From dfbf6991d07f31261c5f10f6729898a169e6bdaa Mon Sep 17 00:00:00 2001 From: WangWeiye Date: Tue, 23 May 2023 17:29:57 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=94=B912610,12627,12631=E9=A2=98?= =?UTF-8?q?=E9=9D=A2?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 工具/latex界面修改题目内容.py | 4 ++-- 题库0.3/Problems.json | 15 +++++++++------ 2 files changed, 11 insertions(+), 8 deletions(-) diff --git a/工具/latex界面修改题目内容.py b/工具/latex界面修改题目内容.py index 099038aa..cce99434 100644 --- a/工具/latex界面修改题目内容.py +++ b/工具/latex界面修改题目内容.py @@ -1,7 +1,7 @@ import os,re,json """这里编辑题号(列表)后将在vscode中打开窗口, 编辑后保存关闭""" -problems = "13506" -editor = "朱敏慧" +problems = "12610,12627,12631" +editor = "王伟叶" def generate_number_set(string,dict): string = re.sub(r"[\n\s]","",string) diff --git a/题库0.3/Problems.json b/题库0.3/Problems.json index 1a70903f..41fe12f0 100644 --- a/题库0.3/Problems.json +++ b/题库0.3/Problems.json @@ -332212,7 +332212,7 @@ }, "012610": { "id": "012610", - "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.", + "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.", "objs": [], "tags": [ "第六单元" @@ -332224,7 +332224,8 @@ "usages": [], "origin": "2023届普陀区一模试题19", "edit": [ - "20221215\t王伟叶" + "20221215\t王伟叶", + "20230523\t王伟叶" ], "same": [], "related": [], @@ -332601,7 +332602,7 @@ }, "012627": { "id": "012627", - "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}", + "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}", "objs": [], "tags": [ "第八单元" @@ -332613,7 +332614,8 @@ "usages": [], "origin": "2023届长宁区一模试题15", "edit": [ - "20221215\t王伟叶" + "20221215\t王伟叶", + "20230523\t王伟叶" ], "same": [], "related": [], @@ -332689,7 +332691,7 @@ }, "012631": { "id": "012631", - "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.", + "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.", "objs": [], "tags": [ "第六单元" @@ -332701,7 +332703,8 @@ "usages": [], "origin": "2023届长宁区一模试题19", "edit": [ - "20221215\t王伟叶" + "20221215\t王伟叶", + "20230523\t王伟叶" ], "same": [], "related": [],