数列-清华北大金秋营试题

(2019 AMC 12B) Define a sequence recursively by $x_0=5$ and\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]for all nonnegative integers $n.$ Let $m$ be the least positive integer such that\[x_m\leq 4+\frac{1}{2^{20}}.\]In which of the following intervals does $m$ lie?

$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$

 

(2019 AMC 12A) A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \frac{3}{7}$, and\[a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}\]for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

 

 

\section{清华大学数学系2020年“大中衔接”研讨与教学活动}

\begin{center}
\textbf{2020年清华大学“大中衔接”试题}
\end{center}

1.设指数$\alpha_1,\alpha_2,\cdots,\alpha_n$两两不同,系数$a_1,a_2,\cdots,a_n$不全为零.证明:
$$f(x)=a_1x^{\alpha_1}+a_2x^{\alpha_2}+\cdots+a_nx^{\alpha_n}$$
在$(0,+\infty)$上至多有$n-1$个零点.

2.设$f:[-1,1]\to \mathbb{R}$是连续函数,证明:
$$\lim_{\lambda\to \infty}
\left[\int_{-1}^{1}\frac{|\lambda|}{\lambda^2+x^2}f(x)\mathrm{d}x\right]=\pi f(0).$$

3.设整数$n>1$,证明:至多只有有限多个正整数$a$,使得方程
$$x_1^2+x_2^2+\cdots+x_n^2=ax_1x_2\cdots x_n$$
有非零整数解.

%\let\oldwidering\widering
%\let\widering\undefined
%\usepackage{yhmath} %弧AB
%\let\widering\oldwidering

4.设$A$、$B$、$C$是单位球面$S$上三个不同点且都位于第一象限中,对于任何两个不同点$P,Q\in S$,只要$PQ$不是$S$的直径,则平面$OPQ$与$S$的交集是$S$上的一个圆周, $P$、$Q$将此圆分成两段弧,将其中较短的那段弧记为$\wideparen{PQ}$.设$\wideparen{BC},\wideparen{CA},\wideparen{AB}$的中点分别为$D$、$E$、$F$.证明: $\wideparen{AD},\wideparen{BE},\wideparen{CF}$经过同一个点.

5.设共有$k\geqslant 4$个不同的字母,可用它们构成单词.给定一族(记为$T$)禁用单词,其中任何两个禁用单词长度不等.称一个单词是“可用的”,如果它不含连续一段字母恰为某禁用单词.证明:至少有$\left( \frac{k+\sqrt{k^2-4k}}{2} \right) ^n$个长为$n$的可用单词.

6.设$n,p$是正整数,集合$A_1,A_2,\cdots,A_k$是$\{1,2,\cdots,n\}$的子集,满足对任何$i\neq j$都有$|A_i\backslash A_j|\geqslant p$.证明:
$$
k\leqslant \frac{\left( p-1 \right) !\cdot n!}{\left( \lfloor \frac{n+p-1}{2} \rfloor \right) !\cdot \left( \lceil \frac{n+p-1}{2} \rceil \right) !}.
$$

\section{2020年北京大学金秋营试题}

\begin{center}
\textbf{2020年北京大学金秋营}

\textbf{第一天}
\end{center}

1.对于非负实数$a_1,a_2,\cdots,a_n$,考虑如下$2^n$个实数
$$i_1a_1+i_2a_2+\cdots+i_na_n,$$其中$i_k=\pm 1,1\leqslant k\leqslant n$,记$S$为这$2^n$个数中所有正数之和,在$\sum_{i=1}^{n}a_n=1$的条件下,求$S$的最小值.

2.在$\triangle ABC$中, $D$为$BC$, $E,F$分别为$\wideparen{BC},\wideparen{BAC}$, 取$\triangle ABC$外接圆$\omega$, $\triangle ADE$外接圆与射线$AB,AC$交于点$J,K$, $\triangle ADF$外接圆与射线$AB,AC$交于点$L,M$,证明:若$AD$、$JK$、$LM$共点,则$JM$、$LK$交点在$\omega$上.

3.数列$\{a_n\}$满足: $a_0=2,a_1=5,a_{n+1}=5a_n-a_{n-1},n\geqslant 1$,已知$a_m\mid a_{2n-2}+ a_{2n-1}$,求证: $3\mid m,3\mid n$.

4.求$k$的最小值,使得将$7\times 7$方格挖去$k$个格后,剩余图形不存在$T$字形. ($T$字形指一个方格与其相邻的三个方格有公共边构成的图形)

\begin{center}
\textbf{第二天}
\end{center}

5. $\triangle ABC$内部取一点$O$,直线$AO,BO,CO$分别交对边于$D,E,F$,若四边形$AEFO,BDFO,CDEO$都有内切圆,求证: $O$在$\triangle ABC$内心和垂心所在直线上

6.若自然数$n$可以写成若干个自己的不同的因数的和,其中有个为$1$,就称$n$为好数,证明:对任意$m$大于1,存在无穷个$m$的正倍数为好数,且最小的倍数不大于$pm$,其中$p$是$m$最大的奇素因数(若$m$为二的幂,则$p$为$3$)

7.求证: $\sum_{r=0}^{p}\sum_{t=1}^{p}(-1)^{t-1}C_t^rC_p^tC_{p(p-t)}^{p(k-r)}\equiv C_{p^2}^{kp}\,(\bmod p^p)$ ,其中$p$为素奇数, $k\in [1,p]$.

8.求所有的$n$,使得平面上有$n$个完全相同的凸多边形,且满足对任意$k$个凸多边形,所有在它们之中且不在其余多边形中的点的集合为凸多边形(非退化).

 

(Austrian Regional Competition For Advanced Students 2019) Let $x,y$ be real numbers such that $(x+1)(y+2)=8.$ Prove that$$(xy-10)^2\ge 64.$$

$$16=(2x+2)(y+2)\le \frac{(2x+2+y+2)^2}{4}=\frac{(10-xy)^2}{4}$$

(Austrian Regional Competition For Advanced Students 2018) If $a, b$ are positive reals such that $a+b<2$. Prove that$$\frac{1}{1+a^2}+\frac{1}{1+b^2} \le \frac{2}{1+ab}$$and determine all $a, b$ yielding equality.

Proposed by Gottfried Perz

Because$$\frac{2}{1+ab}-\frac{1}{1+a^2}-\frac{1}{1+b^2}=\frac{(a-b)^2(1-ab)}{(1+ab)(1+a^2)(1+b^2)}\geq0.$$

(Austria-Poland 2004 system of equations) Solve the following system of equations in $\mathbb{R}$ where all square roots are non-negative:
$$
\begin{matrix}

a - \sqrt{1-b^2} + \sqrt{1-c^2} = d \\
b - \sqrt{1-c^2} + \sqrt{1-d^2} = a \\
c - \sqrt{1-d^2} + \sqrt{1-a^2} = b \\
d - \sqrt{1-a^2} + \sqrt{1-b^2} = c \\

\end{matrix}
$$

Apply the substitution $a=\sin\alpha$, ..., where $\alpha,...\in[-\pi/2,\pi/2]$.
Summing up (1) and (2) we obtain $\sin\beta-\cos\beta=\sin\delta-\cos\delta$. It follows that $\beta=\delta$ OR $\beta+\delta=-\pi/2$. The same for $\alpha$ and $\gamma$. Now it is routine to analyze 3 variants.

(2019 AMC 12B) Define a sequence recursively by $x_0=5$ and\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]for all nonnegative integers $n.$ Let $m$ be the least positive integer such that\[x_m\leq 4+\frac{1}{2^{20}}.\]In which of the following intervals does $m$ lie?

$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$

 

(2019 AMC 12A) A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \frac{3}{7}$, and\[a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}\]for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

(Austria-Poland 1997 Problem) Numbers $\frac{49}{1}, \frac{49}{2}, ... , \frac{49}{97}$ are writen on a blackboard. Each time, we can replace two numbers (like $a, b$) with $2ab-a-b+1$. After $96$ times doing that prenominate action, one number will be left on the board. Find all the possible values fot that number.

Cute; nice to find the invariant $\Pi = \Pi_n = \prod_{i=1}^n (2a_i -1)$, where $n$ is the number of the numbers written on the blackboard! Since $\Pi_{97} = 1$, it follows the last number will also be $\Pi_1 = 1$.
(This is obviously so, because $2(2ab - a - b + 1) - 1 = (2a-1)(2b-1)$).