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ATTII-solution-2.md

---
title: 第二届广州市中学生第二精神病院选拔测试解答 · 中
createTime: 2025/12/30
categories:
    - study
tags:
    - maths
---

## 5

### 5.1

联立得 $A\dfrac{P(t)}{R(t)} + B\dfrac{Q(t)}{R(t)} + C = 0$，整理得

$$
(A p_2 + B q_2 + C r_2) t^2 + (A p_1 + B q_1 + C r_1) t + (A p_0 + B q_0 + C r_0) = 0
$$

由两根存在，根据韦达定理，

$$
t_1 t_2 = \frac{A p_0 + B q_0 + C r_0}{A p_2 + B q_2 + C r_2}, \quad t_1 + t_2 = -\frac{A p_1 + B q_1 + C r_1}{A p_2 + B q_2 + C r_2}
$$

由 $\alpha_2 t_1 t_2 + \alpha_1 t_1 + \alpha_1 t_2 + \alpha_0 = 0$ 整理得

$$
A(\alpha_0 p_0 - \alpha_1 p_1 + \alpha_2 p_2) + B(\alpha_0 q_0 - \alpha_1 q_1 + \alpha_2 q_2) + C(\alpha_0 r_0 - \alpha_1 r_1 + \alpha_2 r_2) = 0
$$

即 $l$ 过定点 $\left( \dfrac{\alpha_2 p_0 - \alpha_1 p_1 + \alpha_2 p_2}{\alpha_2 r_0 - \alpha_1 r_1 + \alpha_2 r_2}, \quad \dfrac{\alpha_2 q_0 - \alpha_1 q_1 + \alpha_2 q_2}{\alpha_2 r_0 - \alpha_1 r_1 + \alpha_2 r_2} \right)$。

### 5.2

联立并整理得

$$
\sum_{i=0}^{n} (A p_i + B q_i + C r_i) t^i = 0
$$

令 $K_i = A p_i + B q_i + C r_i$，因为方程有 $n$ 个根，则方程可化为

$$
F(t) = t^n + \dfrac{K_{n-1}}{K_n} t^{n-1} + \dots + \dfrac{K_1}{K_n} t + \dfrac{K_0}{K_n}= 0
$$

设方程的根为 $t_1, t_2, \dots, t_n$，则方程可表示为

$$
F(t) = \prod_{i=1}^{t} (t-t_i) = \sum_{S \subseteq T} (-1)^{|S|} t^{n-|S|} \prod_{t \in S} t_i
$$

比对系数得

$$
\dfrac{K_{n-i}}{K_n} = (-1)^i \sum_{\substack{S\subseteq T \\ |S|=i}} \prod_{t\in S} t
$$

由

$$
\sum_{S \subseteq T} \alpha_{|S|} \prod_{t \in S} t
= \sum_{i=0}^{n} \alpha_{i} \sum_{\substack{S \subseteq T \\ |S| = i}} \prod_{t \in S} t
= \sum_{i=0}^{n} (-1)^i \alpha_i \dfrac{K_{n-i}}{K_n} = 0
$$

有

$$
\sum_{i=0}^{n} (-1)^i \alpha_i K_{n-i} = A \left( \sum_{i=0}^{n} (-1)^i \alpha_i p_{n-i} \right) + B \left( \sum_{i=0}^{n} (-1)^i \alpha_i q_{n-i} \right) + C \left( \sum_{i=0}^{n} (-1)^i \alpha_i r_{n-i} \right) = 0
$$

于是 $l$ 过定点 $\left( \dfrac{\sum_{i=0}^{n} (-1)^i \alpha_i p_{n-i}}{\sum_{i=0}^{n} (-1)^i \alpha_k r_{n-i}}, \ \dfrac{\sum_{i=0}^{n} (-1)^i \alpha_i q_{n-i}}{\sum_{i=0}^{n} (-1)^i \alpha_k r_{n-i}} \right)$。

## 6

### 6.1

在 $E$ 中选 $m$ 项改变，有 $\mathrm{C}_{n}^{m}$ 种方案，而每项有 $3$ 种改变结果，于是共有 $3^m \mathrm{C}_{n}^{m}$ 种。

### 6.2

注意到 $E \in B_r(F) \iff F \in B_r(E)$，

记 $[a \in S] = \begin{cases} 1,&a\in S \\ 0,&a\notin S\end{cases}$，则有

$$
\begin{align*}
    &\mathrel{\phantom{=}} \dfrac{1}{4^{nm}} \sum_{E_1}\sum_{E_2}\cdots\sum_{E_m} \bigl|B_r(E_1)\cap B_r(E_2) \cap\cdots\cap B_r(E_2) \bigr| \\
    &= \dfrac{1}{4^{nm}} \sum_{E_1}\sum_{E_2}\cdots\sum_{E_m}\sum_{F} \bigl[ F\in B_r(E_1)\cap B_r(E_2) \cap\cdots\cap B_r(E_2) \bigr] \\
    &= \dfrac{1}{4^{nm}} \sum_{F}\sum_{E_1}\sum_{E_2}\cdots\sum_{E_m} [E_1\in B_r(F)][E_2\in B_r(F)]\cdots[E_m\in B_r(F)] \\
    &= \dfrac{1}{4^{nm}} \sum_{F} \left(\sum_{E_1}[E_1\in B_r(F)] \right) \left(\sum_{E_2}[E_2\in B_r(H)] \right) \cdots \left(\sum_{E_m}[E_m\in B_r(F)] \right) \\
    &= \dfrac{1}{4^{nm}} 4^n b_r^m = 4^{n-nm} b_r^m
\end{align*}
$$

### 6.3

设 $[a=b] = \begin{cases} 1, &a=b \\ 0, &a\ne b \end{cases}$，记 $c_{i,j} = \left|\left\{ \displaystyle\sum_{k=1}^{m} [E_{k,i}=j] \right\}\right|, i=1,\dots, n,j\in S$，由

$$
\sum_{i=k}^{m} d(E_k, F) = \sum_{k=1}^{m} \sum_{i=1}^{n} [E_{k,i} = F_i] = \sum_{i=1}^{n} c_{n,F_i}
$$

与任意 $F_i$ 无关，可知 $\forall i,j,k, c_{i,j}=c_{i,k}$。

设 $m=4s$，在 $4s$ 个序列中分别选 $s$ 项各填上 $\texttt{A}, \texttt{T}, \texttt{G}, \texttt{C}$，共有 $\dfrac{(4s)!}{(s!)^4}$ 种填法。于是共有

$$
\begin{cases}
\left(\dfrac{(4s)!}{(s!)^4}\right)^n, & m=4s,s\in\mathbb{N} \\
0, &\text{otherwise}
\end{cases}
$$

种方案。