北京中考数学试卷模板
Author:
何睿涵
Last Updated:
3 days ago
License:
Creative Commons CC BY 4.0
Abstract:
基于GEEexam的北京中考数学试卷模板，应使用XeLaTeX编译，该模板较好地复刻了原卷方正书版的排版风格。
\begin
Discover why over 25 million people worldwide trust Overleaf with their work.
基于GEEexam的北京中考数学试卷模板，应使用XeLaTeX编译，该模板较好地复刻了原卷方正书版的排版风格。
\begin
Discover why over 25 million people worldwide trust Overleaf with their work.
\documentclass[11pt,space]{ctexart}
\usepackage{GEEexam}
%\usepackage{tikz}
\usepackage{tikz}
\usetikzlibrary{arrows,shapes,chains}
\usetikzlibrary{arrows.meta}
\usetikzlibrary{calc}
\usetikzlibrary{patterns,patterns.meta,snakes}
\usepackage{tkz-euclide}
\usetikzlibrary{tikzmark}
\usepackage{upgreek}
% 自定义线条和箭头
\tikzstyle{line} = [thick, line cap = round, line join = round] % 基本线条
\tikzstyle{dline} = [line, dashed] % 虚线
\tikzstyle{tline} = [line cap = round, line join = round] % 细线
\tikzstyle{arrow} = [->,>=Latex, line]
\usepackage{multirow}
\usepackage{graphicx}
\usepackage{subfig}
\usepackage{geometry}
\usepackage{calc}
\usetikzlibrary{calc}
%\newcommand{\gridbox}[1]{
%\begin{tikzpicture}[baseline=(base.base)]
% \node (base) at (0,0) {};
% \foreach \i in {1,...,#1}{
% \draw (\i-1,0) rectangle (\i,0.8);
% }
% \end{tikzpicture}
%}
\begin{document}
\zihao{5}
%\juemi
\biaoti{{\fangsong20XX年北京市初中学业水平考试}\vspace{0.2em}}
\fubiaoti{数~~~~~学~~~~~试~~~~~卷}\vspace{0.5em}
\hspace{0.5em}姓名\underline{\qquad\qquad\qquad\qquad\!\quad}\vspace{2em}~~准考证号\hspace{-0.5em}
\scalebox{1.3}{
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
\hspace{-0.1em} & \hspace{-0.1em} & \hspace{-0.1em}& \hspace{-0.1em} & \hspace{-0.1em} & \hspace{-0.1em} & \hspace{-0.1em} &\hspace{-0.1em} \\
\hline
\end{tabular}
}
\hspace{-0.1em}考场号\hspace{-0.5em}
\scalebox{1.3}{
\begin{tabular}{|c|c|c|}
\hline
\hspace{-0.1em} & \hspace{-0.1em} & \hspace{-0.1em} \\
\hline
\end{tabular}
}
\hspace{-0.1em}座位号\hspace{-0.5em}
\scalebox{1.3}{
\begin{tabular}{|c|c|}
\hline
\hspace{-0.1em} & \hspace{-0.1em} \\
\hline
\end{tabular}
}
\vspace{-1.7em}
% \noindent
\begin{tikzpicture}[baseline]
% ---------------------- 顶部填写区域 ----------------------
% 姓名横线
%\node[anchor=south west,font=\large] (name) at (0,6.2) {姓名 \rule{3cm}{0.4pt}};
% 准考证号+8格方格
%\node[anchor=south west,font=\large] at (name.east) {准考证号 \gridbox{8}};
% 考场号+3格方格
%\node[anchor=south west,font=\large,xshift=0.6cm] at (name.east) {考场号 \gridbox{3}};
% 座位号+2格方格
%\node[anchor=south west,font=\large,xshift=3.8cm] at (name.east) {座位号 \gridbox{2}};
% ---------------------- 外大边框与左右分栏 ----------------------
% 右侧须知文本框
\node[draw,inner sep=7.5pt,align=left,text width=15.1cm,line width=0.5mm] (textright) at (8.8,3) {
\hspace{1cm}1. 本试卷共6页，共两部分，三道大题，28道小题。满分100分。考试时间120分钟。
\hspace{1cm}2. 在试卷和草稿纸上准确填写姓名、准考证号、考场号和座位号。
\hspace{1cm}3. 试题答案一律填涂或书写在答题卡上，在试卷上作答无效。
\hspace{1cm}4. 在答题卡上，选择题、作图题用2B铅笔作答，其他试题用黑色字迹签字笔作答。
\hspace{1cm}5. 考试结束，将本试卷、答题卡和草稿纸一并交回。
};
% 左边界坐标（框内左侧分割线）
\coordinate (midline) at ($(textright.west)+(1cm,0)$);
% 绘制框内竖分隔线（在大边框内部）
\draw (midline |- textright.north) -- (midline |- textright.south);
% 左侧竖排文字「考生须知」，正向从上到下阅读
\node[font=\large\heiti,align=center,text width=1em,inner sep=0pt] at (midline) {\hspace{-1cm}考\vspace{0.25em}\\
\hspace{-1cm}生\vspace{0.25em}\\
\hspace{-1cm}须\vspace{0.25em}\\
\hspace{-1cm}知};
\end{tikzpicture}
\vspace{-1.5cm}
\begin{center}
\Large{\heiti 第一部分~~~~~~选择题}
\end{center}%\vspace{0.02em}
\section{选择题（共16分，每题2分）\vspace{-0.4em}}
\hspace{2.5em}{\heiti 第1－8题均有四个选项，符合题意的选项只有一个.}
\begin{enumerate}[itemsep=0.3em,topsep=0pt]
\item 下列图形中，\vspace{-1.1em}既是轴对称图形又是中心对称图形的是
\begin{center}
\hspace{-1em}\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-0.73,xscale=0.95]
%uncomment if require: \path (0,87); %set diagram left start at 0, and has height of 87
%Shape: Right Triangle [id:dp7117339769397362]
\draw (92.23,6.76) -- (162.87,74.86) -- (92.23,74.86) -- cycle ;
%Shape: Trapezoid [id:dp18963626447473114]
\draw (229.91,73.96) -- (247.63,14.89) -- (292.5,14.89) -- (310.22,73.96) -- cycle ;
%Shape: Regular Polygon [id:dp2232044176776058]
\draw (450.96,35.54) -- (438.21,74.58) -- (397.14,74.52) -- (384.5,35.45) -- (417.76,11.35) -- cycle ;
%Shape: Ellipse [id:dp8845669666353968]
\draw (518.97,39.81) .. controls (518.97,20.52) and (536.77,4.88) .. (558.73,4.88) .. controls (580.68,4.88) and (598.48,20.52) .. (598.48,39.81) .. controls (598.48,59.1) and (580.68,74.74) .. (558.73,74.74) .. controls (536.77,74.74) and (518.97,59.1) .. (518.97,39.81) -- cycle ;
\end{tikzpicture}
\end{center}\vspace{-1.7em}
\hspace{4.5em}$\mathrm{(~\!A~\!)}$\hspace{8.35em}$\mathrm{(~\!B~\!)}$\hspace{8.35em}$\mathrm{(~\!C~\!)}$\hspace{7.6em}$\mathrm{(~\!D~\!)}$\vspace{-0.5em}
\item 据《人民日报》报道，我国正深入开展电动汽车充电设施服务能力“三年倍增”行动，到2027年底将建成$2~800$万个充电设施，预计拉动投资$2~000$亿元以上. 将$28~000~000$用科学记数法表示为
$\mathrm{(~\!A~\!)}~2.8\times 10^6$\hfil$\mathrm{(~\!B~\!)}~2.8\times 10^7$\hfil$\mathrm{(~\!C~\!)}~0.28\times10^8$\hfil$\mathrm{(~\!D~\!)}~28\times10^6$
\item 实数$a,~b,~c$在数轴上的对应点的位置如图所示，下列结论正确的是\vspace{-1.5em}
\begin{center}
\begin{tikzpicture}
% 绘制坐标轴
\draw[arrow](-2.7,0)--(2.8,0)node[left,below]{$\ $};
%\draw[arrow](0,-0.6)--(0,4.6)node[right]{$y$};
%\node at (0,0)[below,xshift=-6]{$O$};
% 绘制刻度线
\foreach \x in {-2,-1, 0, 1, 2} {
\draw(\x,0)--(\x,0.1)node[below,outer sep=1pt]at(\x,0){$\x$};
}
\foreach \x in {-1.5} {
\draw(\x,0)--(\x,0.1)node[above,outer sep=1pt]at(\x,0.05){$a$};
}
\foreach \x in {1.3} {
\draw(\x,0)--(\x,0.1)node[above,outer sep=1pt]at(\x,0.05){$c$};
}
\foreach \x in {-0.4} {
\draw(\x,0)--(\x,0.1)node[above,outer sep=1pt]at(\x,0.05){$b$};
}
%\foreach \y in {1, 2, 3, 4} {
% \draw(0,\y)--(0.1,\y)node[left,outer sep=1pt]at(0,\y){$\y$};
% }
% 绘制函数图像
%\draw [line,color=blue,domain=-2:2, very thick] plot (\x,{(\x)^2});
\end{tikzpicture}
\end{center}\vspace{-1.5em}
$\mathrm{(~\!A~\!)}~b-a<0$\hfil$\mathrm{(~\!B~\!)}~b+c>2$\hfil$\mathrm{(~\!C~\!)}~c-a>2$\hfil$\mathrm{(~\!D~\!)}~|b|>|c|$
\item 在不透明的袋子中仅有3个小球，其中有2个白球、1个红球，这3个小球除颜色外无其他差别. 从袋子中随机摸出2个小球，这2个小球颜色不同的概率是\vspace{0.5em}
$\mathrm{(~\!A~\!)}~\dfrac14$\hfil$\mathrm{(~\!B~\!)}~\dfrac13$\hfil$\mathrm{(~\!C~\!)}~\dfrac12$\hfil$\mathrm{(~\!D~\!)}~\dfrac23$\vspace{0.5em}
\item 如图，在正五边形$ABCDE$中，$\angle~CAD$的大小为
$\mathrm{(~\!A~\!)}~30^\circ$\hspace{7.1em}$\mathrm{(~\!B~\!)}~36^\circ$
$\mathrm{(~\!C~\!)}~45^\circ$\hspace{7.1em}$\mathrm{(~\!D~\!)}~72^\circ$\vspace{-8.2em}
%\item 方程组$\begin{cases}
%3x-4y=3,\\
%7y-2x=11
%\end{cases}$\hspace{-0.5em}的解为\vspace{0.2em}
%$\mathrm{(~\!A~\!)}~\begin{cases}
%x=5,\\[-0.45em]
%y=3
%\end{cases}$\hfil$\mathrm{(~\!B~\!)}~\begin{cases}
% x=3,\\[-0.45em]
%y=-1
%\end{cases}$\hfil$\mathrm{(~\!C~\!)}~\begin{cases}
%x=-5,\\[-0.45em]
%y=3
%\end{cases}$\hfil$\mathrm{(~\!D~\!)}~\begin{cases}
%x=3,\\[-0.45em]
% y=1
%\end{cases}$
\begin{center}
\hspace{15em}\begin{tikzpicture}[x=0.35pt,y=0.35pt,yscale=-1,xscale=1]
%uncomment if require: \path (0,357); %set diagram left start at 0, and has height of 357
%Shape: Regular Polygon [id:dp07434668126275334]
\draw (311.15,248.08) -- (279.42,150.42) -- (362.5,90.07) -- (445.57,150.42) -- (413.84,248.08) -- cycle ;
%Straight Lines [id:da6919003780099726]
\draw (362.5,90.07) -- (413.84,248.08) ;
%Straight Lines [id:da2060900131247314]
\draw (362.5,90.07) -- (311.15,248.08) ;
% Text Node
\draw (354.53,57.52) node [anchor=north west][inner sep=0.75pt] {$A$};
% Text Node
\draw (449.99,143) node [anchor=north west][inner sep=0.75pt] {$B$};
% Text Node
\draw (410.84,250.51) node [anchor=north west][inner sep=0.75pt] {$C$};
% Text Node
\draw (295.16,250.51) node [anchor=north west][inner sep=0.75pt] {$D$};
% Text Node
\draw (249,144.14) node [anchor=north west][inner sep=0.75pt] {$E$};
\end{tikzpicture}
\end{center}\vspace{-2em}
\item 在平面直角坐标系$xOy$中，抛物线$y=x^2+m^2$与直线$y=(6-2m)x$无交点，则整数$m$的
$\mathrm{(~\!A~\!)}~$最小值为$-2$\hfil$\mathrm{(~\!B~\!)}~$最小值为$2$\hfil$\mathrm{(~\!C~\!)}~$最大值为$-2$\hfil$\mathrm{(~\!D~\!)}~$最大值为$2$
\newpage
\item 下面是“过圆外⼀点$P$作$\odot~O$的切线”的尺规作图方法.
{\kaishu $\qquad$如右图，连接$OP$，先作出线段$OP$的中点 $A$, 再以$A$为圆⼼,以$AO$为半径作$\odot ~A$,与 $\odot~O$交于两点$Q$和$R$,连接$PQ,~PR$,则$PQ,~PR$即为所求.} \vspace{-0.6cm}
\begin{center}
\hspace{10cm}\begin{tikzpicture}[x=0.35pt,y=0.35pt,yscale=-1,xscale=1]
%uncomment if require: \path (0,373); %set diagram left start at 0, and has height of 373
%Shape: Ellipse [id:dp8667389076871396]
\draw (319.81,188.46) .. controls (319.81,145.41) and (353.58,110.52) .. (395.24,110.52) .. controls (436.9,110.52) and (470.67,145.41) .. (470.67,188.46) .. controls (470.67,231.51) and (436.9,266.4) .. (395.24,266.4) .. controls (353.58,266.4) and (319.81,231.51) .. (319.81,188.46) -- cycle ;
%Straight Lines [id:da2880919752850841]
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%Shape: Ellipse [id:dp08100462421645072]
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%Straight Lines [id:da5481485126582861]
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%Straight Lines [id:da7027722625998901]
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%Shape: Circle [id:dp8836883479471054]
\draw [fill={rgb, 255:red, 0; green, 0; blue, 0 } ,fill opacity=1 ] (302.94,188.83) .. controls (302.79,188) and (303.34,187.33) .. (304.17,187.33) .. controls (304.99,187.33) and (305.79,188) .. (305.94,188.83) .. controls (306.09,189.66) and (305.54,190.33) .. (304.71,190.33) .. controls (303.88,190.33) and (303.09,189.66) .. (302.94,188.83) -- cycle ;
% Text Node
\draw (397.24,191.86) node [anchor=north west][inner sep=0.75pt] {$O$};
% Text Node
\draw (354.79,82.34) node [anchor=north west][inner sep=0.75pt] {$Q$};
% Text Node
\draw (354.79,263.96) node [anchor=north west][inner sep=0.75pt] {$R$};
% Text Node
\draw (182.7,191.27) node [anchor=north west][inner sep=0.75pt] {$P$};
% Text Node
\draw (289.17,192.53) node [anchor=north west][inner sep=0.75pt] {$A$};
\end{tikzpicture}
\end{center}\vspace{-2.8cm}
上述方法判定$PQ,~PR$与$\odot~O$相切的依据是\vspace{-0.1cm}
$\mathrm{(~\!A~\!)}~$直径所对的圆周角是直角\\
$\mathrm{(~\!B~\!)}~$角平分线上的点到角的两边距离相等\\
$\mathrm{(~\!C~\!)}~$到角的两边距离相等的点在角平分线上\\
$\mathrm{(~\!D~\!)}~$斜边和一条直角边对应相等的两个直角三角形全等
\item 已知函数$y=\dfrac{m}{x}~(m>0)$，其图象为曲线$W$. 下面给出四个结论：
\raisebox{-0.4ex}{\Large \ding{172}}~存在正方形$ABCD$，其四个顶点均在曲线$W$上；
\raisebox{-0.4ex}{\Large \ding{173}}~在曲线$W$上存在两点$A$和$B$，使得以$AB$为直径的圆经过原点$O$；
\raisebox{-0.4ex}{\Large \ding{174}}~若直线$y=kx+b$经过点$(\sqrt m,~\sqrt m)$，且与曲线$W$有且仅有一个交点，则$k=-1$；
\raisebox{-0.4ex}{\Large \ding{175}}~若$A(x_1,y_1),~B(x_2,y_2)$是曲线$W$上的两点，且$x_1x_2<0$，则$AB$长度的最小值为$2\sqrt{2m}$.
上述结论中，所有正确结论的序号是
$\mathrm{(~\!A~\!)}~$\raisebox{-0.4ex}{\Large \ding{175}}\hfil$\mathrm{(~\!B~\!)}~$\raisebox{-0.4ex}{\Large \ding{173}\ding{174}}\hfil$\mathrm{(~\!C~\!)}~$\raisebox{-0.4ex}{\Large \ding{174}\ding{175}}\hfil$\mathrm{(~\!D~\!)}~$\raisebox{-0.4ex}{\Large \ding{172}\ding{173}\ding{174}\ding{175}}
\end{enumerate}
\begin{center}
\Large{\heiti 第二部分~~~~~~非选择题}
\end{center}
\section{填空题（共16分，每题2分）}
\begin{enumerate}[itemsep=0.3em,topsep=0pt]
\setcounter{enumi}{8}
\item 若$\sqrt{2-x}$在实数范围内有意义，则实数$x$的取值范围是\underline{\qquad\qquad}.
\item 分解因式：$4x^3-4x^2-x+1=$\underline{\qquad\qquad}.
%\item 已知复数$z_1$，$z_2$满足不等式$|z-4+\mathrm i|\leqslant1$，下列说法错误的有
%\begin{tasks}(2)
%\task $|z_1+z_2|<2\sqrt{17}$
%\task $|z_1-z_2|<2$
%\task $|z_1|+|z_2|>2$
%\task $|z_1|-|z_2|<2$
%\end{tasks}
%\begin{minipage}[h][25ex][t]{.3\textwidth}
\item 方程$\dfrac{1}{2x-1}-\dfrac{2}{x+1}=0$的解为\underline{\qquad\qquad}.
%\end{minipage}\vspace{2.5cm}
%\end{enumerate}
%\section{填空题：本题共3小题，每小题5分，共15分。}
%\begin{enumerate}[itemsep=0.3em,topsep=0pt,resume]%\setcounter{enumi}{12}
\item 赤道式日晷（如左图）是古代经典的天文仪器，由底座、晷盘、晷针三部分组成. 右图反映了其工作方式，$\odot~O$是地球的示意图，其中$AB$表示赤道，已知赤道式日晷平稳放置于地面上，晷针与地轴平行，晷盘与赤道平行，晷针垂直穿过晷盘. 对于地球表面一点$P$，满足$\angle~AOP=37^\circ$，若小明想在$P$点处设一赤道式日晷，则该赤道式日晷的晷盘与底座所成角$\angle~\alpha=$\underline{\qquad\qquad}.\vspace{1em}
%\begin{figure}
%\centering
\hspace{7.5em} \includegraphics[width=0.28\linewidth]{T12.jpg}
%\caption{Enter Caption}
% \label{fig:placeholder}
%\end{figure}
\vspace{-11.3em}
\hspace{24em}\begin{tikzpicture}[x=0.36pt,y=0.36pt,yscale=-1,xscale=1]
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
%\draw (402.68,373.09) node [anchor=north west][inner sep=0.75pt] [align=left] {图2};
\end{tikzpicture}
\item 已知$A(x_1,y_1),~B(x_2,y_2)$分别是函数$y=\dfrac{4}{x}$与$y=-\dfrac4x$图象上的点，且$y_1=y_2$，则$x_1+x_2=$\underline{\qquad\qquad}.
\newpage
\item 若$(x_1,y_1),(x_2,y_2)$是函数$y=x^2$图象上的两点，则$\dfrac{x_1+x_2}{2}$\underline{\qquad\quad}$\sqrt{\dfrac{y_1+y_2}{2}}$（填“$>$”“$=$”或“$<$”）.
\item 如图，在矩形$ABCD$中，$E$是$AB$的中点，连接$DE$并延长
与射线$CB$交于点$F$，点$M,~N,~P$分别是线段$DE,~BE,~EF$
的中点，连接$MN,~NP$，若$AD=2$，$CD=2\sqrt2$，
则$\triangle MNP$的周长为\underline{\qquad\qquad}.
\vspace{-8.4em}
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
\draw (415.6,32.7) node [anchor=north west][inner sep=0.75pt] {$F$};
\end{tikzpicture}
\vspace{-1em}
\item 已知$4$行$4$列的数表$A$如下所示：\vspace{-1.1em}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
1 & 1 & 1 & 1 \\
\hline
1 & -1 & -1 & 1 \\
\hline
1 & -1 & -1 & 1 \\
\hline
1 & 1 & 1 & 1 \\
\hline
\end{tabular}
% \caption{Caption}
% \label{tab:placeholder}
\end{center}\vspace{-0.8em}
对数表的1次$T$变换是指任选数表的某一列（或某一行），并将该列（或该行）的每个数都变为原来的相反数. 数表$A$经过1次$T$变换得到数表$A_1$，数表$A_1$经过1次$T$变换得到数表$A_2$……以此类推得到数表$A_n$，称$A_n~(n\geqslant1)$是数表$A$的一个“$n$次相伴数表”.\vspace{0.5em}
（1）若数表~~%\centering
\begin{tabular}{|c|c|c|c|}
\hline
-1 & 1 & 1 & 1 \\
\hline
$p$ & 1 & 1 & $q$ \\
\hline
-1 & -1 & -1 & 1 \\
\hline
$r$ & 1 & 1 & 1 \\
\hline
\end{tabular}~~是数表$A$的一个“$2$次相伴数表”，则$p+q+r=$\underline{\qquad\qquad}；
% \caption{Caption}
% \label{tab:placeholder}
%\end{table}
\vspace{0.5em}
（2）下面给出$B_1,~B_2,~B_3$三个数表：\vspace{-0.6em}
\begin{center}
$B_1=$~\begin{tabular}{|c|c|c|c|}
\hline
1 & -1 & -1 & 1 \\
\hline
-1 & -1 & -1 & -1 \\
\hline
-1 & -1 & -1 & -1 \\
\hline
1 & -1 & -1 & 1 \\
\hline
\end{tabular}~~；~~
$B_2=$~\begin{tabular}{|c|c|c|c|}
\hline
-1 & 1 & 1 & -1 \\
\hline
1 & 1 & 1 & 1 \\
\hline
1 & -1 & 1 & 1 \\
\hline
-1 & 1 & 1 & -1 \\
\hline
\end{tabular}~~；~~
$B_3=$~\begin{tabular}{|c|c|c|c|}
\hline
-1 & 1 & -1 & -1 \\
\hline
1 & 1 & 1 & 1 \\
\hline
1 & -1 & 1 & 1 \\
\hline
-1 & 1 & 1 & -1 \\
\hline
\end{tabular}~~.
\end{center}\vspace{-0.6em}
\hspace{1.9em}其中不可能是数表$A$的“$n$次相伴数表”的为~\underline{\qquad\qquad}.
\end{enumerate}
\section{解答题（共68分，~第17－19题每题5分, ~第20题6分, ~第21题5分, ~第22题6分,~ 第23题5分, ~第24题6分, ~第25题5分, ~第26题6分, ~第27－28题每题7分）\\
\hspace{2.4em}解答应写出文字说明、演算步骤或证明过程.}
%\subsection{必考题：60分。}
\begin{enumerate}[itemsep=0.5em,topsep=5pt,resume]
\setcounter{enumi}{16}
%\newpage
\item 计算：$|-\sqrt{27}|-6\cos30^\circ+3\times(\tan45^\circ-\uppi)^0$.
\vspace{0cm}
\item 解不等式组：$\begin{cases}
2(x-3)<4x+1,\\[1em]
\dfrac{7x-1}{3}\geqslant 5x+2.
\end{cases}$%\vspace{0.1em}
%\newpage
%\vspace{-1.5cm}
%\newpage
\item 已知$2x-3y-2=0$，求代数式$\dfrac{2(x-2y)+y}{4x^2-12xy+9y^2}$的值.
%\vspace{1.5cm}
\newpage
%\vspace{1em}
\item 如图，在菱形$ABCD$中，$\angle~ADC=60^\circ$，线段$DC$绕点$D$顺时针
旋转$60^\circ$得到线段$DE$，连接$AC,~CE$. $H,~N$分别为$BC,~CE$的
中点，$AH$与$BD$交于点$M$，$MN$与$AC$交于点$P$.
%已知四边形$ACED$是平行四边形.
（1）求$\cos\angle~MNB$的值；
（2）求证：$P$为$MN$的中点.
\vspace{-10.5em}
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
\draw (368.63,190.33) node [anchor=north west][inner sep=0.75pt] {$P$};
\end{tikzpicture}
\vspace{-1em}
%\newpage
\item 某体育训练基地采购智能篮球、普通篮球两款用球. 拿出等额资金采购，买到的智能篮球数量仅\vspace{0.5em}为普通篮球数量的$\dfrac{2}{3}$；\vspace{0.5em}购买5个智能篮球、2个普通篮球，比购买2个智能篮球、6个普通篮球多花120元.求单个智能篮球的售价.\vspace{-0.8em}
%\vspace{1.5cm}
\item 割圆术是中国古代数学的经典成就，核心思想是通过计算正多边形的周长与其外接圆直径的比值，得到圆周率$\uppi$的近似值，且正多边形的边数越多，所得近似值越精确. 在半径为$1$的圆中，设圆内接正多边形的边数为$x$，周长为$y$，计算$x=3,4,\cdots,8$时对应$y$的取值，在下图中找到$(x,y)$对应各点，并用平滑曲线连接，得到曲线$C_1$.（本题中的$y$均精确到小数点后三位）
\vspace{-1em}
\begin{center}
\begin{tikzpicture}[x=0.52pt,y=0.52pt,yscale=-1,xscale=1]
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
\draw (420.79,580.75) node [anchor=north west][inner sep=0.75pt] {$8$};
% Text Node
\draw (460.22,580.75) node [anchor=north west][inner sep=0.75pt] {$9$};
% Text Node
\draw (495.00,580.75) node [anchor=north west][inner sep=0.75pt] {$10$};
% Text Node
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% Text Node
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% Text Node
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% Text Node
\draw (428.23,340.91) node [anchor=north west][inner sep=0.75pt] {$C_{1}$};
% Text Node
%\draw (438.04,293.12) node [anchor=north west][inner sep=0.75pt] {$C_{2}$};
\end{tikzpicture}
\end{center}\vspace{-1.5em}
%（1）若曲线$C_1$经过点$(8,y_8)$，则$y_8$
（1）曲线$C_1$经过点$(6,6),~(7,6.074)$，为证明$\uppi>3.01$，圆内接正多边形的边数至少为\underline{\qquad\qquad}；\vspace{-0.6em}
（2）当$x=2~027$时，试猜想$y$与$6.3$的大小关系，简要说明理由；\vspace{-0.6em}
（3）小融同学类比割圆术，想到计算圆外切正多边形与圆直径的比值，得到$\uppi$的近似值.在半径为\vspace{-0.6em}
\hspace{2em}$1$的圆中，利用计算机技术，小融得到了圆外切正多边形周长$y$与边数$x$的关系如下表：
\vspace{-1em}
\begin{center}
\centering
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
~~$x$~~ & ~3~ & ~4~ & ~5~ & ~6~ & ~7~ & ~8~ & ~$\cdots$~ \\
\hline
$y$ & 10.392 & 8.000 & 7.265 & 6.928 & 6.742 & 6.627 & $\cdots$ \\
\hline
\end{tabular}
% \caption{Caption}
% \label{tab:placeholder}
\end{center}\vspace{-1em}
\hspace{2em}根据表中数据，在上图中描出$(x,y)$对应的各点，并用平滑曲线连接，得到曲线$C_2$. 观察图象，\vspace{-2em}
\hspace{2em}回答问题：作半径为1的圆的内接、外切正$n$边形，周长分别记为$c_1,~c_2$，则$(c_1+c_2)$的值随\vspace{-0.5em}
\hspace{2em}$n$的增大而\underline{\qquad\qquad}（填“增大”或“减小”）.
\newpage
\item 在平面直角坐标系$xOy$中，已知函数$y=kx+b$的图象经过$(-1,2),~(3,-2)$两点.
（1）求该函数的解析式；
（2）已知$y=mx+n$的图象与$y=kx+b$的图象关于直线$x=\dfrac 12$对称，若当$x>1$时，对于$x$的每
\hspace{1.9em}一个值，函数$y=tx-1$的值大于函数$y=kx+b$的值，且小于函数$y=mx+n$的值，直接写
\hspace{1.9em}出实数$t$的取值范围~\underline{\qquad\qquad}.
\item 如图，在圆内接四边形$ABCD$中，$AC$与$BD$交于
点$E$，$BD$平分$\angle BAD=90^\circ$，$\angle BAC=\angle ADB$.
（1）过点$C$作$AD$的平行线交$AB$延长线于点$F$.
\hspace{1.9em}若$AC=AD,~BF=2$，求此圆的半径长；\vspace{0.5em}
（2）若$\sin\angle ABD~=\dfrac{\sqrt5}{3}$， $AD=4$，求$CE$的长.\vspace{-11.6em}
\begin{center}
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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\end{tikzpicture}
\end{center}
\item 小尹同学某年在某家饮品店购买饮品的种类及数量的情况如下：\vspace{-0.7em}
\begin{center}
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% Text Node
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% Text Node
\draw (41.15,201.12) node [anchor=north west][inner sep=0.75pt] {$15$};
% Text Node
\draw (41.72,171.07) node [anchor=north west][inner sep=0.75pt] {$20$};
% Text Node
\draw (41.93,142.57) node [anchor=north west][inner sep=0.75pt] {$25$};
% Text Node
\draw (42.06,112.71) node [anchor=north west][inner sep=0.75pt] {$30$};
% Text Node
\draw (0,82.3) node [anchor=north west][inner sep=0.75pt] [align=left] {数量/杯};
% Text Node
\draw (333.31,304.23) node [anchor=north west][inner sep=0.75pt] [align=left] {种类名称};
% Text Node
\draw (99.03,299.13) node [anchor=north west][inner sep=0.75pt] [align=left] {\hspace{-0.5em}牛乳茶};
% Text Node
\draw (224.78,298.74) node [anchor=north west][inner sep=0.75pt] [align=left] {\hspace{-0.5em}芒果味\\[-0.5em]\hspace{-0.5em}果茶};
% Text Node
\draw (162.23,299.15) node [anchor=north west][inner sep=0.75pt] [align=left] {\hspace{-0.5em}葡萄味\\[-0.5em]\hspace{-0.5em}果茶};
% Text Node
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% Text Node
\draw (517.11,144.67) node [anchor=north west][inner sep=0.75pt] [align=left] {牛乳茶};
% Text Node
\draw (519.64,168.18) node [anchor=north west][inner sep=0.75pt] {$37.5\%$};
% Text Node
\draw (593.12,193.91) node [anchor=north west][inner sep=0.75pt] {$31.25\%$};
% Text Node
\draw (512.09,219.35) node [anchor=north west][inner sep=0.75pt] [align=left] {\hspace{-0.3em}葡萄味\\[-0.5em]\hspace{-0.3em}果茶};
% Text Node
\draw (590.22,154.84) node [anchor=north west][inner sep=0.75pt] [align=left] {芒果味\\[-0.5em]果茶};
% Text Node
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% Text Node
\draw (613.44,246.25) node [anchor=north west][inner sep=0.75pt] {\hspace{-2em}$(~?~)~\%$};
% Text Node
%\draw (575.67,247.06) node [anchor=north west][inner sep=0.75pt] [align=left] {（？）};
% Text Node
\draw (49.15,286.63) node [anchor=north west][inner sep=0.75pt] {$0$};
\end{tikzpicture}
\end{center}\vspace{-1.5em}
根据以上信息，回答下列问题：
（1）根据左图，写出右图中“$(~?~)$”处缺失的数据\underline{\qquad\qquad}；
（2）为督促小尹健康饮食，小尹的妈妈设计了一种评分机制，已知小尹在一个月中购买饮品的次
\hspace{1.9em}数越少，得分越高.以下是小尹某年各月所得分数：\vspace{-0.7em}%初始分数为0分，在一天中，只要购
%\hspace{1.7em}买饮品则扣1分；若没有购买饮品则加1分. 每个月统计当月总分，若当月购买饮品的天数不
%\hspace{1.7em}大于5，则额外加3分（每月按30天计）.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
月份 & ~1 ~& ~2 ~& ~3 ~& ~4 ~& ~5 ~& ~6 ~& 7 & 8 & 9 & 10 & ~11~ & ~12~~\\
\hline
得分 & ~23~ & ~28~ & ~32~ & ~24~ & ~29~ & ~28 ~& ~30 ~& ~23 ~& ~28 ~& ~28~ & ~31~ & ~24~~\\
\hline
\end{tabular}
\end{center}\vspace{-0.7em}
\hspace{2em}\raisebox{-0.4ex}{\Large \ding{172}}~这一年小尹同学得分的众数为\underline{\qquad\qquad}，中位数为\underline{\qquad\qquad}；
\hspace{2em}\raisebox{-0.4ex}{\Large \ding{173}}~小尹同学每月生活费由她的妈妈根据上个月的得分情况按以下规则发放：若当月小尹同学
\hspace{3.3em}得分在24分及以下，则她的下个月生活费减半；若当月小尹同学得分在30分及以上，则
\hspace{3.3em}她的下个月生活费加50元. 已知每个月初始生活费均为200元，若这一年的后6个月所得
\hspace{3.3em}生活费的方差为$s_1^2$，全年12个月生活费的方差为$s_2^2$，则$s_1^2~\underline{\qquad\qquad}~s_2^2$（填“$>$”
“$=$”或
\hspace{3.3em}“$<$”）.%（用符号“$>$”、“$<$”
% \hspace{3.3em}或“$=$”作答）
%\item
%\newpage
%\item
\newpage
\item 在平面直角坐标系$xOy$中，\vspace{0.5em}已知点$A(-1,0),~B(3,0)$，抛物线$C:y=ax^2+bx+c~(a\neq0)$的顶点坐标为$(\dfrac{3}{2a},a-\dfrac{9}{4a})$.\vspace{0.5em}
（1）抛物线$C$能否同时经过$A,~B$两点？若能，求抛物线$C$的方程；若不能，请给出证明；
（2）若抛物线$C$与射线$AB$有且仅有一个公共点，求$a$的取值范围.
\item 在等腰梯形$ABCD$中，$AB<CD$，$AD=BC$，点$E$在$CD$上，$AE\perp CD$；点$F$在射线$CB$上，$AF\perp CB$.
（1）如图1，比较$\dfrac{AD}{CD}$与$\dfrac{AE}{AF}$的大小，并说明理由；
（2）如图2，点$P$在$EF$上，$M,~N$分别是$AD,~PE$的中点，若$\angle~APB=90^\circ$，求证：$EF\perp MN$.\vspace{-1em}
\begin{center}
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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% Text Node
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\end{tikzpicture}
\end{center}
\vspace{-1.5em}
%\newpage
\item
在平面直角坐标系$xOy$中，$\odot~T$是以点$T$为圆心，半径为$1$的圆. 点$Q(0,~2)$，$P$是平面内的点，点$P^\prime$在射线$QP$上，将线段$QP^\prime$绕点$Q$逆时针旋转角度$\alpha~(0^\circ<\alpha\leqslant180^\circ)$后$P^\prime$的对应点记为$P^{\prime\prime}$，对于点$P,~P^{\prime\prime}$和点$Q$给出如下定义：若$QP^\prime=n~\!QP~(n>0)$，且$P^{\prime\prime}$在$\odot~T$上，则称$P$是$P^{\prime\prime}$关于$Q$的“$n-\alpha$倍旋点”.
%设$x_1$，$x_2$是不大于$p$$\ (p\in\mathbf N,p\geqslant2)$的两个不同自然数.定义“$\Omega(n,p)$-数列”$\Omega{(n,p)}:x_1,x_2,\cdots,x_n$，其中$x_t(t=3,4,\cdots,n)$的值由以下方式确定：取$x_t$前面任意两项$x_i$，$x_j$\ $(i,j=1,2,\cdots,t-1\ \text{且}\ i\neq j)$，计算并比较$|x_i-x_j|$，取其中最小的值作为$x_t$.
%\begin{enumerate}[itemsep=-0.3em,label={（\arabic*）},topsep=0pt,labelsep=.5em,leftmargin=3em]
%\item 判断$2024,2025,1$是否是“$\Omega(3,2024)$-数列”？是否是“$\Omega(3,2025)$-数列”？
%\item 已知$2025,a,b,0$是“$\Omega(4,2025)$-数列”或“$\Omega(4,a)$-数列”，求$a$，$b$的值；
%\item 若$K:x_1,x_2,\cdots,x_{20}$是“$\Omega(20,1000)$-数列”，证明：数列$K$中至少有两项相同.
%\end{enumerate}
%\begin{enumerate}[itemsep=-0.3em,label={（\arabic*）},topsep=0pt,labelsep=.5em,leftmargin=3em]
%\item \hspace{-0.5em}求甲以$2:0$获得最终胜利的概率；
（1）如图，点$T$与原点$O$重合.\vspace{-1.5em}
\begin{center}
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\draw (327,233) node [anchor=north west][inner sep=0.75pt] {{\small $O~~(T)$}};
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\draw (478.07,236) node [anchor=north west][inner sep=0.75pt] {$x$};
% Text Node
\draw (332.43,77.8) node [anchor=north west][inner sep=0.75pt] {$y$};
% Text Node
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\end{tikzpicture}
\end{center}\vspace{-0.5em}
\hspace{2em}\raisebox{-0.4ex}{\Large \ding{172}}~在点$P_1(-\sqrt{3},2),P_2(\sqrt3,2),O(0,0)$中，点$\underline{\qquad\qquad}$是$P^{\prime\prime}$关于$Q$的“$\dfrac{\sqrt3}{3}-90^\circ$倍旋点”；
\hspace{2em}\raisebox{-0.4ex}{\Large \ding{173}}~若直线$y=-x+b$上存在$P^{\prime\prime}$关于$Q$的“$\dfrac{\sqrt3}{3}-90^\circ$倍旋点”，求$b$的取值范围；
（2）设点$T$在$x$轴上，点$W$在$\odot~T$上，$N$是平面内一点且满足$NT\leqslant\dfrac32$，当存在点$N$为$W$关于$Q$
\hspace{1.6em} 的“$2-60^\circ$倍旋点”时，直接写出$T$的横坐标$x_T$的取值范围.
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% 2026.07.09 - 2026.07.26