{"ok":true,"question":{"id":21015,"question_uuid":"ca1df7c7-5766-4e5d-845f-7a72124ce49e","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案820.pdf","source_page_start":8,"source_page_end":9,"source_bbox":"[30.0, 387.9, 557.04, 796.15]","subject_code":"math","grade_level":"JUNIOR","tree_code":"KT_JUNIOR_MATH_STANDARD","question_type":"single_choice","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 708, \"height\": 545, \"sha256\": \"2663b39cf5ce4ef2dc72022be14a419db4ffd7ea9a08b2728a52cb7de7bd3fb7\", \"caption\": \"\", \"asset_key\": \"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png\", \"source_bbox\": [30.0, 408.52001953125, 171.69000244140625, 519.52001953125], \"source_page\": 8}, {\"type\": \"option\", \"label\": \"A\", \"blocks\": [{\"text\": \"2\", \"type\": \"text\"}], \"source_bbox\": [30.0, 559.52, 72.62, 572.49], \"source_page\": 8}, {\"type\": \"option\", \"label\": \"B\", \"blocks\": [{\"type\": \"math_inline\", \"latex\": \"8\\\\sqrt{3}\\\\frac{3}{3}\"}], \"source_bbox\": [48.0, 580.5, 63.52, 589.51], \"source_page\": 8}, {\"type\": \"option\", \"label\": \"C\", \"blocks\": [{\"text\": \"√3\", \"type\": \"text\"}], \"source_bbox\": [30.0, 611.7, 58.57, 627.11], \"source_page\": 8}, {\"type\": \"option\", \"label\": \"D\", \"blocks\": [{\"text\": \"2√2\", \"type\": \"text\"}], \"source_bbox\": [30.0, 645.83, 87.14, 659.28], \"source_page\": 8}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"B\", \"type\": \"text\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"设圆O半径为R。正三角形ABC内接于圆O，弦BC=8。AD⊥BC于D，则D为BC中点，BD=4。在Rt△ABD中，AB=BC=8（正三角形边长相等），故AD=√(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"AB^{2}\"}, {\"text\": \"-\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"BD^{2}\"}, {\"text\": \")=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{64-16}\"}, {\"text\": \"=√48=4√3。圆心O在AD上（正三角形外接圆圆心在对称轴上），且AO=R，OD=AD-AO=4√3-R。在Rt△OBD中，OB=R，BD=4，OD=4√3-R，由勾股定理：\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"R^{2}\"}, {\"text\": \"=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"4^2\"}, {\"text\": \"+(4√3-R)^2 ⇒\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"R^{2}\"}, {\"text\": \"=16+48-8√3 R+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"R^{2}\"}, {\"text\": \"⇒ 0=64-8√3 R ⇒ R=64/(8√3)=8/√3=8√3/3。\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [30.0, 387.9, 557.04, 796.15], \"file\": \"A中考/2026江苏中考期末数学试卷+答案820.pdf\", \"regions\": [{\"bbox\": [30.0, 387.9, 557.04, 796.15], \"page\": 8}, {\"bbox\": [498.1, 782.87, 557.04, 796.15], \"page\": 9}], \"section\": \"match\", \"page_end\": 9, \"page_start\": 8, \"section_label\": \"智适化匹配练习   ★ 交互式系统学习\", \"answer_regions\": [], \"answer_original\": \"\", \"solution_original\": \"\"}, \"version\": 1, \"grade_level\": \"JUNIOR\", \"number_label\": \"6\", \"subject_code\": \"math\", \"question_type\": \"single_choice\", \"content_sha256\": \"4c1556f8a248cf6550ec5c5ec84af45b9de6fe6a449f21f62869d6a2221ebeb9\"}","body_html":"<div>如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22<img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png\" style=\"width:188.9px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><br>A. 2<br>B. \\(8\\sqrt{3}\\frac{3}{3}\\)<br>C. √3<br>D. 2√2</div>","answer_html":"B","ways_html":"设圆O半径为R。正三角形ABC内接于圆O，弦BC=8。AD⊥BC于D，则D为BC中点，BD=4。在Rt△ABD中，AB=BC=8（正三角形边长相等），故AD=√(\\(AB^{2}\\)-\\(BD^{2}\\))=\\(\\sqrt{64-16}\\)=√48=4√3。圆心O在AD上（正三角形外接圆圆心在对称轴上），且AO=R，OD=AD-AO=4√3-R。在Rt△OBD中，OB=R，BD=4，OD=4√3-R，由勾股定理：\\(R^{2}\\)=\\(4^2\\)+(4√3-R)^2 ⇒\\(R^{2}\\)=16+48-8√3 R+\\(R^{2}\\)⇒ 0=64-8√3 R ⇒ R=64/(8√3)=8/√3=8√3/3。","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.913,"error_message":"QA failed: VISION","published_question_id":null,"content_sha256":"4c1556f8a248cf6550ec5c5ec84af45b9de6fe6a449f21f62869d6a2221ebeb9","created_at":"2026-09-24T21:48:30","updated_at":"2026-09-24T22:16:24","answer_missing":false,"formula_fallback":false,"_number":"6","_section":"match","_section_label":"智适化匹配练习   ★ 交互式系统学习"},"assets":[{"id":4589,"asset_uuid":"3208beb20abb4a40bee135a5957f589c","job_id":4,"import_question_id":21015,"asset_type":"embedded_image","source_file":"2026江苏中考期末数学试卷+答案820.pdf","source_page":8,"source_bbox":"[30.0, 408.52, 171.69, 519.52]","sha256":"2663b39cf5ce4ef2dc72022be14a419db4ffd7ea9a08b2728a52cb7de7bd3fb7","local_path":"/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png","asset_key":"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png","public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png","width":708,"height":545,"qa_status":"PENDING","created_at":"2026-09-24T22:04:17","safe_public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/ca1df7c7-5766-4e5d-845f-7a72124ce49e/2663b39cf5ce.png"}],"knowledge":[],"audits":[{"id":413906,"import_question_id":21015,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22 [图] A. 2 B. 8\\sqrt{3}\\frac{3}{3} C. √3 D. 2√2","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T22:13:21"},{"id":413907,"import_question_id":21015,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"8\\sqrt{3}\\frac{3}{3}","result_json":"{\"risky\": 1, \"issues\": [], \"checked\": 1, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-24T22:13:21"},{"id":413908,"import_question_id":21015,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22 [图] A. 2 B. 8\\sqrt{3}\\frac{3}{3} C. √3 D. 2√2","result_json":"{\"picks\": [], \"tree_code\": \"KT_JUNIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T22:13:21"},{"id":413909,"import_question_id":21015,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22 [图] A. 2 B. 8\\sqrt{3}\\frac{3}{3} C. √3 D. 2√2","result_json":"{\"basis\": \"设圆O半径为R。正三角形ABC内接于圆，边长BC=8。由正弦定理，BC / sin(∠BAC) = 2R。因△ABC为正三角形，∠BAC=60°，故8 / sin(60°) = 2R。sin(60°)=√3/2，代入得8/(√3/2)=2R ⇒ 16/√3=2R ⇒ R=8/√3=8√3/3。\", \"answer\": \"B\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-24T22:13:42"},{"id":413910,"import_question_id":21015,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案: || 盲解:B","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-24T22:13:42"},{"id":413911,"import_question_id":21015,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.95,"input_snapshot":"题干：\n如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22 A. 2 B. \\(8\\sqrt{3}\\frac{3}{3}\\) C. √3 D. 2√2\n答案：\n\n解析：\n","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.95, \"suggestions\": [\"题干中‘且于AD垂直’表述略含糊，建议明确为‘AD⊥BC’或‘AD垂直于BC’，以避免歧义。\", \"选项B的排版‘8√3 3/3’易被误读，建议写作‘\\\\frac{8\\\\sqrt{3}}{3}’并加括号或说明，确保清晰。\"]}","created_at":"2026-09-24T22:14:05"},{"id":413912,"import_question_id":21015,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.5,"input_snapshot":"答案: || 盲解:B","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"未提取到来源答案\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-24T22:14:05"},{"id":413913,"import_question_id":21015,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.8,"input_snapshot":"答案: || 解析:","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析缺失，无法验证推导过程；答案选项B格式混乱（8√3与3/3并列），疑似排版错误；题干中‘且于AD垂直’表述不清，缺少主语或条件说明。\"], \"confidence\": 0.8, \"equation_checks\": []}","created_at":"2026-09-24T22:14:16"},{"id":413914,"import_question_id":21015,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.913,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"COHERENCE\", \"ANSWER\"]}","created_at":"2026-09-24T22:14:16"},{"id":413917,"import_question_id":21015,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22 [图] A. 2 B. 8\\sqrt{3}\\frac{3}{3} C. √3 D. 2√2","result_json":"{\"basis\": \"设圆O半径为R。正三角形ABC内接于圆，边长BC=8。由正弦定理，BC / sin(∠BAC) = 2R。正三角形内角∠BAC=60°，sin60°=√3/2，故8 / (√3/2) = 2R → R = 8√3 / 3。\", \"answer\": \"B\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-24T22:15:33"},{"id":413918,"import_question_id":21015,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案:B || 盲解:B","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-24T22:15:33"},{"id":413919,"import_question_id":21015,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"PASS","score":1.0,"input_snapshot":"答案:B || 解析:设圆O半径为R。正三角形ABC内接于圆O，弦BC=8。AD⊥BC于D，则D为BC中点，BD=4。在Rt△ABD中，AB=BC=8（正三角形边长相等），故AD=√(\\(AB^{2}\\)-\\(BD^{2}\\))=\\(\\sqrt{64-16}\\)=√48=4√3。圆心O在AD上（正三角形外接圆圆心在对称轴上），且AO=R，OD=AD-AO=4√3-R。在Rt△OBD中，OB=R，BD=4，OD=4√3-","result_json":"{\"ok\": true, \"which\": [], \"issues\": [], \"confidence\": 1.0, \"equation_checks\": [{\"valid\": true, \"reason\": \"在Rt△ABD中，AB=8，BD=4，勾股定理计算正确。\", \"equation\": \"AD = \\\\sqrt{AB^2 - BD^2} = \\\\sqrt{64 - 16} = \\\\sqrt{48} = 4\\\\sqrt{3}\"}, {\"valid\": true, \"reason\": \"圆心O在AD上，AO=R，故OD=AD-R，正确。\", \"equation\": \"OD = AD - AO = 4\\\\sqrt{3} - R\"}, {\"valid\": true, \"reason\": \"在Rt△OBD中，OB=R，BD=4，OD=4\\\\sqrt{3}-R，勾股定理列式正确。\", \"equation\": \"R^2 = 4^2 + (4\\\\sqrt{3} - R)^2\"}, {\"valid\": true, \"reason\": \"展开(4\\\\sqrt{3}-R)^2 = 48 - 8\\\\sqrt{3}R + R^2，正确。\", \"equation\": \"R^2 = 16 + 48 - 8\\\\sqrt{3}R + R^2\"}, {\"valid\": true, \"reason\": \"两边消去R^2，得16+48=64，移项正确。\", \"equation\": \"0 = 64 - 8\\\\sqrt{3}R\"}, {\"valid\": true, \"reason\": \"代数运算正确，有理化分母正确。\", \"equation\": \"R = 64/(8\\\\sqrt{3}) = 8/\\\\sqrt{3} = 8\\\\sqrt{3}/3\"}]}","created_at":"2026-09-24T22:15:54"},{"id":413920,"import_question_id":21015,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"PASS","score":1.0,"input_snapshot":"答案:B || 盲解:B","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案一致\"}, \"status\": \"PASS\", \"applicable\": true}","created_at":"2026-09-24T22:15:54"},{"id":413921,"import_question_id":21015,"audit_type":"AUTO_REPAIR_ATTEMPT","auditor_name":"answer-repair","status":"PASS","score":1.0,"input_snapshot":null,"result_json":"{\"critic\": true, \"symbolic\": true, \"coherence\": true, \"consensus\": \"two_solvers+symbolic+critic+coherence\", \"confidence\": 1.0, \"verifier_base\": \"http://127.0.0.1:8091/v1\", \"verifier_confidence\": 1.0, \"critic_explicit_pass\": true, \"symbolic_literal_warn\": false, \"symbolic_explicit_pass\": true, \"coherence_explicit_pass\": true}","created_at":"2026-09-24T22:15:54"},{"id":413922,"import_question_id":21015,"audit_type":"AUTO_REPAIR","auditor_name":"answer-repair","status":"PASS","score":1.0,"input_snapshot":null,"result_json":"{\"basis\": \"设圆O半径为R。正三角形ABC内接于圆，边长BC=8。由正弦定理，BC / sin(∠BAC) = 2R。因△ABC为正三角形，∠BAC=60°，故8 / sin(60°) = 2R。sin(60°)=√3/2，代入得8/(√3/2)=2R ⇒ 16/√3=2R ⇒ R=8/√3=8√3/3。\", \"answer\": \"B\", \"critic\": {\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}, \"reason\": \"当前题干与来源答案冲突，使用高置信盲解修复并通过二次校验\", \"symbolic\": {\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案一致\"}, \"status\": \"PASS\", \"applicable\": true}, \"coherence\": {\"ok\": true, \"which\": [], \"issues\": [], \"confidence\": 1.0, \"equation_checks\": [{\"valid\": true, \"reason\": \"在Rt△ABD中，AB=8，BD=4，勾股定理计算正确。\", \"equation\": \"AD = \\\\sqrt{AB^2 - BD^2} = \\\\sqrt{64 - 16} = \\\\sqrt{48} = 4\\\\sqrt{3}\"}, {\"valid\": true, \"reason\": \"圆心O在AD上，AO=R，故OD=AD-R，正确。\", \"equation\": \"OD = AD - AO = 4\\\\sqrt{3} - R\"}, {\"valid\": true, \"reason\": \"在Rt△OBD中，OB=R，BD=4，OD=4\\\\sqrt{3}-R，勾股定理列式正确。\", \"equation\": \"R^2 = 4^2 + (4\\\\sqrt{3} - R)^2\"}, {\"valid\": true, \"reason\": \"展开(4\\\\sqrt{3}-R)^2 = 48 - 8\\\\sqrt{3}R + R^2，正确。\", \"equation\": \"R^2 = 16 + 48 - 8\\\\sqrt{3}R + R^2\"}, {\"valid\": true, \"reason\": \"两边消去R^2，得16+48=64，移项正确。\", \"equation\": \"0 = 64 - 8\\\\sqrt{3}R\"}, {\"valid\": true, \"reason\": \"代数运算正确，有理化分母正确。\", \"equation\": \"R = 64/(8\\\\sqrt{3}) = 8/\\\\sqrt{3} = 8\\\\sqrt{3}/3\"}]}}","created_at":"2026-09-24T22:15:54"},{"id":413923,"import_question_id":21015,"audit_type":"VISION","auditor_name":"vision-auditor","status":"WARN","score":0.2,"input_snapshot":"当前答案解析经过独立复算修正了来源错误。题干、条件、选项和配图仍必须忠实原图；答案解析与来源不同时须独立验证修正是否成立，不能仅因不同就否决，也不能因已修正就通过。\n题干：\n如图, 正三角形ABC是圆O的内接三角形, 弦BC = 8, 且于AD垂直, 则圆O的半径等于( ) 2026-06-22 2026-06-22 A. 2 B. \\(8\\sqrt{3}\\frac{3}{3}\\) C. √3 D. 2√2\n答案：\nB\n解析：\n设圆O半径为R。正三角形ABC内接于圆O，弦BC=8。AD⊥BC于D，则D为BC中点，BD=4。在Rt△ABD中，AB=BC=8（正三角形边长相等），故AD=√(\\(AB^{2}\\)-\\(BD^{2}\\))=\\(\\sqrt{64-16}\\)=√48=4√3。圆心O在AD上（正三角形外接圆圆心在对称轴上），且AO=R，OD=AD-AO=4√3-R。在Rt△OBD中，OB=R，BD=4，OD=4√3-R，由勾股定理：\\(R^{2}\\)=\\(4^2\\)+(4√3-R)^2 ⇒\\(R^{2}\\)=16+48-8√3 R+\\(R^{2}\\)⇒ 0=64-8√3 R ⇒ R=64/(8√3)=8/√3=8√3/3。","result_json":"{\"note\": \"视觉比对不可用\", \"error\": \"invalid_json_root\", \"unavailable\": true}","created_at":"2026-09-24T22:16:24"}]}