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5b:T564,Обозначим через $S$ вершину пирамиды и через $H$ центр основания. Пусть $AB$ — одно из рёбер основания и $M$ — его середина. По условию $\angle ASB = \angle SAH$. Обозначим этот угол $\varphi$. Тогда \[\angle SAM=\frac{\pi}{2}-\angle ASM=\frac{\pi}{2}-\frac{1}{2}\angle ASB=\frac{\pi}{2}-\frac{1}{2}\varphi.\] С другой стороны, \[\cos\angle SAM=\frac{AM}{AS}=\frac{AM}{AH}\cdot\frac{AH}{AS}=\cos\angle HAM\cdot\cos\angle SAH=\frac{1}{\sqrt{2}}\cos\varphi.\] Получаем $\cos\varphi=\sqrt{2}\sin\frac{\varphi}{2}$, т.е. \[2\sin^2\frac{\varphi}{2}+\sqrt{2}\sin\frac{\varphi}{2}-1=0.\] Отсюда $\sin\frac{\varphi}{2}=\frac{-\sqrt{2}+\sqrt{10}}{4}$ (ибо $\frac{-\sqrt{2}-\sqrt{10}}{4}<-1$), $\cos\varphi=\sqrt{2}\sin\frac{\varphi}{2}=\frac{\sqrt{5}-1}{2}$, $\sin\varphi=\sqrt{\frac{\sqrt{5}-1}{2}}$ (берём положительный, ибо $0<\varphi<\frac{\pi}{2}$), $\operatorname{tg}\varphi=\sqrt{\frac{2}{\sqrt{5}-1}}=\sqrt{\frac{\sqrt{5}+1}{2}}$. Стало быть, объём пирамиды равен \[\frac{1}{3}AB^2 SH=\frac{1}{3}AB^2 AH\operatorname{tg}\varphi=\frac{1}{3\sqrt{2}}AB^3\operatorname{tg}\varphi=\frac{1}{3\sqrt{2}}\sqrt{\frac{\sqrt{5}+1}{2}}=\frac{\sqrt{1+\sqrt{5}}}{6}.\]41:[["$","script",null,{"id":"collection-schema","type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Course\",\"name\":\"ДВИ МГУ 2023 — вариант 233\",\"description\":\"Коллекция \\\"ДВИ МГУ 2023 — вариант 233\\\" для подготовки к экзамену. Содержит 7 задач. 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\$\\dfrac{3}{\\sqrt[4]{16}}+\\dfrac{5}{\\sqrt[3]{27}}+\\dfrac{11}{\\sqrt{36}}\$.","answer_text":"\$5\$","answer":"\$5\$","solution_text":"\\[\\dfrac{3}{\\sqrt[4]{16}}+\\dfrac{5}{\\sqrt[3]{27}}+\\dfrac{11}{\\sqrt{36}}=\\dfrac{3}{2}+\\dfrac{5}{3}+\\dfrac{11}{6}=\\dfrac{9+10+11}{6}=\\dfrac{30}{6}=5.\\]","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":1,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"a4c5b1b9-850a-422b-b0b4-723ad67fab1c","name":"Вычисления и оценки","description":null},"tags":[{"id":"da67531d-0011-489f-adab-84744c7aa977","code":"tag-208","name":"Преобразования выражений, включающих корни натуральной 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Найдите \$a_{2023}\$.","answer_text":"\$2068\$","answer":"\$2068\$","solution_text":"Для каждых натуральных \$n, m\$, таких что \$m^2 < n < (m+1)^2\$, справедливо \$n = a_{n-m}\$. Стало быть, для каждого \$n\$, удовлетворяющего условию \$45^2 = 2025 < n < 2116 = 46^2\$, справедливо \$n = a_{n-45}\$. Поскольку \$n - 45 = 2023\$ при \$n = 2068\$, получаем \$a_{2023} = 2068\$.","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":2,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"aaf1e2e8-a5c8-4dae-880f-0853b25f4b34","name":"Последовательности","description":null},"tags":[{"id":"a564b21e-7ea8-4bd4-a203-fbb13b2c7d7c","code":"tag-152","name":"Числа и их свойства"},{"id":"cac715c7-756d-4aa2-91ef-8ac2fd2653b6","code":"tag-193","name":"Последовательности и прогрессии"}]}},{"problem_id":"bbad6ca4-99a0-426e-9b14-6c125e554339","collection_id":"9d74f424-e3ac-4383-975f-1156e9bfbc4b","display_order":3,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"bbad6ca4-99a0-426e-9b14-6c125e554339","short_id":17622,"task_number":16,"topic_id":"72c9fcf0-a99f-4128-b5bb-edf7f779b53c","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":5,"content_text":"Решите неравенство\n\\[\\log_{\\sqrt{3-x}}\\left(3+x\\right)\\leqslant 2.\\]","answer_text":"\$x \\in (-3, 0] \\cup (2, 3)\$","answer":"\$x \\in (-3, 0] \\cup (2, 3)\$","solution_text":"\\[\\log_{\\sqrt{3-x}}\\left(3+x\\right)\\leqslant 2 \\iff \\log_{3-x}\\left(3+x\\right)\\leqslant 1 \\iff\\]\n\\[\\iff\\left[\\begin{array}{l}\\begin{cases}3-x>1\\\\ 0<3+x\\leqslant 3-x\\end{cases}\\\\ \\begin{cases}0<3-x<1\\\\ 3+x\\geqslant 3-x\\end{cases}\\end{array}\\right.\\iff\\left[\\begin{array}{l}\\begin{cases}x<2\\\\ x>-3\\\\ x\\leqslant 0\\end{cases}\\\\ \\begin{cases}x<3\\\\ x>2\\\\ x\\geqslant 0\\end{cases}\\end{array}\\right.\\iff x\\in(-3,0]\\cup(2,3).\\]","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":3,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"72c9fcf0-a99f-4128-b5bb-edf7f779b53c","name":"Логарифмические неравенства","description":null},"tags":[{"id":"a13da04a-1585-4241-aa6f-14021c064334","code":"tag-145","name":"Неравенства с логарифмами по переменному основанию"},{"id":"e7c29632-1fe7-4246-8016-d9ef8e73c2c8","code":"tag-255","name":"Разбор случаев"}]}},{"problem_id":"8e32e351-3e16-4343-970b-6b8226075ad7","collection_id":"9d74f424-e3ac-4383-975f-1156e9bfbc4b","display_order":4,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"8e32e351-3e16-4343-970b-6b8226075ad7","short_id":17623,"task_number":17,"topic_id":"56ee582a-3c52-4fd3-85c9-966671ec7454","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":5,"content_text":"Решите уравнение\n\\[\\cos^4 x-\\cos\\left(x+\\frac{\\pi}{3}\\right)\\cos\\left(x-\\frac{\\pi}{3}\\right)=2\\sin\\left(x+\\frac{\\pi}{6}\\right)\\sin\\left(x-\\frac{\\pi}{6}\\right).\\]","answer_text":"\$x=\\frac{\\pi}{4}+\\frac{k\\pi}{2},\\ k\\in\\mathbb{Z}\$","answer":"\$x=\\frac{\\pi}{4}+\\frac{k\\pi}{2},\\ k\\in\\mathbb{Z}\$","solution_text":"\\[\\cos^4 x-\\cos\\left(x+\\frac{\\pi}{3}\\right)\\cos\\left(x-\\frac{\\pi}{3}\\right)=2\\sin\\left(x+\\frac{\\pi}{6}\\right)\\sin\\left(x-\\frac{\\pi}{6}\\right)\\iff\\]\n\\[\\iff\\cos^4 x-\\cos\\left(x+\\frac{\\pi}{3}\\right)\\cos\\left(x-\\frac{\\pi}{3}\\right)=2\\cos\\left(\\frac{\\pi}{3}-x\\right)\\cos\\left(\\frac{2\\pi}{3}-x\\right)\\iff\\]\n\\[\\iff\\cos^4 x-\\cos\\left(x+\\frac{\\pi}{3}\\right)\\cos\\left(x-\\frac{\\pi}{3}\\right)=-2\\cos\\left(x-\\frac{\\pi}{3}\\right)\\cos\\left(x+\\frac{\\pi}{3}\\right)\\iff\\]\n\\[\\iff\\cos^4 x+\\cos\\left(x+\\frac{\\pi}{3}\\right)\\cos\\left(x-\\frac{\\pi}{3}\\right)=0\\iff\\cos^4 x+\\frac{1}{2}\\left(\\cos 2x+\\cos\\frac{2\\pi}{3}\\right)=0\\iff\\]\n\\[\\iff\\cos^4 x+\\cos^2 x-\\frac{3}{4}=0\\iff\\left(\\cos^2 x-\\frac{1}{2}\\right)\\left(\\cos^2 x+\\frac{3}{2}\\right)=0\\iff\\]\n\\[\\iff\\cos x=\\pm\\frac{1}{\\sqrt{2}}\\iff x=\\frac{\\pi}{4}+\\frac{k\\pi}{2},\\ k\\in\\mathbb{Z}.\\]","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":4,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"56ee582a-3c52-4fd3-85c9-966671ec7454","name":"Тригонометрические уравнения","description":null},"tags":[{"id":"37945dd4-6787-495c-a3b9-4524691f6dd7","code":"tag-51","name":"Тригонометрические формулы для произведения функций"},{"id":"6352f8cc-638a-4f87-beeb-1608b2daf4f4","code":"tag-249","name":"Замена переменной"}]}},{"problem_id":"c0f1200d-1708-4a94-8b3f-35ae43b7fe01","collection_id":"9d74f424-e3ac-4383-975f-1156e9bfbc4b","display_order":5,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"c0f1200d-1708-4a94-8b3f-35ae43b7fe01","short_id":17624,"task_number":18,"topic_id":"1148d06d-7ecc-4ba9-b2d9-947ac91b9c9d","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":7,"content_text":"Прямая \$\\ell\$ касается окружности, описанной около треугольника \$ABC\$, в точке \$A\$. Известно, что \$AB > AC\$ и что \$AC = 1\$. На стороне \$AB\$ отмечена точка \$D\$ так, что \$AD = AC\$. Прямая, проходящая через точку \$D\$ и через центр окружности, вписанной в треугольник \$ABC\$, пересекает прямую \$\\ell\$ в точке \$E\$. Найдите длину отрезка \$AE\$.","answer_text":"\$1\$","answer":"\$1\$","solution_text":"Пусть \$I\$ — центр окружности, вписанной в треугольник \$ABC\$. Поскольку \$AD = AC\$, точка \$I\$ лежит и на биссектрисе, и на высоте треугольника \$ADC\$. Следовательно, \$\\angle ADI = \\angle ACI = \\frac{1}{2}\\angle ACB\$. Далее, поскольку \$\\ell\$ — касательная, \$\\angle EAC = \\angle ABC\$. Отсюда следует, что \$\\angle AED+\\angle ADE = 180^\\circ-\\angle DAE = 180^\\circ-(\\angle DAC+\\angle EAC) = 180^\\circ-(\\angle BAC+\\angle ABC) = \\angle ACB = 2\\angle ACI = 2\\angle ADE\$. Стало быть, \$\\angle AED = \\angle ADE\$, то есть треугольник \$AED\$ равнобедренный и \$AE = AD = AC = 1\$.","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":5,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"1148d06d-7ecc-4ba9-b2d9-947ac91b9c9d","name":"Окружность","description":null},"tags":[{"id":"017b058d-5964-4739-b710-9610e2d94723","code":"tag-257","name":"Свойства биссектрисы и медианы"},{"id":"4b1ff467-d0bf-44bd-913c-7ae8d7e220ef","code":"tag-72","name":"Окружность, вписанная в треугольник"},{"id":"d3b75ad2-c6a5-4ba6-85c8-0cbefe7f8516","code":"tag-199","name":"Окружность, описанная вокруг треугольника"}]}},{"problem_id":"c2818341-f8ab-4ac7-89f6-29afa667c855","collection_id":"9d74f424-e3ac-4383-975f-1156e9bfbc4b","display_order":6,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"c2818341-f8ab-4ac7-89f6-29afa667c855","short_id":17625,"task_number":19,"topic_id":"e463f8eb-fd15-4d94-acb7-4092de9c6b54","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":8,"content_text":"Положительные числа \$a\$, \$b\$, \$c\$ удовлетворяют соотношению\n\\[a^2+b^2+c^2=1.\\]\nНайдите наибольшее возможное значение выражения \$ab+bc\\sqrt{3}\$.","answer_text":"\$1\$","answer":"\$1\$","solution_text":"В силу неравенства между средним арифметическим и средним геометрическим\n\\[ab+bc\\sqrt{3}=\\sqrt{a^2b^2}+\\sqrt{3b^2c^2}=2\\cdot\\sqrt{a^2\\cdot\\frac{1}{4}b^2}+2\\cdot\\sqrt{\\frac{3}{4}b^2\\cdot c^2}\\leqslant\\left(a^2+\\frac{1}{4}b^2\\right)+\\left(\\frac{3}{4}b^2+c^2\\right)=a^2+b^2+c^2=1.\\]\nПри этом равенство достигается при \$a^2=\\frac{1}{4}b^2\$ и \$\\frac{3}{4}b^2=c^2\$, то есть при \$a=\\frac{1}{2}b\$ и \$c=\\frac{\\sqrt{3}}{2}b\$. При выполнении этих равенств имеем \$1=a^2+b^2+c^2=2b^2\$, откуда \$b=\\frac{1}{\\sqrt{2}}\$, \$a=\\frac{1}{2\\sqrt{2}}\$, \$c=\\frac{\\sqrt{3}}{2\\sqrt{2}}\$ и \$ab+bc\\sqrt{3}=\\frac{1}{4}+\\frac{3}{4}=1\$.","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":6,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"e463f8eb-fd15-4d94-acb7-4092de9c6b54","name":"Экстремальные задачи и оценки","description":null},"tags":[{"id":"33fabf53-5145-4325-a6fc-cb998f32f4d5","code":"tag-252","name":"Оценки и случай равенства"},{"id":"aba0d5e0-77b3-42f2-b014-b9857f26a702","code":"tag-246","name":"АМ-ГМ и неравенства о средних"}]}},{"problem_id":"e0ea91a7-ae61-4641-8479-35853501ebfb","collection_id":"9d74f424-e3ac-4383-975f-1156e9bfbc4b","display_order":7,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"e0ea91a7-ae61-4641-8479-35853501ebfb","short_id":17626,"task_number":20,"topic_id":"ff580604-591d-4621-bd18-2997cb5ee2a4","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":7,"content_text":"Найдите объём правильной четырёхугольной пирамиды со стороной основания равной 1, если известно, что плоские углы при вершине равны углам наклона боковых рёбер к плоскости основания.","answer_text":"\$\\frac{\\sqrt{1+\\sqrt{5}}}{6}\$","answer":"\$\\frac{\\sqrt{1+\\sqrt{5}}}{6}\$","solution_text":"$5b","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2020-2025-v1","position_in_variant":7,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2023_ans.pdf","variant_key":"2023-v2","variant_label":"233","year":2023},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"ff580604-591d-4621-bd18-2997cb5ee2a4","name":"Многогранники","description":null},"tags":[{"id":"21d0f6ea-1a09-4672-be9b-dcc5d1bc43a9","code":"tag-31","name":"Правильная четырёхугольная пирамида"},{"id":"fccaf147-c5ab-4708-b5a2-315cacf9246c","code":"tag-242","name":"Объём куба, прямоугольного параллелепипеда, пирамиды, призмы"}]}}],"total":7},"serverExams":[{"id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","code":"dvi-math","name":"ДВИ МГУ (математика)","subject_id":"d1be5154-7be0-4c8c-92d3-c486ea0a1a57","subject_name":"Математика","subject_code":"math","program_id":"1aed6121-06cc-4a5a-87f5-77de484ebb36","program_code":"dvi","exam_type":"ДВИ","description":"Дополнительное вступительное испытание МГУ им. М.В. Ломоносова по математике. С 2026 года экзамен состоит из двух частей: лёгкий тест с кратким ответом и традиционная письменная часть с полным решением.","has_part2":true,"max_task_number":20,"exam_date":"2026-07-11","metadata":{"duration_minutes":180,"max_secondary_score":100,"mock_exam_enabled":true,"part2_tasks":6,"parts":[{"answer_format":"detailed","part":1,"scale":"linear","slots_in_variant":10,"tasks":[1,2,3,4,5,6,7,8,9,10,11,12,13],"weight":30},{"answer_format":"detailed","part":2,"scale":"score_conversion","slots_in_variant":6,"tasks":[14,15,16,17,18,19,20],"weight":70}],"score_conversion":[2,30,30,35,35,35,35,35,40,40,40,40,40,45,45,45,45,45,50,50,50,50,50,55,55,55,55,55,60,60,60,60,60,65,65,65,65,65,70,70,70,70,70,75,75,75,75,75,80,80,80,80,80,85,85,85,85,85,90,90,90,90,90,95,95,95,95,95,100,100,100],"scoring_table":{"1":1,"2":1,"3":1,"4":1,"5":1,"6":1,"7":1,"8":1,"9":1,"10":1,"11":1,"12":1,"13":1,"14":10,"15":10,"16":10,"17":10,"18":10,"19":10,"20":10},"structure_2026":{"consultation_video_url":"https://disk.360.yandex.ru/d/NbP139HUZpCPDA","exam_2026":{"exam_date":"2026-07-11","note":"Реальный основной день; в образце Ш12622 тест был из 15 задач.","test_task_count":10,"written_task_count":6},"exam_portal":"https://exam.msu.ru/","official_exam_page":"https://cpk.msu.ru/exam/101","parts":["base","written"],"test_task_count":15},"task_count":7,"task_numbering":"part_relative","year":2026},"created_at":"2026-06-10T20:49:34.08312Z","updated_at":"2026-09-01T00:33:23.788959Z","problem_count":1008,"theory_count":36,"system_collection_count":190},{"id":"3a8471c5-2a11-4932-ba3e-7ab7d1ae7a1d","code":"ege-base","name":"Математика (база) ЕГЭ","subject_id":"d1be5154-7be0-4c8c-92d3-c486ea0a1a57","subject_name":"Математика","subject_code":"math","old_code":"math_base","program_id":"175d2aa1-bdcb-4df2-9710-7a61057ddd66","program_code":"ege","exam_type":"ЕГЭ","description":"Базовый ЕГЭ по математике для получения аттестата","has_part2":false,"max_task_number":21,"exam_date":"2026-06-04","metadata":{"duration_minutes":180,"max_score":21,"max_secondary_score":5,"mock_exam_enabled":true,"part1_tasks":21,"part2_tasks":0,"score_conversion":[2,2,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5],"scoring_table":{"1":1,"2":1,"3":1,"4":1,"5":1,"6":1,"7":1,"8":1,"9":1,"10":1,"11":1,"12":1,"13":1,"14":1,"15":1,"16":1,"17":1,"18":1,"19":1,"20":1,"21":1},"year":2026},"created_at":"2025-11-09T12:48:12.440483Z","updated_at":"2025-11-09T12:48:12.440483Z","problem_count":8561,"theory_count":85,"system_collection_count":111},{"id":"23f00ba7-bd19-4ca8-9349-5134d88397cf","code":"ege-prof","name":"Математика (профиль) ЕГЭ","subject_id":"d1be5154-7be0-4c8c-92d3-c486ea0a1a57","subject_name":"Математика","subject_code":"math","old_code":"math_prof","program_id":"175d2aa1-bdcb-4df2-9710-7a61057ddd66","program_code":"ege","exam_type":"ЕГЭ","description":"Профильный ЕГЭ по математике для поступающих в технические вузы","has_part2":true,"max_task_number":20,"exam_date":"2026-06-08","metadata":{"duration_minutes":235,"max_score":33,"max_secondary_score":100,"max_task_number":20,"mock_exam_enabled":true,"part1_tasks":13,"part2_tasks":7,"score_conversion":[0,6,11,17,22,27,34,40,46,52,58,64,70,72,74,76,78,80,82,84,86,88,90,92,94,95,96,97,98,99,100,100,100,100],"scoring_table":{"1":1,"2":1,"3":1,"4":1,"5":1,"6":1,"7":1,"8":1,"9":1,"10":1,"11":1,"12":1,"13":1,"14":2,"15":3,"16":2,"17":2,"18":3,"19":4,"20":4},"year":2026},"created_at":"2025-11-09T12:48:12.440483Z","updated_at":"2026-08-30T23:51:20.728833Z","problem_count":2491,"theory_count":123,"system_collection_count":255},{"id":"b0db9506-ddcf-4200-8815-7a709ea0dc4e","code":"oge-math","name":"ОГЭ (математика)","subject_id":"d1be5154-7be0-4c8c-92d3-c486ea0a1a57","subject_name":"Математика","subject_code":"math","program_id":"ff8f63b8-bfcc-47f3-9d55-bc666ee30237","program_code":"oge","exam_type":"ОГЭ","description":"Основной государственный экзамен по математике (9 класс): 25 заданий, 235 минут. 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