录入2025届周末卷补充题至07并建立题目关联
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"20230707\t王伟叶"
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"20230707\t王伟叶"
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@ -621239,6 +621241,288 @@
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"space": "4em",
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"space": "4em",
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"unrelated": []
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},
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"023193": {
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"id": "023193",
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"content": "圆柱的底面半径为 $4 \\mathrm{cm}$, 它的侧面积为 $64 \\pi \\mathrm{cm}^2$, 那么圆柱的母线长为\\blank{50}$\\mathrm{cm}$.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023194": {
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"id": "023194",
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"content": "正三棱锥的侧面与底面所成的角为 $60^{\\circ}$, 则棱锥的侧棱与底面所成角的大小为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023195": {
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"id": "023195",
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"content": "已知斜棱柱高为 $4$, 侧棱与底面成 $30^{\\circ}$ 角, 若用垂直于侧棱的平面截斜棱柱所得截面周长为 $8$ (截面与底面无公共点), 则其侧面积为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023196": {
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"id": "023196",
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"content": "在三棱锥 $S-ABC$ 中, $\\angle ASB=\\angle ASC=\\angle BSC=60^{\\circ}$, 则侧棱 $SA$ 与侧面 $SBC$ 所成的角的大小是\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023197": {
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"id": "023197",
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"content": "直平行六面体的底面是菱形, 分别过两对不柤邻侧棱的两个截面面积是 $3 \\mathrm{cm}^2$ 和 $4 \\mathrm{cm}^2$,则它的侧面积为\\blank{50} $\\mathrm{cm}^2$.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023198": {
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"id": "023198",
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"content": "已知三棱锥 $P-ABC$, 用``内心''、``外心''或``垂心''填空:\\\\\n(1) 若 $PA, PB, PC$ 与底面所成角都相等, 则顶点 $P$ 在底面的射影是三角形 $ABC$ 的\\blank{50};\\\\\n(2) 若二面角 $P-AB-C$, 二面角 $P-BC-A$, 二面角 $P-AC-B$ 的大小都相等, 且$P$在平面$ABC$上的射影在$\\triangle ABC$的内部, 则顶点 $P$ 在底面的射影是三角形 $ABC$ 的\\blank{50};\\\\\n(3) $PA \\perp BC$, $PB \\perp AC$, $PC \\perp AB$, 则顶点 $P$ 在底面的射影是三角形 $ABC$ 的\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023199": {
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"id": "023199",
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"content": "一个正四面体的顶点是一个正方体的顶点, 则正方体的表面积是该正四面体表面积的\\blank{50}倍.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023200": {
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"id": "023200",
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"content": "若正三棱柱的所有棱长均为 $a$, 且其体积为 $16 \\sqrt{3}$, 则侧面积为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023201": {
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"id": "023201",
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"content": "在长方体 $ABCD-A_1B_1C_1D_1$ 中, $AB=1$, $AD=2$, $AA_1=3$, 则直线 $A_1C_1$ 到平面 $ACB_1$ 的距离为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023202": {
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"id": "023202",
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"content": "定义八个顶点都在某圆柱的底面圆周上的长方体叫做圆柱的内接长方体, 圆柱也叫该长方体的外接圆柱, 设长方体 $ABCD-A_1B_1C_1D_1$ 的长、宽、高分别为 $a$、$b$、$c$ (其中 $a>b>c$ ), 那么该长方体的外接圆柱侧面积的最大值是\\bracket{20}.\n\\fourch{$\\pi a \\sqrt{b^2+c^2}$}{$\\pi b \\sqrt{a^2+c^2}$}{$\\pi c \\sqrt{a^2+b^2}$}{$\\pi a b c$}",
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"objs": [],
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"tags": [],
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"genre": "选择题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "",
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"unrelated": []
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},
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"023203": {
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"id": "023203",
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"content": "正四棱柱 $ABCD-A_1B_1C_1D_1$ 的底面边长 $AB=2$, 若异面直线 $A_1A$ 与 $B_1C$ 所成角的大小为 $\\arctan \\dfrac{1}{2}$, 求正四棱柱 $ABCD-A_1B_1C_1D_1$ 的侧面积和体积.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\def\\l{1}\n\\def\\m{1}\n\\def\\n{2}\n\\draw (0,0,0) node [below left] {$A$} coordinate (A);\n\\draw (A) ++ (\\l,0,0) node [below right] {$B$} coordinate (B);\n\\draw (A) ++ (\\l,0,-\\m) node [right] {$C$} coordinate (C);\n\\draw (A) ++ (0,0,-\\m) node [left] {$D$} coordinate (D);\n\\draw (A) -- (B) -- (C);\n\\draw [dashed] (A) -- (D) -- (C);\n\\draw (A) ++ (0,\\n,0) node [left] {$A_1$} coordinate (A_1);\n\\draw (B) ++ (0,\\n,0) node [above] {$B_1$} coordinate (B_1);\n\\draw (C) ++ (0,\\n,0) node [above right] {$C_1$} coordinate (C_1);\n\\draw (D) ++ (0,\\n,0) node [above left] {$D_1$} coordinate (D_1);\n\\draw (A_1) -- (B_1) -- (C_1) -- (D_1) -- cycle;\n\\draw (A) -- (A_1) (B) -- (B_1) (C) -- (C_1);\n\\draw [dashed] (D) -- (D_1);\n\\draw (B_1)--(C);\n\\end{tikzpicture}\n\\end{center}",
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"objs": [],
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"tags": [],
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"genre": "解答题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "4em",
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"unrelated": []
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},
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"023204": {
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"id": "023204",
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"content": "如图, 四边形 $ABCD$ 为边长是 $4$ 的正方形, 点 $E$、$F$ 分别是 $AB$、$AD$ 的中点, $GC$ 垂直于 $ABCD$ 所在的平面, 且 $GC=2$, 求点 $B$ 到平面 $EFG$ 的距离.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [left] {$C$} coordinate (C);\n\\draw (4,0,0) node [right] {$B$} coordinate (B);\n\\draw (4,0,4) node [below] {$A$} coordinate (A);\n\\draw (0,0,4) node [below] {$D$} coordinate (D);\n\\draw (0,2,0) node [above] {$G$} coordinate (G);\n\\draw (G)--(C)--(D)--(A)--(B);\n\\path [name path = BC] (B)--(C);\n\\filldraw ($(A)!0.5!(B)$) circle (0.03) node [right] {$E$} coordinate (E);\n\\filldraw ($(A)!0.5!(D)$) circle (0.03) node [below] {$F$} coordinate (F);\n\\draw (E)--(F)--(G)--cycle;\n\\path [name path = EG] (E)--(G);\n\\path [name path = FG] (F)--(G);\n\\draw [name intersections = {of = EG and BC, by = U}];\n\\draw [name intersections = {of = FG and BC, by = V}];\n\\draw (C)--(V)(U)--(B);\n\\draw [dashed] (U)--(V);\n\\end{tikzpicture}\n\\end{center}",
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"objs": [],
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"tags": [],
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"genre": "解答题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [
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"019035"
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],
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"remark": "",
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"space": "4em",
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"unrelated": []
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},
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"023205": {
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"id": "023205",
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"content": "如图, 已知三棱锥 $P-ABC$ 的侧面 $PAC$ 是底角为 $45^{\\circ}$ 的等腰三角形, $PA=PC$, 且该侧面垂直于底面, $\\angle ACB=90^{\\circ}$, $AB=10$, $BC=6$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.3]\n\\draw (0,0,0) node [left] {$A$} coordinate (A);\n\\draw ({sqrt(136)},0,0) node [right] {$C$} coordinate (C);\n\\draw ({sqrt(34)},{sqrt(34)},0) node [above] {$P$} coordinate (P);\n\\draw ({100/sqrt(136)},0,{60/sqrt(136)}) node [below] {$B$} coordinate (B);\n\\draw (P)--(A)--(B)--(C)--cycle(P)--(B);\n\\draw [dashed] (A)--(C);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: 二面角 $A-PB-C$ 是直二面角;\\\\\n(2) 求二面角 $P-AB-C$ 的正切值;",
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"objs": [],
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"tags": [],
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"genre": "解答题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"20240106\t杨懿荔"
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"space": "4em",
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"unrelated": []
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},
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"023206": {
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"id": "023206",
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"content": "如图, 在直三棱柱 $ABC-A_1B_1C_1$ 中, 底面是等腰直角三角形, $\\angle ACB=90^{\\circ}$ 且 $AC=a$, 侧棱 $AA_1=2$, 点 $D$ 和 $E$ 分别是 $CC_1$ 和 $A_1B_1$ 的中点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw ({-sqrt(5)},0,0) node [left] {$A$} coordinate (A);\n\\draw ({sqrt(5)},0,0) node [right] {$B$} coordinate (B);\n\\draw (0,0,{-sqrt(5)}) node [below] {$C$} coordinate (C);\n\\draw (A) ++ (0,2,0) node [left] {$A_1$} coordinate (A_1);\n\\draw (B) ++ (0,2,0) node [right] {$B_1$} coordinate (B_1);\n\\draw (C) ++ (0,2,0) node [above] {$C_1$} coordinate (C_1);\n\\draw ($(A_1)!0.5!(B_1)$) node [above] {$E$} coordinate (E);\n\\draw ($(C)!0.5!(C_1)$) node [right] {$D$} coordinate (D);\n\\draw ($1/3*(A)+1/3*(B)+1/3*(D)$) node [below] {$G$} coordinate (G);\n\\draw (A)--(B)--(B_1)--(A_1)--cycle(A_1)--(C_1)--(B_1)(A)--(E)--(B);\n\\draw [dashed] (C)--(C_1)(A)--(D)--(B)(D)--(E)(E)--(G)(A)--(C)--(B);\n\\filldraw (G) circle (0.03);\n\\end{tikzpicture}\n\\end{center}\n(1) 求直三棱柱 $ABC-A_1B_1C_1$ 的体积 (用字母 $a$ 表示);\\\\\n(2) 若点 $E$ 在平面 $ABD$ 上的射影是三角形 $ABD$ 的重心 $G$,\\\\\n\\textcircled{1} 求直线 $EB$ 与平面 $ABD$ 所成角的余弦值;\\\\\n\\textcircled{2} 求点 $A_1$ 到平面 $ABD$ 的距离.",
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"objs": [],
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"tags": [],
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"genre": "解答题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "25届周末卷补充题目",
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"edit": [
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"20240106\t杨懿荔"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": "4em",
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"unrelated": []
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},
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"030001": {
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"030001": {
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"id": "030001",
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"id": "030001",
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"content": "若$x,y,z$都是实数, 则:(填写``\\textcircled{1} 充分非必要、\\textcircled{2} 必要非充分、\\textcircled{3} 充要、\\textcircled{4} 既非充分又非必要''之一)\\\\\n(1) ``$xy=0$''是``$x=0$''的\\blank{50}条件;\\\\\n(2) ``$x\\cdot y=y\\cdot z$''是``$x=z$''的\\blank{50}条件;\\\\\n(3) ``$\\dfrac xy=\\dfrac yz$''是``$xz=y^2$''的\\blank{50}条件;\\\\\n(4) ``$|x |>| y|$''是``$x>y>0$''的\\blank{50}条件;\\\\\n(5) ``$x^2>4$''是``$x>2$'' 的\\blank{50}条件;\\\\\n(6) ``$x=-3$''是``$x^2+x-6=0$'' 的\\blank{50}条件;\\\\\n(7) ``$|x+y|<2$''是``$|x|<1$且$|y|<1$'' 的\\blank{50}条件;\\\\\n(8) ``$|x|<3$''是``$x^2<9$'' 的\\blank{50}条件;\\\\\n(9) ``$x^2+y^2>0$''是``$x\\ne 0$'' 的\\blank{50}条件;\\\\\n(10) ``$\\dfrac{x^2+x+1}{3x+2}<0$''是``$3x+2<0$'' 的\\blank{50}条件;\\\\\n(11) ``$0<x<3$''是``$|x-1|<2$'' 的\\blank{50}条件.",
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"content": "若$x,y,z$都是实数, 则:(填写``\\textcircled{1} 充分非必要、\\textcircled{2} 必要非充分、\\textcircled{3} 充要、\\textcircled{4} 既非充分又非必要''之一)\\\\\n(1) ``$xy=0$''是``$x=0$''的\\blank{50}条件;\\\\\n(2) ``$x\\cdot y=y\\cdot z$''是``$x=z$''的\\blank{50}条件;\\\\\n(3) ``$\\dfrac xy=\\dfrac yz$''是``$xz=y^2$''的\\blank{50}条件;\\\\\n(4) ``$|x |>| y|$''是``$x>y>0$''的\\blank{50}条件;\\\\\n(5) ``$x^2>4$''是``$x>2$'' 的\\blank{50}条件;\\\\\n(6) ``$x=-3$''是``$x^2+x-6=0$'' 的\\blank{50}条件;\\\\\n(7) ``$|x+y|<2$''是``$|x|<1$且$|y|<1$'' 的\\blank{50}条件;\\\\\n(8) ``$|x|<3$''是``$x^2<9$'' 的\\blank{50}条件;\\\\\n(9) ``$x^2+y^2>0$''是``$x\\ne 0$'' 的\\blank{50}条件;\\\\\n(10) ``$\\dfrac{x^2+x+1}{3x+2}<0$''是``$3x+2<0$'' 的\\blank{50}条件;\\\\\n(11) ``$0<x<3$''是``$|x-1|<2$'' 的\\blank{50}条件.",
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Reference in New Issue