高中奥数 2021-12-18

2021-12-18-01

（来源: 数学奥林匹克小丛书第二版高中卷复数与向量张思汇复数的模与幅角（二） P061 例6）

设、、是3个模不大于1的复数,、是方程的两个根.证明:对,都有

分析与解

由对称性,只需证明:

不妨设.令,由

得

因此,若,结论成立.

另一方面,由,,

又

所以

$\begin{aligned} \dfrac{1}{z-\omega_{1}}+\dfrac{1}{z-\omega_{2}}&=\dfrac{3\left(2z_{1}-\dfrac{2}{3}\left(z_{1}+z_{2}+z_{3}\right)\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)}\\ &=\dfrac{2\left(2z_{1}-z_{2}-z_{3}\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)}, \end{aligned}$

因此,当|时,结论成立.

下设,.

如图,考虑以、、为顶点的三角形.记和分别是三角形的边上的中线和高,则,.

图1

由于、,所以,,由此推出、都小于.

又因为,所以,即为锐角三角形.

所以,为单位圆内的锐角三角形.平移使、在单位圆周内(或圆周上),延长交单位圆于,则由得,所以.即外接圆半径,于是,矛盾!

因此这种情况不可能发生.

综上所述,原命题成立,证毕.

2021-12-18-02

（来源: 数学奥林匹克小丛书第二版高中卷复数与向量张思汇复数的模与幅角（二） P062 例7）

设是个实数,满足

求证:对平面上的任意个向量,存在的一个排列使得

分析与解

设,.我们只须证明

其中为的排列的集合不妨设

$\begin{gathered} \left|x_{n}-x_{1}\right|=\max _{1 \leqslant i<j \leqslant n}\left|x_{j}-x_{i}\right|=B, \\ \left|\alpha_{n}-\alpha_{1}\right|=\max _{1 \leqslant i<j \leqslant n}\left|\alpha_{j}-\alpha_{i}\right| . \end{gathered}$

考虑两个向量

$\begin{aligned} &\beta_{1}=x_{1} \alpha_{1}+x_{2} \alpha_{2}+\cdots+x_{n-1} \alpha_{n-1}+x_{n} \alpha_{n}, \\ &\beta_{2}=x_{n} \alpha_{1}+x_{2} \alpha_{2}+\cdots+x_{n-1} \alpha_{n-1}+x_{1} \alpha_{n}, \end{aligned}$

则

$\begin{aligned} & \max _{\left(k_{1}, k_{2}, \cdots, k_{n}\right) \in S_{n}}\left|\sum_{i=1}^{n} x_{k_{i}} a_{i}\right| \geqslant \max \left\{\left|\beta_{1}\right|,\left|\beta_{2}\right|\right\} \\ \geqslant & \frac{1}{2}\left(\left|\beta_{1}\right|+\left|\beta_{2}\right|\right) \geqslant \frac{1}{2}\left|\beta_{2}-\beta_{1}\right| \\ =& \frac{1}{2}\left|x_{1} \alpha_{n}+x_{n} \alpha_{1}-x_{1} \alpha_{1}-x_{n} \alpha_{n}\right| \\ =& \frac{1}{2}\left|x_{n}-x_{1}\right|\left|\alpha_{n}-\alpha_{1}\right| \\ =& \frac{1}{2} B\left|\alpha_{n}-\alpha_{1}\right| .(1) \end{aligned}$

设由三角形不等式易知.因此(1)中的不等式可写为

另一方面,考虑个向量

$\begin{gathered} \gamma_{1}=x_{1} \alpha_{1}+x_{2} \alpha_{2}+\cdots+x_{n-1} \alpha_{n-1}+x_{n} \alpha_{n} ,\\ \gamma_{2}=x_{2} \alpha_{1}+x_{3} \alpha_{2}+\cdots+x_{n} \alpha_{n-1}+x_{1} \alpha_{n} ,\\ \gamma_{3}=x_{3} \alpha_{1}+x_{4} \alpha_{2}+\cdots+x_{1} \alpha_{n-1}+x_{2} \alpha_{n} ,\\ \cdots \cdots \\ \gamma_{n}=x_{n} \alpha_{1}+x_{1} \alpha_{2}+\cdots+x_{n-2} \alpha_{n-1}+x_{n-1} \alpha_{n}. \end{gathered}$

则

$\begin{aligned} & \max _{\left(k_{1}, k_{2}, \cdots, k_{n}\right) \in S_{n}}\left|\sum_{i=1}^{n} x_{k_{i}} a_{i}\right| \\ \geqslant & \max _{1 \leqslant i \leqslant n}\left|\gamma_{i}\right| \\ \geqslant &\frac{1}{n}\left(\left|\gamma_{1}\right|+\left|\gamma_{2}\right|+\cdots+\left|\gamma_{n}\right|\right) \\ \geqslant & \frac{1}{n}\left|\gamma_{1}+\gamma_{2}+\cdots+\gamma_{n}\right|\\ =&\frac{A}{n}\left|\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\right| \\ =& \frac{A}{n}\left|n \alpha_{k}-\sum_{j \neq k}\left(\alpha_{k}-\alpha_{j}\right)\right| \\ \geqslant &\frac{A}{n}\left\{n\left|\alpha_{k}\right|-\sum_{j \neq k}\left|\alpha_{k}-\alpha_{j}\right|\right\} \\ \geqslant & \frac{A}{n}\left\{n\left|\alpha_{k}\right|-(n-1)\left|\alpha_{n}-\alpha_{1}\right|\right\}\\ =&\frac{A}{n}\left\{n\left|\alpha_{k}\right|-(n-1) x\left|\alpha_{k}\right|\right\} \\ =& A\left(1-\frac{n-1}{n} x\right)\left|\alpha_{k}\right| .(3) \end{aligned}$

结合(2)、(3),可得

$\begin{aligned} \max _{\left(k_{1}, k_{2}, \cdots, k_{n}\right) \in S_{n}}\left|\sum_{i=1}^{n} x_{k_{i}} \alpha_{i}\right| & \geqslant \max \left\{\dfrac{B x}{2}, A\left(1-\dfrac{n-1}{n} x\right)\right\}\left|\alpha_{k}\right| \\ & \geqslant \dfrac{\dfrac{B x}{2} \cdot A \cdot \dfrac{n-1}{n}+A\left(1-\dfrac{n-1}{n} x\right) \dfrac{B}{2}}{A \dfrac{n-1}{n}+\dfrac{B}{2}}\left|\alpha_{k}\right| \\ &=\dfrac{A B}{2 A+B-\dfrac{2 A}{n}}\left|\alpha_{k}\right| \\ &\geqslant \dfrac{A B}{2 A+B}\left|\alpha_{k}\right| . \end{aligned}$

证毕.

2021-12-18-03

（来源: 数学奥林匹克小丛书第二版高中卷复数与向量张思汇复数的模与幅角（二） P063 例8）

复系数多项式的个根为,且,求证:.

分析与解

引理1若正实数都不大于1(或都不小于1),则

引理的证明

因为

故引理1成立.

引理2

若(均为复系数多项式)满足

$\begin{gathered} f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}, \\ g(x)=b_{0} x^{k}+b_{1} x^{k-1}+\cdots+b_{k-1} x+b_{k},(n=k+l) \\ h(x)=c_{0} x^{l}+c_{1} x^{l-1}+\cdots+c_{l-1} x+c_{l}, \end{gathered}$

则

引理的证明

由已知可知(规定或时,;或时,).

考虑 $\begin{aligned} & \sum_{m^{\prime} \in \mathbf{Z}}\left|\sum_{i} b_{i} \overline{c_{m^{\prime}+i}}\right|^{2} \\ =& \sum_{m^{\prime} \in \mathbf{Z}}\left(\sum_{i} b_{i} \overline{c_{m^{\prime}+i}}\right)\left(\sum_{i} \overline{b_{j}} c_{m^{\prime}+j}\right) \\ =& \sum_{m^{\prime} \in \mathbf{Z}}\left(\sum_{i, j} b_{i} c_{m^{\prime}+j}\right) \overline{b_{j} c_{m^{\prime}+i}} \\ =& \sum_{m^{\prime} \in \mathbf{Z}, i, j} b_{i} c_{m^{\prime}+j} \overline{b_{j} c_{m^{\prime}+i}} \\ =& \sum_{(m-i-j), i, j} b_{i} c_{m-i} \overline{b_{j} c_{m-j}} \\ =& \sum_{m} \sum_{i, j} b_{i} c_{m-i} \overline{b_{j} c_{m-j}} \\ =& \sum_{m}\left(\sum_{i} b_{i} c_{m-i}\right)\left(\sum_{j} \overline{b_{j} c_{m-j}}\right) \\ =& \sum_{m}\left|a_{m}\right|^{2}\left(m \leqslant-1 \text { 或 } m \geqslant n+1 \text { 时 } a_{m}=0\right), \end{aligned}$

所以

$\begin{aligned} \sum_{m=0}^{n}\left|a_{m}\right|^{2} &=\sum_{m^{\prime} \in \mathbf{Z}}\left|\sum_{i} b_{i} \overline{c_{m^{\prime}+i}}\right|^{2} \\ & \geqslant\left|\sum_{i} b_{i} \overline{c_{i-k}}\right|^{2}+\left|\sum_{i} b_{i} \overline{c_{i+l}}\right|^{2} \\ &=\left|b_{k} \overline{c_{0}}\right|^{2}+\left|b_{0} \overline{c_{l}}\right|^{2} \\ &=\left|b_{k} c_{0}\right|^{2}+\left|b_{0} c_{l}\right|^{2}, \end{aligned}$

引理2得证.

下面看原命题:设中,的模小于1,的模不小于1.则由引理1可得

$\begin{aligned} \sum_{i=1}^{n}\left|z_{i}\right|^{2} &=\sum_{i=1}^{k}\left|z_{i}\right|^{2}+\sum_{j=k+1}^{n}\left|z_{j}\right|^{2} \\ & \leqslant k-1+\left|z_{1} z_{2} \cdots z_{k}\right|^{2}+(n-k-1)+\left|z_{k+1} z_{k+2} \cdots z_{n}\right|^{2}. \end{aligned}$

在引理2中令,,则有

$\begin{aligned} &\left|z_{1} z_{2} \cdots z_{k}\right|^{2}+\left|z_{k+1} z_{k+2} \cdots z_{n}\right|^{2} \\ \leqslant & 1^{2}+\left|a_{1}\right|^{2}+\cdots+\left|a_{n}\right|^{2} \leqslant 2 \end{aligned}$

所以,证毕.

高中奥数 2021-12-18

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