{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"omni-math","formal_name":"Omni-MATH","introduction":"Omni-MATH evaluates mathematical reasoning on Olympiad-level problems. Its official dataset contains 4,428 problems accompanied by domain and difficulty information.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/KbsdJames/Omni-MATH","indexing_mode":"noindex","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[]}},"task_id":"e8387dd5-3fea-51e7-ad32-dc86cd5db0bf","task_key":"test--e8387dd5-3fea-51e7-ad32-dc86cd5db0bf","task_revision_id":"4","upstream_id":"","short_description":"In a right angled-triangle $ABC$, $\\angle{ACB} = 90^o$. Its incircle $O$ meets…","config":"","split":"test","body":"{\"problem\":\"In a right angled-triangle $ABC$, $\\\\angle{ACB} = 90^o$. Its incircle $O$ meets $BC$, $AC$, $AB$ at $D$,$E$,$F$ respectively. $AD$ cuts $O$ at $P$. If $\\\\angle{BPC} = 90^o$, prove $AE + AP = PD$.\"}","display_format":"text","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/KbsdJames/Omni-MATH","history":"initial import","indexing_mode":"noindex","subproblems":[],"grids":[]}