Find all positive integer $n$, such that there exists $n$ points $P_1,\ldots,P_n$ on the unit circle , satisfying the condition that for any point $M$ on the unit circle, $\sum_{i=1}^n MP_i^k$ is a fixed value for a) $k=2018$ b) $k=2019$.
Problem
Source: 2019 China TST Day 1 Q3
Tags: algebra
CatalinBordea
05.03.2019 11:52
It may help looking at this problems, which are equivalent: https://artofproblemsolving.com/community/q3h1575121p9691943 https://artofproblemsolving.com/community/c7h1574436
62861
06.03.2019 00:14
Can someone confirm if the problem is correct as stated? I am getting that no $n$ can even allow for $\sum MP_i^{2019}$ to be fixed. Perhaps this is supposed to be a two-part problem?
mzy
09.03.2019 21:24
This looks like a two-part problem. We shall solve for the odd and even cases. (a) $k=2m$.
We show all $n>m$ works.
Let $\text{arg}(M)=x, \text{arg}(P_i)=y_i$, and we deduce that $MP_i=| 2\sin(\frac{x-y_i}{2}) |$.
Note $|2\sin(x)|^{2m}=(e^{ix}+e^{-ix})^{2m}=\sum_{j=-m}^m a_j e^{2ijx}$, where $a_j$ are non-zero binomial coefficients.
Hence, $f(x)=\sum_i MP_i^{2m}=\sum_i (2\sin(\frac{x-y_i}{2}))^{2m}=\sum_i \sum_{j=-m}^m a_j e^{ij(x-y_i)}$.
Note that $1,e^{ix},e^{-ix},e^{2ix},e^{-2ix},\cdots$ form a linearly independent basis of $f(x)$. For $f(x)=\sum_j c_j e^{ijx}$ to be constant, it is equivalent to have $c_j=0$ for all nonzero $j$. Hence, for fixed $-m\le j\le m$ and $j\neq 0$, $\sum_i a_j e^{ij(x-y_i)}=a_j e^{ijx} \sum_i e^{-ijy_i}=0$ for all $x$. Letting $e^{-iy_i}=z_i$, it is clear that $\sum_i e^{-ijy_i}=\sum_i z_i^j=0$.
If $n>m$, we can construct $y_k=\frac{2\pi (k-1)}{n}$ from $1\le k\le n$, so $\sum_k e^{-ijy_k}=\frac{e^{-2\pi ij}-1}{e^{-\frac{2\pi ij}{n}}-1}=0$, since $|j|\le m<n$.
If $n\le m$, by the theory of symmetric polynomials, the set of power sum symmetric polynomials $p_k=\sum_i z_i^k$ of degrees $1\le k\le n$ in $n$ variables generates the ring of symmetric polynomials in $n$ variables. This implies that all symmetric polynomials $P(z_i)=0$. Now, $Q(z)=\prod(z-z_i)=z^n+P_1(z_i)z^{n-1}+\cdots +P_n(z_i)=z^n$ since $P_j(z_i)$ are all symmetric polynomials. Hence, all $z_i=0$, a contradiction.
(b) $k=2m+1$
We show no $n$ work.
WLOG, we can assume that $1,P_1,P_2,\cdots,P_n$ are arranged counterclockwise on the unit circle, with $P_n\neq 1$. We only consider $M$ on the arc between $P_n$ and $1$, so that $2\pi> x> y_i$ for all $i$.
In this domain, $MP_i=| 2\sin(\frac{x-y_i}{2}) |=2\sin(\frac{x-y_i}{2})$ since $0< \frac{x-y_i}{2}< \pi$.
As in part (a), we show that $(2\sin(x))^{2m+1}$ can be expressed as $\sum_{j=-m}^{m+1}a_je^{(2j-1)ix}$, for some non-zero binomial coefficients $a_j$.
Substituting, $f(x)=\sum_i MP_i^{2m+1}=\sum_i \sum_{j=-m}^{m+1} a_j e^{\frac{(2j-1)i(x-y_i)}{2}}$
Considering the basis of $e^{\frac{ix}{2}}$ when substituting $j=1$, we have $\sum_i a_1 e^{\frac{i(x-y_i)}{2}}=0$. However, each individual $\text{Im}(e^{\frac{i(x-y_i)}{2}})=\sin(\frac{(x-y_i)}{2})$ is positive, a contradiction.
bumjoooon
31.03.2020 09:39
mzy wrote: This looks like a two-part problem. We shall solve for the odd and even cases. Note that $1,e^{ix},e^{-ix},e^{2ix},e^{-2ix},\cdots$ form a linearly independent basis of $f(x)$. Maybe more elementary alternative for this part. (still using Vandermonde's determinant though) Assume for some $N>0$, $a_{-N}, ..., a_{N}$, $\sum_{-N}^{N} a_{k}e^{ik \theta} = 0$ for every $\theta$. Differentiating the equation $m$ times, we get $\sum_{-N}^{N} (ik)^m a_{k}e^{ik \theta} = 0$ for every $\theta$. Using the fact that Vandermonde's determinant, we get $a_{k}e^{ik \theta} = 0$ for every $\theta$ and thus $a_k = 0$. For part b), using the argument with $\theta$ on nice interval, we get the constant value should be zero, which is impossible. Therefore, none of $n$ satisfies the condition.
CANBANKAN
25.05.2022 09:42
The answer is (1) $N\ge 1010$ and (2) no $N$ works. Let $p_i=q_i^2$ 1) I claim $\sum_{i=1}^N MP_i^{2k}$ can be possibly constant as M varies around the unit circle iff $N>k$. We first rewrite this in polynomial condition. Let $x$ denote $M$ and $$\sum_{i=1}^{N} (|x-p_i|^2)^k$$ $$=\sum_{i=1}^N \left((x-p_i)(\frac 1x -\frac{1}{p_i}) \right)^k=\sum_{i=1}^N \left(2-\frac{x}{q_i^2}-\frac{q_i^2}{x} \right)^k = x^{-k} \sum_{i=1}^N \left(2x-(\frac{x}{q_i})^2-q_i^2 \right)^k$$ $$= x^{-k}\sum_{i=1}^N (\frac{x}{q_i}-q_i)^{2k}$$ We are given this is constant whenever $|x|=1$, and call this constant $c$. This means $P(x)=\sum\limits_{i=1}^N (\frac{x}{q_i}-q_i)^{2k}-cx^k$ vanishes whenever $|x|=1$, so it is the zero polynomial. By comparing the $x^j$ coefficient for $j\in \{0, \cdots, k-1\}$, we get $\sum_{i=1}^N q_i^{2t} = \sum_{i=1}^N p_i^t=0$ for all $1\le t\le k$ is equivalent. Let $Q(x)=\prod\limits_{i=1}^N (x-p_i)$. We can use the theorem of fundamental polynomials (symmetric sums to newton sums) to prove $[x^t]Q(x)=0$ for all $1\le t\le k$. This implies $N>k$. A construction is $p_t=e^{\frac{2\pi ti}{N}}$, which works since $Q(x)=x^N-1$ 2) I claim $\sum\limits_{j=1}^N MP_j^{2k+1}$ is not fixed for all integers $k$. Let $x=v^2$ Then $|v^2-q^2|^2=(v^2-q^2)(v^{-2}-q^{-2})=2-(q/v)^2-(v/q)^2=-(v/q-q/v)^2 $ Therefore, either $|v^2-q_i^2|^{2k+1} = i(\frac{q_i}{v}-\frac{v}{q_i})^{2k+1}$ or $|v^2-q_i^2|^{2k+1} = i(\frac{-q_i}{v}-\frac{v}{-q_i})^{2k+1}$ For a choice of signs of $q_i$, we have infinitely many $|v|=1$ satisfies $$i\sum\limits_{i=1}^N \left( \frac{q_i}{v} - \frac{v}{q_i} \right)^{2k+1}=iv^{-(2k+1)} \sum\limits_{i=1}^N \left(q_i-\frac{v^2}{q_i} \right)^{2k+1}=c$$where $c$ is a constant independent of $v$. This means $P(v)=i\sum\limits_{i=1}^N \left(q_i-\frac{v^2}{q_i} \right)^{2k+1}-cv^{2k+1}$ is $0$ at infinitely many points where $|v|=1$. This implies $P(v)\equiv 0$. Notice $c=[v^{2k+1}]P(v)=0$, which contradicts the fact that the sum of a bunch of nonzero distances is positive.
Joider
05.08.2023 17:59
mzy wrote: This looks like a two-part problem. We shall solve for the odd and even cases. (a) $k=2m$.
We show all $n>m$ works.
Let $\text{arg}(M)=x, \text{arg}(P_i)=y_i$, and we deduce that $MP_i=| 2\sin(\frac{x-y_i}{2}) |$.
Note $|2\sin(x)|^{2m}=(e^{ix}+e^{-ix})^{2m}=\sum_{j=-m}^m a_j e^{2ijx}$, where $a_j$ are non-zero binomial coefficients.
Hence, $f(x)=\sum_i MP_i^{2m}=\sum_i (2\sin(\frac{x-y_i}{2}))^{2m}=\sum_i \sum_{j=-m}^m a_j e^{ij(x-y_i)}$.
Note that $1,e^{ix},e^{-ix},e^{2ix},e^{-2ix},\cdots$ form a linearly independent basis of $f(x)$. For $f(x)=\sum_j c_j e^{ijx}$ to be constant, it is equivalent to have $c_j=0$ for all nonzero $j$. Hence, for fixed $-m\le j\le m$ and $j\neq 0$, $\sum_i a_j e^{ij(x-y_i)}=a_j e^{ijx} \sum_i e^{-ijy_i}=0$ for all $x$. Letting $e^{-iy_i}=z_i$, it is clear that $\sum_i e^{-ijy_i}=\sum_i z_i^j=0$.
If $n>m$, we can construct $y_k=\frac{2\pi (k-1)}{n}$ from $1\le k\le n$, so $\sum_k e^{-ijy_k}=\frac{e^{-2\pi ij}-1}{e^{-\frac{2\pi ij}{n}}-1}=0$, since $|j|\le m<n$.
If $n\le m$, by the theory of symmetric polynomials, the set of power sum symmetric polynomials $p_k=\sum_i z_i^k$ of degrees $1\le k\le n$ in $n$ variables generates the ring of symmetric polynomials in $n$ variables. This implies that all symmetric polynomials $P(z_i)=0$. Now, $Q(z)=\prod(z-z_i)=z^n+P_1(z_i)z^{n-1}+\cdots +P_n(z_i)=z^n$ since $P_j(z_i)$ are all symmetric polynomials. Hence, all $z_i=0$, a contradiction.
(b) $k=2m+1$
We show no $n$ work.
WLOG, we can assume that $1,P_1,P_2,\cdots,P_n$ are arranged counterclockwise on the unit circle, with $P_n\neq 1$. We only consider $M$ on the arc between $P_n$ and $1$, so that $2\pi> x> y_i$ for all $i$.
In this domain, $MP_i=| 2\sin(\frac{x-y_i}{2}) |=2\sin(\frac{x-y_i}{2})$ since $0< \frac{x-y_i}{2}< \pi$.
As in part (a), we show that $(2\sin(x))^{2m+1}$ can be expressed as $\sum_{j=-m}^{m+1}a_je^{(2j-1)ix}$, for some non-zero binomial coefficients $a_j$.
Substituting, $f(x)=\sum_i MP_i^{2m+1}=\sum_i \sum_{j=-m}^{m+1} a_j e^{\frac{(2j-1)i(x-y_i)}{2}}$
Considering the basis of $e^{\frac{ix}{2}}$ when substituting $j=1$, we have $\sum_i a_1 e^{\frac{i(x-y_i)}{2}}=0$. However, each individual $\text{Im}(e^{\frac{i(x-y_i)}{2}})=\sin(\frac{(x-y_i)}{2})$ is positive, a contradiction.
Nice but how $|2\sin(x)|^{2m}=(e^{ix}+e^{-ix})^{2m}$.... Correct me if i wrong $e^{ix} +e^{-ix} = 2cos(x)$