20221018 afternoon
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 17,
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"0"
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]
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},
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"execution_count": 17,
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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@ -21,7 +21,7 @@
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"\n",
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"\"\"\"---设置关键字, 同一field下不同选项为or关系, 同一字典中不同字段间为and关系, 不同字典间为or关系, _not表示列表中的关键字都不含, 同一字典中的数字用来供应同一字段不同的条件之间的and---\"\"\"\n",
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"keywords_dict_table = [\n",
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" {\"tags\":[\"五\"]}\n",
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" {\"tags\":[\"六\"],\"content\":[\"所成的角\",\"perp\",\"垂直\",\"平行\",\"parallel\"]}\n",
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"]\n",
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"\"\"\"---关键字设置完毕---\"\"\"\n",
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"# 示例： keywords_dict_table = [\n",
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@ -85,7 +85,7 @@
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3.8.8 ('base')",
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"display_name": "Python 3.9.7 ('base')",
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"language": "python",
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"name": "python3"
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},
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@ -99,12 +99,12 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.8"
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"version": "3.9.7"
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},
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"orig_nbformat": 4,
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"vscode": {
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"interpreter": {
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"hash": "d311ffef239beb3b8f3764271728f3972d7b090c974f8e972fcdeedf230299ac"
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"hash": "e4cce46d6be9934fbd27f9ca0432556941ea5bdf741d4f4d64c6cd7f8dfa8fba"
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}
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},
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@ -11,7 +11,7 @@
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"text": [
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"首个空闲id: 11988 , 直至 020000\n",
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"首个空闲id: 20227 , 直至 030000\n",
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"首个空闲id: 30154 , 直至 999999\n"
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"首个空闲id: 30169 , 直至 999999\n"
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]
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}
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],
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@ -45,7 +45,7 @@
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3.8.8 ('base')",
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"display_name": "Python 3.9.7 ('base')",
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"language": "python",
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"name": "python3"
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},
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.8"
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"version": "3.9.7"
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},
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"orig_nbformat": 4,
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"vscode": {
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"interpreter": {
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"hash": "d311ffef239beb3b8f3764271728f3972d7b090c974f8e972fcdeedf230299ac"
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"hash": "e4cce46d6be9934fbd27f9ca0432556941ea5bdf741d4f4d64c6cd7f8dfa8fba"
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}
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}
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},
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@ -2,15 +2,15 @@
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 2,
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"execution_count": 1,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"首行题目数量: 1236\n",
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"剩余题目数量: 1092\n"
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"首行题目数量: 393\n",
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"剩余题目数量: 337\n"
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]
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}
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],
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@ -22,7 +22,7 @@
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"\"\"\"---设置题号列表文件结束---\"\"\"\n",
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"\n",
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"\"\"\"---设置要排除的题号所在的绝对路径---\"\"\"\n",
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"mainpath = r\"C:/Users/Weiye/Documents/wwy sync/23届/\"\n",
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"mainpath = r\"C:/Users/wang Weiye/Documents/wwy sync/23届/\"\n",
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"\n",
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"used_path_list = [\n",
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"mainpath + \"第一轮复习讲义/\",\n",
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@ -84,7 +84,7 @@
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3.8.8 ('base')",
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"display_name": "Python 3.9.7 ('base')",
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"language": "python",
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"name": "python3"
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},
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@ -98,12 +98,12 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.8"
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"version": "3.9.7"
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},
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"orig_nbformat": 4,
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"vscode": {
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"interpreter": {
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"hash": "d311ffef239beb3b8f3764271728f3972d7b090c974f8e972fcdeedf230299ac"
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"hash": "e4cce46d6be9934fbd27f9ca0432556941ea5bdf741d4f4d64c6cd7f8dfa8fba"
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}
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}
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},
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File diff suppressed because one or more lines are too long
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@ -9,8 +9,8 @@
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"import os,re,json,time\n",
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"\n",
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"\"\"\"---设置原题目id与新题目id---\"\"\"\n",
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"old_id = \"9191\"\n",
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"new_id = \"30159\"\n",
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"old_id = \"167\"\n",
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"new_id = \"30169\"\n",
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"\"\"\"---设置完毕---\"\"\"\n",
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"\n",
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"old_id = old_id.zfill(6)\n",
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@ -2,7 +2,7 @@
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 1,
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"execution_count": 2,
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"metadata": {},
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"outputs": [
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{
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@ -13,9 +13,9 @@
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"题块 1 处理完毕.\n",
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"正在处理题块 2 .\n",
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"题块 2 处理完毕.\n",
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"开始编译教师版本pdf文件: 临时文件/24_体积及表面积的计算_教师_20221017.tex\n",
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"开始编译教师版本pdf文件: 临时文件/24_体积及表面积的计算_教师_20221018.tex\n",
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"0\n",
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"开始编译学生版本pdf文件: 临时文件/24_体积及表面积的计算_学生_20221017.tex\n",
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"开始编译学生版本pdf文件: 临时文件/24_体积及表面积的计算_学生_20221018.tex\n",
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"0\n"
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]
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}
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@ -55,8 +55,8 @@
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"\n",
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"\"\"\"---设置题号数据---\"\"\"\n",
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"problems = [\n",
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"\"10500,3475,9207,9211,9731,9400,9399,4994,9868,10524,9210,9720,202,10517,4196\",\n",
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"\"10498,4084,215,411,11332,212,10499,10519,10522,9209,10515,201,10521\"\n",
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"\"10500,3475,30160,30161,30162,30163,30164,4994,9868,10524,9210,9720,202,30165,4196\",\n",
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"\"10498,4084,215,411,11332,212,10499,30166,30167,30168,10515,201,10521\"\n",
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"]\n",
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"\"\"\"---设置题号数据结束---\"\"\"\n",
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"\n",
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@ -2,16 +2,16 @@
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 5,
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"开始编译教师版本pdf文件: 临时文件/复数向量_教师用_20221016.tex\n",
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"开始编译教师版本pdf文件: 临时文件/周末卷05_预选_教师用_20221018.tex\n",
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"0\n",
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"开始编译学生版本pdf文件: 临时文件/复数向量_学生用_20221016.tex\n",
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"开始编译学生版本pdf文件: 临时文件/周末卷05_预选_学生用_20221018.tex\n",
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"0\n"
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]
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}
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@ -26,14 +26,14 @@
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"\"\"\"---设置题目列表---\"\"\"\n",
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"#留空为编译全题库, a为读取临时文件中的题号筛选.txt文件生成题库\n",
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"problems = r\"\"\"\n",
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"173,382,1884,1900,1908,1915,1997,2011,2022,2024,2081,2094,3335,3342,3350,3354,3531,3702,3841,4337,4754,7018,7042,7050,7065,7068\n",
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"30169,173,182,187,298,1677,3531,3533,4092,3891,1643,3455,3462,3495,4180,3500\n",
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"\n",
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"\"\"\"\n",
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"\"\"\"---设置题目列表结束---\"\"\"\n",
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"\n",
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"\"\"\"---设置文件名---\"\"\"\n",
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"#目录和文件的分隔务必用/\n",
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"filename = \"临时文件/复数向量\"\n",
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"filename = \"临时文件/周末卷05_预选\"\n",
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"\"\"\"---设置文件名结束---\"\"\"\n",
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"\n",
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"\n",
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3.8.8 ('base')",
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"display_name": "Python 3.9.7 ('base')",
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"language": "python",
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"name": "python3"
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},
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@ -188,12 +188,12 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.8"
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"version": "3.9.7"
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},
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"orig_nbformat": 4,
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"vscode": {
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"interpreter": {
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"hash": "d311ffef239beb3b8f3764271728f3972d7b090c974f8e972fcdeedf230299ac"
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"hash": "e4cce46d6be9934fbd27f9ca0432556941ea5bdf741d4f4d64c6cd7f8dfa8fba"
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}
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}
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},
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@ -4315,7 +4315,9 @@
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"20220624\t王伟叶, 余利成"
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],
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"same": [],
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"related": [],
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"related": [
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"030169"
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],
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"remark": "",
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"space": "12ex"
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},
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@ -105274,7 +105276,7 @@
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},
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"004196": {
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"id": "004196",
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"content": "课本中介绍了应用祖暅原理推导棱锥体积公式的做法. 祖暅原理也可用来求旋转体的体积. 现介绍用祖暅原理求球体体积公式的做法: 可构造一个底面半径和高都与球半径相等的圆柱, 然后在圆柱内挖去一个以圆柱下底面圆心为顶点, 圆柱上底面为底面的圆锥, 用这样一个几何体与半球应用祖暅原理(左图), 即可求得球的体积公式. 请研究和理解球的体积公式求法的基础上, 解答以下问题: 已知椭圆的标准方程为$\\dfrac{x^2}4+\\dfrac{y^2}{25}=1$, 将此椭圆绕$y$轴旋转一周后, 得一橄榄状的几何体(右图), 其体积等于\\blank{50}.\n\\begin{center}\n \\begin{tikzpicture}\n \\draw (0,0) arc (180:0:2) arc (0:-180:2 and 0.5);\n \\draw [dashed] (0,0) arc (180:0:2 and 0.5) -- (0,0);\n \\fill [color = gray!30] (2,1) ellipse ({sqrt(3)} and {sqrt(3)/4});\n \\draw ({2-sqrt(3)},{1}) arc (180:360:{sqrt(3)} and {sqrt(3)/4});\n \\draw [dashed] ({2-sqrt(3)},{1}) arc (180:0:{sqrt(3)} and {sqrt(3)/4});\n \\draw [dashed] (2,0) -- (2,1) (2,0.2) node [left] {$h$};\n \\draw [dashed] (2,0) -- ({2+sqrt(3)},1) (3,0) node [below] {$R$};\n \\filldraw [even odd rule, gray!30] (7,1) ellipse (2 and 0.5) (7,1) ellipse (1 and 0.25);\n \\draw (5,0) arc (180:360:2 and 0.5) (5,2) arc (180:-180:2 and 0.5) (5,0) -- (5,2) (9,0) -- (9,2);\n \\draw [dashed] (5,0) -- (9,0) (7,0) -- (7,1) (7,0) -- (5,2) (7,0) -- (9,2) (8,0) node [below] {$R$} (7,0.4) node [left] {$h$};\n \\draw (5,1) arc (180:360:2 and 0.5);\n \\draw [dashed] (5,1) arc (180:0:2 and 0.5) (6,1) arc (180:-180:1 and 0.25);\n \\end{tikzpicture}\n \\begin{tikzpicture}{>=latex}\n \\draw [->] (-1.5,0) -- (1.5,0) node [below] {$x$};\n \\draw [->] (0,-2) -- (0,2) node [left] {$y$};\n \\draw (0,0) node [below left] {$O$};\n \\draw (0,0) ellipse (1 and 1.5);\n \\draw [dashed] (0,0) ellipse (0.5 and 1.5);\n \\end{tikzpicture}\n\\end{center}",
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"content": "课本中介绍了应用祖暅原理推导棱锥体积公式的做法. 祖暅原理也可用来求旋转体的体积. 现介绍用祖暅原理求球体体积公式的做法: 可构造一个底面半径和高都与球半径相等的圆柱, 然后在圆柱内挖去一个以圆柱下底面圆心为顶点, 圆柱上底面为底面的圆锥, 用这样一个几何体与半球应用祖暅原理(左图), 即可求得球的体积公式. 请研究和理解球的体积公式求法的基础上, 解答以下问题: 已知椭圆的标准方程为$\\dfrac{x^2}4+\\dfrac{y^2}{25}=1$, 将此椭圆绕$y$轴旋转一周后, 得一橄榄状的几何体(右图), 其体积等于\\blank{50}.\n\\begin{center}\n \\begin{tikzpicture}\n \\draw (0,0) arc (180:0:2) arc (0:-180:2 and 0.5);\n \\draw [dashed] (0,0) arc (180:0:2 and 0.5) -- (0,0);\n \\fill [color = gray!30] (2,1) ellipse ({sqrt(3)} and {sqrt(3)/4});\n \\draw ({2-sqrt(3)},{1}) arc (180:360:{sqrt(3)} and {sqrt(3)/4});\n \\draw [dashed] ({2-sqrt(3)},{1}) arc (180:0:{sqrt(3)} and {sqrt(3)/4});\n \\draw [dashed] (2,0) -- (2,1) (2,0.2) node [left] {$h$};\n \\draw [dashed] (2,0) -- ({2+sqrt(3)},1) (3,0) node [below] {$R$};\n \\filldraw [even odd rule, gray!30] (7,1) ellipse (2 and 0.5) (7,1) ellipse (1 and 0.25);\n \\draw (5,0) arc (180:360:2 and 0.5) (5,2) arc (180:-180:2 and 0.5) (5,0) -- (5,2) (9,0) -- (9,2);\n \\draw [dashed] (5,0) -- (9,0) (7,0) -- (7,1) (7,0) -- (5,2) (7,0) -- (9,2) (8,0) node [below] {$R$} (7,0.4) node [left] {$h$};\n \\draw (5,1) arc (180:360:2 and 0.5);\n \\draw [dashed] (5,1) arc (180:0:2 and 0.5) (6,1) arc (180:-180:1 and 0.25);\n \\end{tikzpicture}\n \\begin{tikzpicture}[>=latex]\n \\draw [->] (-1.5,0) -- (1.5,0) node [below] {$x$};\n \\draw [->] (0,-2) -- (0,2) node [left] {$y$};\n \\draw (0,0) node [below left] {$O$};\n \\draw (0,0) ellipse (1 and 1.5);\n \\draw [dashed] (0,0) ellipse (0.5 and 1.5);\n \\end{tikzpicture}\n\\end{center}",
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"objs": [
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|
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"content": "如图, 平面$\\alpha$上的斜线$l$与平面$\\alpha$所成的角为$\\theta$, $l'$是$l$在平面$\\alpha$上的投影, $O$是$l$与平面$\\alpha$的交点, 点$B$是$l$上一点$A$在$\\alpha$上的投影, $OC$是$\\alpha$上的任意一条直线. 如果$\\theta =45^\\circ$, $\\angle BOC=45^\\circ$, 求$\\angle AOC$, 并验证$\\angle AOC>\\theta$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (-1,0,0) -- (3,0,0) node [right] {$l'$} (2,2,0) node [above] {$A$} coordinate (A) -- (2,0,0) coordinate (B) node [below] {$B$} (2.5,2.5,0) node [right] {$l$} -- (0,0,0) coordinate (O) node [below] {$O$};\n\\draw (1,0,1) node [below] {$C$} coordinate (C);\n\\draw ($(O)!-0.5!(C)$) -- ($(O)!1.8!(C)$) node [right] {$l''$};\n\\draw [name path = edge] (-1.5,0,-2.5) coordinate (L) -- (-1.5,0,2.5) --++ (5,0,0) --++ (0,0,-5) coordinate (R);\n\\path [name path = LR] (L) -- (R);\n\\path [name path = OA] (O) -- (A);\n\\path [name path = AB] (A) -- (B);\n\\path [name intersections = {of = OA and LR, by = A1}];\n\\path [name intersections = {of = AB and LR, by = B1}];\n\\draw (L) -- (A1) (B1) -- (R);\n\\draw [dashed] (A1) -- (B1);\n\\path [name path = down] ($(O)!-0.6!(A)$) -- (O);\n\\path [name intersections = {of = down and edge, by = T}];\n\\draw (T) -- ($(O)!-0.6!(A)$);\n\\draw [dashed] (T) -- (O);\n\\draw (O) pic [\"$\\theta$\",draw,angle eccentricity = 1.5] {angle = B--O--A};\n\\draw (O) pic [\"$45^\\circ$\",scale = 1.1,draw,angle eccentricity = 1.7]{angle = C--O--B};\n\\end{tikzpicture}\n\\end{center}",
|
||||
"content": "如图, 平面$\\alpha$上的斜线$l$与平面$\\alpha$所成的角为$\\theta$, $l'$是$l$在平面$\\alpha$上的投影, $O$是$l$与平面$\\alpha$的交点, 点$B$是$l$上一点$A$在$\\alpha$上的投影, $OC$是$\\alpha$上的任意一条与$l'$不重合的直线. 如果$\\theta =45^\\circ$, $\\angle BOC=45^\\circ$, 求$\\angle AOC$, 并验证$\\angle AOC>\\theta$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (-1,0,0) -- (3,0,0) node [right] {$l'$} (2,2,0) node [above] {$A$} coordinate (A) -- (2,0,0) coordinate (B) node [below] {$B$} (2.5,2.5,0) node [right] {$l$} -- (0,0,0) coordinate (O) node [below] {$O$};\n\\draw (1,0,1) node [below] {$C$} coordinate (C);\n\\draw ($(O)!-0.5!(C)$) -- ($(O)!1.8!(C)$) node [right] {$l''$};\n\\draw [name path = edge] (-1.5,0,-2.5) coordinate (L) -- (-1.5,0,2.5) --++ (5,0,0) --++ (0,0,-5) coordinate (R);\n\\path [name path = LR] (L) -- (R);\n\\path [name path = OA] (O) -- (A);\n\\path [name path = AB] (A) -- (B);\n\\path [name intersections = {of = OA and LR, by = A1}];\n\\path [name intersections = {of = AB and LR, by = B1}];\n\\draw (L) -- (A1) (B1) -- (R);\n\\draw [dashed] (A1) -- (B1);\n\\path [name path = down] ($(O)!-0.6!(A)$) -- (O);\n\\path [name intersections = {of = down and edge, by = T}];\n\\draw (T) -- ($(O)!-0.6!(A)$);\n\\draw [dashed] (T) -- (O);\n\\draw (O) pic [\"$\\theta$\",draw,angle eccentricity = 1.5] {angle = B--O--A};\n\\draw (O) pic [\"$45^\\circ$\",scale = 1.1,draw,angle eccentricity = 1.7]{angle = C--O--B};\n\\end{tikzpicture}\n\\end{center}",
|
||||
"objs": [
|
||||
"K0610005B"
|
||||
],
|
||||
|
|
@ -292658,5 +292660,32 @@
|
|||
],
|
||||
"remark": "",
|
||||
"space": ""
|
||||
},
|
||||
"030169": {
|
||||
"id": "030169",
|
||||
"content": "已知复数$z=\\dfrac{(-3-\\mathrm{\\mathrm{i}})^2(2-\\mathrm{\\mathrm{i}})}{(1+2\\mathrm{i})^3}$, 则$|z|=$\\blank{50}.",
|
||||
"objs": [
|
||||
"K0514004B",
|
||||
"K0514001B"
|
||||
],
|
||||
"tags": [
|
||||
"第五单元"
|
||||
],
|
||||
"genre": "填空题",
|
||||
"ans": "",
|
||||
"solution": "",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "教材复习题-20221018修改",
|
||||
"edit": [
|
||||
"20220624\t王伟叶, 余利成",
|
||||
"20221018\t王伟叶"
|
||||
],
|
||||
"same": [],
|
||||
"related": [
|
||||
"000167"
|
||||
],
|
||||
"remark": "",
|
||||
"space": ""
|
||||
}
|
||||
}
|
||||
Reference in New Issue