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本帖最后由 hbghlyj 于 2023-3-31 17:46 编辑 罗马尼亚数学竞赛2004 NMO 预赛 - March 16th 2004
11th Grade
Problem 1. 设$x_0>0$. $\forall n\in\mathbb N,\;x_{n+1}=x_{n}+\frac{1}{\sqrt{x_{n}}}$. 求证$$\lim _{n \rightarrow \infty} \frac{x_n^{3/2}}{n}=\frac32$$
证明$x_{n+1}=x_{n}$且$x=x+\frac1{\sqrt x}$无解, 所以$x_n\to\infty$.$$\frac{x_{n+1}}{x_{n}}=1+\frac{1}{x_{n} \sqrt{x_{n}}}\implies \frac{x_{n+1}}{x_{n}}\to1$$
应用Stolz定理、二项式展开\begin{aligned} \lim _{n \rightarrow \infty} \frac{\sqrt{x_{n}^{3}}}{n} & =\lim _{n \rightarrow \infty}\left(\sqrt{x_{n+1}^{3}}-\sqrt{x_{n}^{3}}\right) \\ & =\lim _{y \rightarrow \infty}\left(\sqrt{\left(y+y^{\frac{-1}{2}}\right)^{3}}-\sqrt{y^{3}}\right)=\frac{3}{2}\end{aligned}
Problem 2. 非零复数 $z_1, z_2, \ldots, z_{2 n}, n \geqslant 3$, $\left|z_1\right|=\left|z_2\right|=\cdots=\left|z_{n+3}\right|$, 且 $\arg z_1 \geqslant \dots \geqslant \arg \left(z_{n+3}\right)$.
对 $i, j \in\{1,2, \ldots, n\}$定义: $b_{i j}=\left|z_i-z_{j+n}\right|$, 令 $B=\left(b_{i j}\right) \in \mathcal{M}_n$. 证明 $\det B=0$.
证明Consider $z=r(\cos 2 x+\mathrm{i} \sin 2 x)$ and $w=r(\cos 2 y+\mathrm{i} \sin 2 y)$, then $|z-w|=2 r\abs{\sin (x-y)}$. Using the hypothesis we can reduce $\operatorname{det} B$ to the form
$$
\left|\begin{array}{cccc}
\sin \left(x_1-x_{n+1}\right) & \sin \left(x_1-x_{n+2}\right) & \sin \left(x_1-x_{n+3}\right) & \ldots \\
\cdots & \ldots & \ldots & \ldots \\
\sin \left(x_n-x_{n+1}\right) & \sin \left(x_n-x_{n+2}\right) & \sin \left(x_n-x_{n+3}\right) & \ldots
\end{array}\right| .
$$
Writing it as a sum of determinants, we reach the desired conclusion.
Problem 3. 连续函数 $f:\mathbb R\to\mathbb R$ 满足$\forall a,b:f\left(\frac{a+b}{2}\right) \in\{f(a), f(b)\}$. 证明$f$为常数.
证明只需证明对于任何$c,d,c<d$有$f(c)=f(d)$就能证明$f$为常数. 事实上, 设 $m_1$ 为 $[c, d]$ 中点, $m_2,m_3$ 为 $\left[c, m_1\right],\left[m_1, d\right]$ 中点, 依此类推. 这些点在 $[c, d]$ 稠密, 因为 $f$ 连续, 它是常数.
Now, suppose there exist $a<b$ such that $f(a) \neq f(b)$, and let $a_1=\sup \{x \in[a, b] \mid f(a)=f(x)\}$ and $b_1=\inf \{x \in[a, b] \mid f(b)=f(x)\}$. Then $f(a)=f\left(a_1\right), f(b)=f\left(b_1\right)$, hence $f$ is constant on both intervals $\left[a, a_1\right]$ and $\left[b_1, b\right]$. But $f\left(\frac{a_1+b_1}{2}\right) \notin\left\{f\left(a_1\right), f\left(b_1\right)\right\}$, a contradiction.
Problem 4. $A=\left(a_{i j}\right) \in \mathcal{M}_{p}(\mathbb{C})$定义为$a_{12}=a_{23}=\dots=a_{p-1,p}=1$, 其余$a_{ij}=0$.
证明不存在非零的$B, C \in \mathcal{M}_p(\mathbb{C})$使得对任意正整数$n$$$\left(I_{p}+A\right)^{n}=B^{n}+C^{n}$$
证明From $(I_p+A)^2=B^2+C^2$ and $(I_p+A)^3=B^3+C^3$ we obtain $B C=C B=0_p$.
The equality $I_p+A=B+C$ implies $B+A B=B^2+B C=B^2$ and also $B+B A=B^2+C B=B^2$. Hence $A B=B A$ and, similarily, $A C=C A$. It is easy to see that it follows that there exist $b, c$ such that
$$
B=\left(\begin{array}{ccc}
b & & * \\
0 & b & \\
0 & & b
\end{array}\right), \quad C=\left(\begin{array}{ccc}
c & & * \\
0 & c & \\
0 & & c
\end{array}\right)
$$
with $b+c=1$. The equality $B C=0_p$ implies $b c=0$, so either $b$ or $c$ equals 0 . If, for instance, $b=0$, then it follows that $c=1$, hence $C$ is invertible. In this case, the equality $B C=0_p$ implies $B=0_p$, as desired.
12th Grade
Problem 1. Let $n \geqslant 2$ be an integer and $r \in\{1,2, \ldots, n\}$. Consider the set
$$
S_r=\left\{A \in \mathcal{M}_n\left(\mathbb{Z}_2\right) \mid \operatorname{rank} A=r\right\}
$$
a) Prove that for any $A \in S_n$ and $B \in S_r$, $A B$ is in $S_r$;
b) Calculate $\sum_{X \in S_r} X$.
Solution.a) $\operatorname{rank} A=n$ implies $A$ invertible. As $\operatorname{rank}A B \leqslant \operatorname{rank}(B)=r$ and $\operatorname{rank} A B \geqslant \operatorname{rank}\left(A^{-1} A B\right)=\operatorname{rank} B=r$, we get $\operatorname{rank}(A B)=r$.
b) Consider $A \in S_n$ and the function $f: S_r \rightarrow S_r, f(X)=A X$. It easy to see that $A$ is one-to-one (and thus onto) if and only if $f$ is one-to-one and onto (the set $S_r$ is finite).
We thus get $S=\sum_{X \in S_r} X=\sum_{X \in S_r} A X=A S$, which can be written $(A-I) S=0$.
In order to prove $S=0$, it suffices to show that there is an $A_n \in S_n$, such that $A_n-I$ is invertible. Take for example, for $n=2$ and $n=3$: $A_2=\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right)$, $A_3=\left(\begin{array}{lll}0 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 0\end{array}\right)$. In general consider matrices of the form
$$
A_{2 n}=\left(\begin{array}{cccc}
A_2 & 0 & \cdots & 0 \\
0 & A_2 & \cdots & 0 \\
\cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & A_2
\end{array}\right), \quad A_{2 n+1}=\left(\begin{array}{cccc}
A_3 & 0 & \cdots & 0 \\
0 & A_2 & \cdots & 0 \\
\cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & A_2
\end{array}\right)
$$
Problem 2. Prove that constant functions are the only continuous functions $f:[0,1] \rightarrow \mathbb{R}$, such that
$$
\int_0^1 f(x) g(x) \mathrm{d} x=\int_0^1 f(x) \mathrm{d} x \cdot \int_0^1 g(x) \mathrm{d} x
$$
for any non-derivable continuous function $g:[0,1] \rightarrow \mathbb{R}$.
Solution.If $f$ verifies the given property, then, for every real $c$, the relation is also verified by $f-c$. Thus, the function
$$
h(x)=f(x)-\int_0^1 f(t) \mathrm{d} t
$$
verifies the equality
(1) $\quad \int_0^1 h(x) g(x) \mathrm{d} x=0$, for all $g$ continuous and non-derivable.
If, by contradiction, $h$ is not the zero function, take $x_0 \in(0,1)$, such that $h\left(x_0\right) \neq 0$. Consider an interval of the form $V=\left[x_0-\varepsilon, x_0+\varepsilon\right] \subset(0,1)$ where $\varepsilon>0$, on which $h$ has constant sign. Such an interval exists by the intermediate value property. The function defined on $[0,1]$ by $g(x)=\left(x-x_0\right)^2-\varepsilon^2$ for $x \in V$ and $g(x)=0$ otherwise, is not derivable. The integral in (1) is not null, a contradiction.
Problem 3. A ring $A$ satisfies the following properties:
i) Its unit $1_A$ has order $p$, a prime number;
ii) There is a set $B \subset A$ with $p$ elements such that: for any $x, y \in A$, there is $b \in B$ satisfying $x y=b y x$.
Prove that $A$ is commutative.
Solution.As $k$ and $p$ are mutually prime, by Euclid theorem there are $a, b \in \mathbb{Z}$ such that $k a+p b=1$, that is $\left(k 1_A\right)\left(a 1_A\right)=1_A$. For $x=y=1_A$ we get $1_A=b \in B$
Suppose $A$ is non-commutative, that is, there are $x, y \in A, b \in B \backslash\left\{1_A\right\}$ such that $x y=b y x$. Thus $x+k 1_A$ and $y$ do not commute for $k=0,1, \ldots, p-1$. As $B \backslash\left\{1_A\right\}$ consists of $p-1$ elements, there are $s<r$ in $\{1,2, \ldots, p-1\}$ and $a \in B \backslash\left\{1_A\right\}$, such that $\left(x+r 1_A\right) y=a y\left(x+r 1_A\right)$ and $\left(x+s 1_A\right) y=a y\left(x+s 1_A\right)$. Denoting by $z=x+s 1_A$ and $t=r-s \in\{1,2, \ldots, p-1\}$ we get $z y=a y z$ and $\left(z+t 1_A\right) y=a y\left(z+t 1_A\right)$. This gives $y=a y$, or $z y=a y z=y z$. In conclusion $\left(x+s 1_A\right) y=y\left(x+s 1_A\right)$ implying $x y=y x$, a contradiction.
Problem 4. Let $a, b \in(0,1)$ and let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function, such that
$$
\int_0^x f(t) \mathrm{d} t=\int_0^{a x} f(t) \mathrm{d} t+\int_0^{b x} f(t) \mathrm{d} t, \text { for any } x \in[0,1]
$$
a) Prove that $a+b<1$ implies $f=0$.
b) Prove that $a+b=1$ implies that $f$ is constant.
Solution.a) As $f$ is continuous, the functions in $x$ defined by the given equality, have finite derivatives. Taking derivatives on both sides, we get
$$
f(x)=a f(a x)+b f(b x)
$$
for all $x \in[0,1]$. Consider $M=\sup _{t \in[0,1]}|f(t)|$. By the mean theorem, we have $|f(x)| \leqslant(a+b) M$, that is $M \leqslant(a+b) M$. Thus $M=0$ implying $f=0$.
b) Let $x_0$ be a maximum point of $f$. Then
$$
f\left(x_0\right)=a f\left(a x_0\right)+b f\left(b x_0\right) \leqslant(a+b) f\left(x_0\right)
$$
It follows that $a x_0$ is a maximum point too. By iteration, $a^n x_0$ is a maximum point, for any $n$. As $f$ is continuous, and $\lim _{n \rightarrow \infty} a^n x_0=0$, we infer that 0 is a maximum point. In thesame way, if $x_1$ is a minimum point for $f$, we get by induction that $a^n x_1$ is a minimum point, for any $n$. Considering limits, we conclude that 0 is a minimum point. As 0 is in the same time maximum and minimum point for $f$, we get $f(x)=f(0)$, for all $x \in[0,1]$, thus $f$ is constant. |
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