? Anyway, I think I have a solution to your problem. Not so difficult after all. https://artofproblemsolving.com/community/c6h135914_zero_tolerance Please at least give some hint then Anyone could say something like IMO 2015 Shortlist G8 or A6 is not difficult (Don't say this ever please ._.). With that being said, I'd like to see your solution (and hopefully different from my solution). mihaig wrote: Authors,actually. But this is not important. Important thing that this is a real problem,and it represented Romania. Not anyone represents Romania,even if it has the citizenship. Do you mean that it's proposed by Romania? I think this is an appropriate question (?), so if you don't know, you should say so. Nice problem though.
$n$ satisfies the condition if and only if $n$ is even.
Working around the equalities, we get:
$$ \sum_{k = 1}^n \left( a_k + \frac{1}{\sqrt{n}} \right)^2 = \sum_{k = 1}^n \left( a_k^2 + \frac{2}{\sqrt{n}} a_k + \frac{1}{n} \right) = 2 $$
$$ \sum_{k = 1}^n a_k \left( a_k + \frac{1}{\sqrt{n}} \right)^2 = \sum_{k = 1}^n \left( a_k^3 + \frac{2}{\sqrt{n}} a_k^2 + \frac{1}{n} a_k \right) = 2\left(b - \frac{1}{n}\right) + \frac{2}{\sqrt{n}} = 2b = \sum_{k = 1}^n b\left( a_k + \frac{1}{\sqrt{n}} \right)^2 $$
But for each $k = 1, 2, \ldots, n$, due to definition of $b$, $a_k \left( a_k + \frac{1}{\sqrt{n}} \right)^2 \leq b\left( a_k + \frac{1}{\sqrt{n}} \right)^2$, with equality if and only if $a_k = b$ or $a_k = -\frac{1}{\sqrt{n}}$. For the second equality to hold, this equality must hold for all $k = 1, 2, \ldots, n$. Note that $b\geq 0$ as at least one of the $a_k$'s must be non-negative. Due to this, the $a_k$'s can only take two possible values; $-\frac{1}{\sqrt{n}}$ or a non-negative real number, say $x$ (which then is $b$).
Now, for the major step: since $a_k\in \{-\frac{1}{\sqrt{n}}, x\}$ for all $k = 1, 2, \ldots, n$ and $\sum_{k = 1}^n a_k = 0$, not all of $a_k$'s equal $-\frac{1}{\sqrt{n}}$. By easy argument with $\sum_{k = 1}^n a_k^2 = 1$, then we must have $x = \frac{1}{\sqrt{n}}$ (as $x\geq 0$). Back to $\sum_{k = 1}^n a_k = 0$, the number of $a_k$'s with value $-\frac{1}{\sqrt{n}}$ must be the same with those of value $\frac{1}{\sqrt{n}}$.
This implies that $n$ is even. Conversely, it can be easily observed that when $n$ is even, we can just pick $a_1 = a_2 = \ldots = a_{\frac{n}{2}} = -\frac{1}{\sqrt{n}}$ and the rest equal $\frac{1}{\sqrt{n}}$ (and the argument above actually shows that it and it's permutation are the only possible solutions in case $n$ is even).
mihaig
09.05.2019 05:14
Bravo!!!
mihaig
09.05.2019 14:55
2016 BMO Shortlist Country: Romania Authors: Leonard Giugiuc and Daniel Sitaru Author's solution (2016,February) Find all integers $n\geq 2$ for which there exist the real numbers $a_k, 1\leq k \leq n$, which are satisfying the following conditions: \[\sum_{k=1}^n a_k=0, \sum_{k=1}^n a_k^2=1 \text{ and } \sqrt{n}\cdot \Bigr(\sum_{k=1}^n a_k^3\Bigr)=2(b\sqrt{n}-1), \text{ where } b=\max_{1\leq k\leq n} \{a_k\}. \] We have: $\Bigr(a_k+\frac{1}{\sqrt{n}}\Bigr)^2(a_k-b)\leq 0\Rightarrow \Bigr(a_k^2+\frac{2}{\sqrt{n}}\cdot a_k+\frac{1}{n}\Bigr)(a_k-b)\leq 0\Rightarrow$ \[a_k^3\leq \Bigr(b-\frac{2}{\sqrt{n}}\Bigr)\cdot a_k^2+\Bigr(\frac{2b}{\sqrt{n}}-\frac{1}{n}\Bigr)\cdot a_k +\frac{b}{n} \forall \in \{1,2,\cdot, n\}. (k) \]Adding up the inequalities $(k)$ we get \[\sum_{k=1}^n a_k^3\leq \Biggr(b-\frac{2}{\sqrt{n}}\Bigr)\cdot \Biggl(\sum_{k=1}^n a_k^2\Biggl)+\Bigr(\frac{2b}{\sqrt{n}}-\frac{1}{n}\Bigr)\cdot \Biggl(\sum_{k=1}^n a_k\Biggl)+b\leftrightarrow \]\[\sum_{k=1}^n a_k^3\leq b-\frac{2}{\sqrt{n}}+b \leftrightarrow \sqrt{n}\cdot \Biggl(\sum_{k=1}^n a_k^3\Biggl)\leq 2(b\sqrt{n}-1). \]But according to hypothesis, \[\sqrt{n}\cdot \Biggl(\sum_{k=1}^n a_k^3\Biggl)=2(b\sqrt{n}-1). \]Hence is necessarily that \[a_k^3=\Biggl(b-\frac{2}{\sqrt{n}}\Biggl)\cdot a_k^2 +\Bigr(\frac{2b}{\sqrt{n}}-\frac{1}{n}\Bigr)\cdot a_k+\frac{b}{n} \forall k\in \{1,2,\cdots,n\} \leftrightarrow \]\[\Bigr(a_k+\frac{1}{\sqrt{n}}\Bigr)^2(a_k-b)=0 \forall k\in \{1,2,\cdots, n\} \leftrightarrow a_k \in \Bigr\{-\frac{1}{\sqrt{n}}, b\Bigr\} \forall k\in \{1,2,\cdots, n\} \]\[\text{We'll prove that } b>0. \text{ Indeed, if } b<0 \text{ then } 0=\sum_{k=1}^n a_k \leq nb<0, \text{ which is absurd.} \]\[\text{If } b=0, \text{ since } \sum_{k=1}^n a_k=0, \text{ then } a_k=0 \forall k\in \{1,2,\cdots, n\}\Rightarrow 1=\sum_{k=1}^n a_k^2=0, \text{ which is absurd}. \]In conclusion $b>0$. \[\text{If } a_k=-\frac{1}{\sqrt{n}} \forall k\in \{1,2,\cdots, n\} \text{ then } \sum_{k=1}^n a_k=-\sqrt{n}<0, \text{ which is absurd and similarly if} \]\[a_k=b \forall k\in \{1,2,\cdots, n\} \text{ then } \sum_{k=1}^n a_k=nb>0, \text{ which is absurd. Hence }\exists m\in \{1,2,\cdots, n-1\} \]such that among the numbers $a_k$ we have $n-m$ equal to $-\frac{1}{\sqrt{n}}$ and $m$ equal to $b$. We get $\begin{cases} -\frac{n-m}{\sqrt{n}}+mb=0\\ \frac{n-m}{n}+mb^2=1 \end{cases}$. From here, $b=\frac{n-m}{m\sqrt{n}}\Rightarrow \frac{n-m}{n}+\frac{(n-m)^2}{mn}=1\Rightarrow\\ \Rightarrow n-m=m\Rightarrow m=\frac{n}{2}$. Hence $n$ is even. Conversely, for any even integer $n\geq 2$ we get that there exist the real numbers $a_k, 1\leq k \leq n$, such that \[\sum_{k=1}^n a_k=0, \sum_{k=1}^n a_k^2=1 \text{ and } \sqrt{n}\cdot \Biggl(\sum_{k=1}^n a_k^3\Biggl)=2(b\sqrt{n}-1), \text{ where } b=\max_{1\leq k\leq n}\{a_k\} \](We may choose for example $a_1=\cdots =a_{\frac{n}{2}}=-\frac{1}{\sqrt{n}}$ and $a_{\frac{n}{2}+1}=\cdots = a_n=\frac{1}{\sqrt{n}}$).
mihaig
09.05.2019 15:43
BlazingMuddy wrote: MathIsNicee wrote: What is this hea? And what are all these anecdotes? Anyway, I think I have a solution to your problem. Not so difficult after all. https://artofproblemsolving.com/community/c6h135914_zero_tolerance Please at least give some hint then Anyone could say something like IMO 2015 Shortlist G8 or A6 is not difficult (Don't say this ever please ._.). With that being said, I'd like to see your solution (and hopefully different from my solution). mihaig wrote: Authors,actually. But this is not important. Important thing that this is a real problem,and it represented Romania. Not anyone represents Romania,even if it has the citizenship. Do you mean that it's proposed by Romania? I think this is an appropriate question (?), so if you don't know, you should say so. Nice problem though. @above are all your answers.
MathIsNicee
09.05.2019 16:18
BlazingMuddy wrote: MathIsNicee wrote: What is this hea? And what are all these anecdotes? Anyway, I think I have a solution to your problem. Not so difficult after all. https://artofproblemsolving.com/community/c6h135914_zero_tolerance Please at least give some hint then Anyone could say something like IMO 2015 Shortlist G8 or A6 is not difficult (Don't say this ever please ._.). With that being said, I'd like to see your solution (and hopefully different from my solution). mihaig wrote: Authors,actually. But this is not important. Important thing that this is a real problem,and it represented Romania. Not anyone represents Romania,even if it has the citizenship. Do you mean that it's proposed by Romania? I think this is an appropriate question (?), so if you don't know, you should say so. Nice problem though.
$n$ satisfies the condition if and only if $n$ is even.
Working around the equalities, we get:
$$ \sum_{k = 1}^n \left( a_k + \frac{1}{\sqrt{n}} \right)^2 = \sum_{k = 1}^n \left( a_k^2 + \frac{2}{\sqrt{n}} a_k + \frac{1}{n} \right) = 2 $$
$$ \sum_{k = 1}^n a_k \left( a_k + \frac{1}{\sqrt{n}} \right)^2 = \sum_{k = 1}^n \left( a_k^3 + \frac{2}{\sqrt{n}} a_k^2 + \frac{1}{n} a_k \right) = 2\left(b - \frac{1}{n}\right) + \frac{2}{\sqrt{n}} = 2b = \sum_{k = 1}^n b\left( a_k + \frac{1}{\sqrt{n}} \right)^2 $$
But for each $k = 1, 2, \ldots, n$, due to definition of $b$, $a_k \left( a_k + \frac{1}{\sqrt{n}} \right)^2 \leq b\left( a_k + \frac{1}{\sqrt{n}} \right)^2$, with equality if and only if $a_k = b$ or $a_k = -\frac{1}{\sqrt{n}}$. For the second equality to hold, this equality must hold for all $k = 1, 2, \ldots, n$. Note that $b\geq 0$ as at least one of the $a_k$'s must be non-negative. Due to this, the $a_k$'s can only take two possible values; $-\frac{1}{\sqrt{n}}$ or a non-negative real number, say $x$ (which then is $b$).
Now, for the major step: since $a_k\in \{-\frac{1}{\sqrt{n}}, x\}$ for all $k = 1, 2, \ldots, n$ and $\sum_{k = 1}^n a_k = 0$, not all of $a_k$'s equal $-\frac{1}{\sqrt{n}}$. By easy argument with $\sum_{k = 1}^n a_k^2 = 1$, then we must have $x = \frac{1}{\sqrt{n}}$ (as $x\geq 0$). Back to $\sum_{k = 1}^n a_k = 0$, the number of $a_k$'s with value $-\frac{1}{\sqrt{n}}$ must be the same with those of value $\frac{1}{\sqrt{n}}$.
This implies that $n$ is even. Conversely, it can be easily observed that when $n$ is even, we can just pick $a_1 = a_2 = \ldots = a_{\frac{n}{2}} = -\frac{1}{\sqrt{n}}$ and the rest equal $\frac{1}{\sqrt{n}}$ (and the argument above actually shows that it and it's permutation are the only possible solutions in case $n$ is even).
Finally a solution. Thank you!
mihaig
09.05.2019 16:39
Finally two solutions.
mihaig
20.03.2022 17:10
A new post, here https://artofproblemsolving.com/community/c6t243f6h2801403_find_all_n_meeting_conditions