修改12610,12627,12631题面
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@ -1,7 +1,7 @@
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import os,re,json
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"""这里编辑题号(列表)后将在vscode中打开窗口, 编辑后保存关闭"""
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problems = "13506"
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editor = "朱敏慧"
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problems = "12610,12627,12631"
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editor = "王伟叶"
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def generate_number_set(string,dict):
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string = re.sub(r"[\n\s]","",string)
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@ -332212,7 +332212,7 @@
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},
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"012610": {
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"id": "012610",
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"content": "如图, ``复兴''桥为人行天桥, 其主体结构是由两根等长的半圆型主梁和四根坚直的立柱吊起一块圆环状的桥面. 主梁在桥面上方相交于点$S$且它们所在的平面互相垂直, $S$在桥面上的射影为桥面的中心$O$. 主梁连接桥面大圆, 立柱连接主梁和桥面小圆, 地面有$4$条可以通往桥面的上行步道. 设$CD$为其中的一根立柱, $A$为主梁与桥面大圆的连接点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [dashed] (-2,-1,0) -- (2,-1,0) (0,-1,-2) -- (0,-1,2);\n\\draw (-2,-1,-0.3) -- (-0.3,-1,-0.3) -- (-0.3,-1,-2);\n\\draw (-2,-1,0.3) -- (-0.3,-1,0.3) -- (-0.3,-1,2);\n\\draw (2,-1,-0.3) -- (0.3,-1,-0.3) -- (0.3,-1,-2);\n\\draw (2,-1,0.3) -- (0.3,-1,0.3) -- (0.3,-1,2);\n\\fill [domain = 0:360, white] plot ({2*cos(\\x)},0,{2*sin(\\x)});\n\\fill [domain = 0:360, pattern = north east lines] plot ({2*cos(\\x)},0,{2*sin(\\x)});\n\\fill [domain = 0:360, white] plot ({cos(\\x)},0,{sin(\\x)});\n\\draw [domain = 0:360,ultra thick,samples = 100] plot ({2*cos(\\x)},0,{2*sin(\\x)});\n\\draw [domain = 0:360,thick] plot ({cos(\\x)},0,{sin(\\x)});\n\\filldraw (0,0) circle (0.03) node [left] {$O$} coordinate (O);\n\\draw [domain = 0:180,thick] plot ({-sqrt(2)*cos(\\x)},{2*sin(\\x)},{sqrt(2)*cos(\\x)});\n\\draw [domain = 0:180,thick] plot ({sqrt(2)*cos(\\x)},{2*sin(\\x)},{sqrt(2)*cos(\\x)});\n\\draw (0,2,0) node [above] {$S$} coordinate (S);\n\\draw ({cos(45)},0,{sin(45)}) --++ (0,{sqrt(3)},0);\n\\draw ({cos(135)},0,{sin(135)}) node [right] {$D$} coordinate (D) --++ (0,{sqrt(3)},0) node [above left] {$C$} coordinate (C);\n\\draw ({cos(225)},0,{sin(225)}) --++ (0,{sqrt(3)},0);\n\\draw ({cos(315)},0,{sin(315)}) --++ (0,{sqrt(3)},0);\n\\draw [thick] ({2*cos(45)},0,{2*sin(45)}) node [below left] {$A$} coordinate (A) -- ({3*cos(45)},-1,{3*sin(45)}) node [below] {$B$} coordinate (B);\n\\draw [thick] ({2*cos(45)},0,{2*sin(45)}) node [below left] {$A$} coordinate (A) -- ({3*cos(45)},-1,{3*sin(45)}) node [below] {$B$} coordinate (B);\n\\draw [thick] ({2*cos(135)},0,{2*sin(135)}) -- ({3*cos(135)},-1,{3*sin(135)});\n\\draw [thick] ({2*cos(315)},0,{2*sin(315)}) -- ({3*cos(315)},-1,{3*sin(315)});\n\\draw [thick,dashed] ({2*cos(225)},0,{2*sin(225)}) -- ({3*cos(225)},-1,{3*sin(225)});\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $CD\\parallel$平面$SOA$;\\\\\n(2) 设$AB$为经过$A$的一条步道, 其长度为$12$米且与地面所成角的大小为$30^{\\circ}$. 桥面小圆与大圆的半径之比为$4: 5$, 当桥面大圆半径为$20$米时, 求点$C$到地面的距离.",
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"content": "如图, ``复兴''桥为人行天桥, 其主体结构是由两根等长的半圆型主梁和四根竖直的立柱吊起一块圆环状的桥面. 主梁在桥面上方相交于点$S$且它们所在的平面互相垂直, $S$在桥面上的射影为桥面的中心$O$. 主梁连接桥面大圆, 立柱连接主梁和桥面小圆, 地面有$4$条可以通往桥面的上行步道. 设$CD$为其中的一根立柱, $A$为主梁与桥面大圆的连接点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [dashed] (-2,-1,0) -- (2,-1,0) (0,-1,-2) -- (0,-1,2);\n\\draw (-2,-1,-0.3) -- (-0.3,-1,-0.3) -- (-0.3,-1,-2);\n\\draw (-2,-1,0.3) -- (-0.3,-1,0.3) -- (-0.3,-1,2);\n\\draw (2,-1,-0.3) -- (0.3,-1,-0.3) -- (0.3,-1,-2);\n\\draw (2,-1,0.3) -- (0.3,-1,0.3) -- (0.3,-1,2);\n\\fill [domain = 0:360, white] plot ({2*cos(\\x)},0,{2*sin(\\x)});\n\\fill [domain = 0:360, pattern = north east lines] plot ({2*cos(\\x)},0,{2*sin(\\x)});\n\\fill [domain = 0:360, white] plot ({cos(\\x)},0,{sin(\\x)});\n\\draw [domain = 0:360,ultra thick,samples = 100] plot ({2*cos(\\x)},0,{2*sin(\\x)});\n\\draw [domain = 0:360,thick] plot ({cos(\\x)},0,{sin(\\x)});\n\\filldraw (0,0) circle (0.03) node [left] {$O$} coordinate (O);\n\\draw [domain = 0:180,thick] plot ({-sqrt(2)*cos(\\x)},{2*sin(\\x)},{sqrt(2)*cos(\\x)});\n\\draw [domain = 0:180,thick] plot ({sqrt(2)*cos(\\x)},{2*sin(\\x)},{sqrt(2)*cos(\\x)});\n\\draw (0,2,0) node [above] {$S$} coordinate (S);\n\\draw ({cos(45)},0,{sin(45)}) --++ (0,{sqrt(3)},0);\n\\draw ({cos(135)},0,{sin(135)}) node [right] {$D$} coordinate (D) --++ (0,{sqrt(3)},0) node [above left] {$C$} coordinate (C);\n\\draw ({cos(225)},0,{sin(225)}) --++ (0,{sqrt(3)},0);\n\\draw ({cos(315)},0,{sin(315)}) --++ (0,{sqrt(3)},0);\n\\draw [thick] ({2*cos(45)},0,{2*sin(45)}) node [below left] {$A$} coordinate (A) -- ({3*cos(45)},-1,{3*sin(45)}) node [below] {$B$} coordinate (B);\n\\draw [thick] ({2*cos(45)},0,{2*sin(45)}) node [below left] {$A$} coordinate (A) -- ({3*cos(45)},-1,{3*sin(45)}) node [below] {$B$} coordinate (B);\n\\draw [thick] ({2*cos(135)},0,{2*sin(135)}) -- ({3*cos(135)},-1,{3*sin(135)});\n\\draw [thick] ({2*cos(315)},0,{2*sin(315)}) -- ({3*cos(315)},-1,{3*sin(315)});\n\\draw [thick,dashed] ({2*cos(225)},0,{2*sin(225)}) -- ({3*cos(225)},-1,{3*sin(225)});\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $CD\\parallel$平面$SOA$;\\\\\n(2) 设$AB$为经过$A$的一条步道, 其长度为$12$米且与地面所成角的大小为$30^{\\circ}$. 桥面小圆与大圆的半径之比为$4: 5$, 当桥面大圆半径为$20$米时, 求点$C$到地面的距离.",
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"objs": [],
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"tags": [
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"第六单元"
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@ -332224,7 +332224,8 @@
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"usages": [],
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"origin": "2023届普陀区一模试题19",
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"edit": [
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"20221215\t王伟叶"
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"20221215\t王伟叶",
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"20230523\t王伟叶"
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],
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"same": [],
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"related": [],
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@ -332601,7 +332602,7 @@
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},
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"012627": {
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"id": "012627",
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"content": "掷两颗骰子, 观察掷得的点数. 设事件$A$为: 至少一个点数是奇数; 事件$B$为: 点数之和是偶数, 事件$A$的概率为$P(A)$, 事件$B$的概率为$P(B)$. 则$1-P(A \\cap B)$是下列哪个事件的概率\\bracket{20}.\n\\twoch{两个点数都是偶}{至多有一个点数是偶数}{两个点数都是奇数}{至多有一个点数是奇数}",
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"content": "掷两颗骰子, 观察掷得的点数. 设事件$A$为: 至少一个点数是奇数; 事件$B$为: 点数之和是偶数, 事件$A$的概率为$P(A)$, 事件$B$的概率为$P(B)$. 则$1-P(A \\cap B)$是下列哪个事件的概率\\bracket{20}.\n\\twoch{两个点数都是偶数}{至多有一个点数是偶数}{两个点数都是奇数}{至多有一个点数是奇数}",
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"objs": [],
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"tags": [
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"第八单元"
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@ -332613,7 +332614,8 @@
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"usages": [],
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"origin": "2023届长宁区一模试题15",
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"edit": [
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"20221215\t王伟叶"
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"20221215\t王伟叶",
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"20230523\t王伟叶"
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],
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"same": [],
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"related": [],
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@ -332689,7 +332691,7 @@
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},
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"012631": {
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"id": "012631",
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"content": "如图, 在三棱锥$D-ABC$中, 平面$ACD \\perp$平面$ABC$, $AD \\perp AC$,$AB \\perp BC$, $E$、$F$分别为棱$BC$、$CD$的中点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [left] {$A$} coordinate (A);\n\\draw (2,0,0) node [right] {$C$} coordinate (C);\n\\draw (0,1.6,0) node [above] {$D$} coordinate (D);\n\\draw (1,0,1) node [below] {$B$} coordinate (B);\n\\draw ($(B)!0.5!(C)$) node [below right] {$E$} coordinate (E);\n\\draw ($(C)!0.5!(D)$) node [above] {$F$} coordinate (F);\n\\draw (A) -- (B) -- (C) -- (D) -- cycle;\n\\draw (D) -- (B) (F) -- (E);\n\\draw [dashed] (A) -- (C);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: 直线$EF\\parallel$平面$ABD$;\\\\\n(2) 求证: 直线$BC \\perp$平面$ABD$;\\\\\n(3) 若直线$CD$与平面$ABC$所成的角为$45^{\\circ}$, 直线$CD$与平面$ABD$所成角为$30^{\\circ}$, 求二面角$B-AD-C$的大小.",
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"content": "如图, 在三棱锥$D-ABC$中, 平面$ACD \\perp$平面$ABC$, $AD \\perp AC$, $AB \\perp BC$, $E$、$F$分别为棱$BC$、$CD$的中点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [left] {$A$} coordinate (A);\n\\draw (2,0,0) node [right] {$C$} coordinate (C);\n\\draw (0,1.6,0) node [above] {$D$} coordinate (D);\n\\draw (1,0,1) node [below] {$B$} coordinate (B);\n\\draw ($(B)!0.5!(C)$) node [below right] {$E$} coordinate (E);\n\\draw ($(C)!0.5!(D)$) node [above] {$F$} coordinate (F);\n\\draw (A) -- (B) -- (C) -- (D) -- cycle;\n\\draw (D) -- (B) (F) -- (E);\n\\draw [dashed] (A) -- (C);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: 直线$EF\\parallel$平面$ABD$;\\\\\n(2) 求证: 直线$BC \\perp$平面$ABD$;\\\\\n(3) 若直线$CD$与平面$ABC$所成的角为$45^{\\circ}$, 直线$CD$与平面$ABD$所成角为$30^{\\circ}$, 求二面角$B-AD-C$的大小.",
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"objs": [],
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"tags": [
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"第六单元"
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@ -332701,7 +332703,8 @@
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"usages": [],
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"origin": "2023届长宁区一模试题19",
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"edit": [
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"20221215\t王伟叶"
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"20221215\t王伟叶",
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"20230523\t王伟叶"
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],
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"same": [],
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"related": [],
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