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5b:T679,Положим $ \alpha = \angle BCA $, $ \beta = \angle ABD $. Тогда $ \angle BAE = 2\beta $. Обозначим через $ F $ точку пересечения $ AE $ и $ BD $ и через $ O $ — центр окружности, описанной около треугольника $ ABC $. Тогда $ \angle AOB = 2\alpha $. Поскольку точки $ A $, $ D $, $ O $, $ E $, $ B $ лежат на окружности, $ \angle ADB = \angle AEB = \angle AOB = 2\alpha $. Далее, из четырёхугольника $ CDFE $ \[ \angle DFE = 360^\circ - \angle FDC - \angle DCE - \angle CEF = 360^\circ - (180^\circ - \angle ADF) - \angle DCE - (180^\circ - \angle BEF) = \angle ADF - \angle DCE + \angle BEF = 2\alpha - \alpha + 2\alpha = 3\alpha. \] Поскольку же угол $ DFE $ является внешним углом треугольников $ BEF $ и $ ADF $, получаем $ \angle FBE = 3\alpha - 2\alpha = \alpha $ и $ \angle FAD = 3\alpha - 2\alpha = \alpha $. Стало быть, сумма углов треугольника $ ABC $ равна $ \alpha + (\alpha + \beta) + (\alpha + 2\beta) = 3(\alpha + \beta) $. Откуда следует, что $ \angle ABC = \alpha + \beta = \frac{1}{3} \cdot 180^\circ = 60^\circ $. Применяя теорему синусов, получаем, что $ \sin \angle BCA = \sin \angle ABC \cdot AB/AC = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{1}{\sqrt{2}} $. Поскольку же $ \alpha < \alpha + \beta = 60^\circ $, заключаем, что $ \alpha = 45^\circ $.41:[["$","script",null,{"id":"collection-schema","type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Course\",\"name\":\"ДВИ МГУ 2022 — вариант 225\",\"description\":\"Коллекция \\\"ДВИ МГУ 2022 — вариант 225\\\" для подготовки к экзамену. Содержит 7 задач. 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\\dfrac{13}{18} \$.","answer_text":"\$4\$","answer":"\$4\$","solution_text":"\\[ \\left( \\frac{2\\sqrt{2}}{27} \\right)^{2/3} + \\left( \\frac{27}{2\\sqrt{2}} \\right)^{2/3} - \\frac{13}{18} = \\frac{2}{9} + \\frac{9}{2} - \\frac{13}{18} = \\frac{4 + 81 - 13}{18} = \\frac{72}{18} = 4. \\]","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_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"6d33b13a-8bf1-4ff0-846e-908c093f56fc","code":"tag-98","name":"Преобразования выражений, включающих операцию возведения в степень"},{"id":"e32afd53-73a2-409e-80f1-08c7daeb1e7f","code":"tag-219","name":"Свойства степени с действительным показателем"}]}},{"problem_id":"f544958c-aecc-40da-a818-5d6aa5ff3136","collection_id":"942c14e8-6d80-4e81-bace-c21913394659","display_order":2,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"f544958c-aecc-40da-a818-5d6aa5ff3136","short_id":17593,"task_number":15,"topic_id":"df51d0dd-1877-4622-8eb8-50a8151bbd7d","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":3,"content_text":"Сумма второго и восьмого членов возрастающей геометрической прогрессии равна \$ 9\\sqrt{2} \$. Произведение четвёртого, пятого и шестого членов этой прогрессии равно 64. Найдите разность между девятым и первым членами этой прогрессии.","answer_text":"\$15\$","answer":"\$15\$","solution_text":"Заметим, что поскольку прогрессия возрастает, её знаменатель положителен. Следовательно, все члены прогрессии имеют одинаковые знаки. Поскольку же сумма второго и восьмого членов положительна, все члены прогрессии являются положительными числами. Пусть \$ b_1, b_2, \\ldots, b_9 \$ — исследуемая прогрессия. Тогда \$ b_2 + b_8 = 9\\sqrt{2} \$ и \$ b_4 b_5 b_6 = 64 \$. Поскольку \$ b_2 b_8 = b_4 b_6 = b_5^2 \$, получаем \$ b_2 b_8 = 64^{2/3} = 16 \$. Стало быть, либо \$ b_2 = \\sqrt{2} \$, \$ b_8 = 8\\sqrt{2} \$, либо \$ b_2 = 8\\sqrt{2} \$, \$ b_8 = \\sqrt{2} \$. Поскольку \$ b_8 > b_2 \$, справедлив первый вариант. Знаменатель прогрессии равен \$ (b_8/b_2)^{1/6} = 8^{1/6} = \\sqrt{2} \$. Тогда \$ b_1 = 1 \$ и \$ b_9 = 16 \$. Искомая разность равна 15.","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_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"df51d0dd-1877-4622-8eb8-50a8151bbd7d","name":"Прогрессии","description":null},"tags":[{"id":"e7c29632-1fe7-4246-8016-d9ef8e73c2c8","code":"tag-255","name":"Разбор случаев"},{"id":"ec133604-ce66-4c15-84f7-9262bb200a7a","code":"tag-227","name":"Задачи на прогрессии"}]}},{"problem_id":"6de57076-e16e-460e-a6f6-82b7653f35e9","collection_id":"942c14e8-6d80-4e81-bace-c21913394659","display_order":3,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"6de57076-e16e-460e-a6f6-82b7653f35e9","short_id":17594,"task_number":17,"topic_id":"56ee582a-3c52-4fd3-85c9-966671ec7454","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":4,"content_text":"Решите уравнение \$ \\sin^4 x + \\sin\\left( x + \\dfrac{\\pi}{3} \\right) \\sin\\left( x - \\dfrac{\\pi}{3} \\right) = 0 \$.","answer_text":"\$x = \\pi/4 + k\\pi/2,\\ k \\in \\mathbb{Z}\$","answer":"\$x = \\pi/4 + k\\pi/2,\\ k \\in \\mathbb{Z}\$","solution_text":"\\[ \\sin^4 x + \\sin\\left( x + \\frac{\\pi}{3} \\right) \\sin\\left( x - \\frac{\\pi}{3} \\right) = 0 \\iff \\sin^4 x + \\frac{1}{2}\\left( \\cos\\frac{2\\pi}{3} - \\cos 2x \\right) = 0 \\iff 4\\sin^4 x - 1 - 2\\cos 2x = 0 \\iff 4\\sin^4 x + 4\\sin^2 x - 3 = 0 \\iff (2\\sin^2 x - 1)(2\\sin^2 x + 3) = 0 \\iff \\sin x = \\pm\\frac{1}{\\sqrt{2}} \\iff x = \\pi/4 + 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":3,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"6352f8cc-638a-4f87-beeb-1608b2daf4f4","code":"tag-249","name":"Замена переменной"},{"id":"9f2c32f9-1c81-45d4-bfd9-70a4620bebf8","code":"tag-142","name":"Тригонометрические уравнения, решаемые разложением на множители"}]}},{"problem_id":"41e77a50-f568-4fa2-ad21-34dab6a7223d","collection_id":"942c14e8-6d80-4e81-bace-c21913394659","display_order":4,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"41e77a50-f568-4fa2-ad21-34dab6a7223d","short_id":17595,"task_number":16,"topic_id":"72c9fcf0-a99f-4128-b5bb-edf7f779b53c","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":6,"content_text":"Решите неравенство \$ \\log_{\\sqrt{6-x}}(6+x) + \\log_{\\sqrt{6+x}}(6-x) \\leqslant 5 \$.","answer_text":"\$x \\in (-6, -5) \\cup [-3, 3] \\cup (5, 6)\$","answer":"\$x \\in (-6, -5) \\cup [-3, 3] \\cup (5, 6)\$","solution_text":"\\[ \\log_{\\sqrt{6-x}}(6+x) + \\log_{\\sqrt{6+x}}(6-x) \\leqslant 5 \\iff \\frac{2\\ln(6+x)}{\\ln(6-x)} + \\frac{2\\ln(6-x)}{\\ln(6+x)} \\leqslant 5 \\iff \\frac{2\\ln^2(6+x) - 5\\ln(6+x)\\ln(6-x) + 2\\ln^2(6-x)}{\\ln(6+x)\\ln(6-x)} \\leqslant 0 \\iff \\frac{\\left( \\ln\\left( (6+x)^2 \\right) - \\ln(6-x) \\right)\\left( \\ln(6+x) - \\ln\\left( (6-x)^2 \\right) \\right)}{\\ln(6+x)\\ln(6-x)} \\leqslant 0 \\iff \\begin{cases} -6 < x < 6 \\\\ \\dfrac{\\left( (6+x)^2 - (6-x) \\right)\\left( (6+x) - (6-x)^2 \\right)}{(5+x)(5-x)} \\leqslant 0 \\end{cases} \\iff \\begin{cases} -6 < x < 6 \\\\ \\dfrac{(x^2 + 13x + 30)(-x^2 + 13x - 30)}{(5+x)(5-x)} \\leqslant 0 \\end{cases} \\iff \\begin{cases} -6 < x < 6 \\\\ \\dfrac{(x+10)(x+3)(x-10)(x-3)}{(x+5)(x-5)} \\leqslant 0 \\end{cases} \\iff \\begin{cases} -6 < x < 6 \\\\ \\left[ \\begin{array}{l} -10 \\leqslant x < -5 \\\\ -3 \\leqslant x \\leqslant 3 \\\\ 5 < x \\leqslant 10 \\end{array} \\right. \\end{cases} \\iff \\left[ \\begin{array}{l} -6 < x < -5 \\\\ -3 \\leqslant x \\leqslant 3 \\\\ 5 < x < 6 \\end{array} \\right. \\]","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_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"2c55495d-5b54-461b-8a9b-3567966b9d28","code":"tag-40","name":"Метод интервалов"},{"id":"6dd5bf32-95ce-48b6-924b-b4ffe052f767","code":"tag-250","name":"Логарифм с переменным основанием"},{"id":"e222ad96-7947-470e-8af5-8c116d9e6e84","code":"tag-256","name":"Рационализация в неравенствах"}]}},{"problem_id":"b4fd9d4d-5789-44a6-ac9f-f97e7a68e4bc","collection_id":"942c14e8-6d80-4e81-bace-c21913394659","display_order":5,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"b4fd9d4d-5789-44a6-ac9f-f97e7a68e4bc","short_id":17596,"task_number":18,"topic_id":"1148d06d-7ecc-4ba9-b2d9-947ac91b9c9d","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":7,"content_text":"Окружность, проходящая через вершины \$ A \$, \$ B \$ треугольника \$ ABC \$ и центр описанной около этого треугольника окружности, пересекает стороны \$ AC \$ и \$ BC \$ в точках \$ D \$ и \$ E \$ соответственно. Найдите угол \$ \\angle BCA \$, если известно, что \$ AB = \\sqrt{2} \$, \$ AC = \\sqrt{3} \$ и что угол \$ \\angle BAE \$ в два раза больше угла \$ \\angle ABD \$.","answer_text":"\$45^\\circ\$","answer":"\$45^\\circ\$","solution_text":"$5b","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_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"6918af04-1043-4bc4-abbf-b4bed84c5797","code":"tag-248","name":"Дополнительные построения"},{"id":"c289978d-0619-415d-b931-ca8286c4546b","code":"tag-182","name":"Окружности и треугольники"}]}},{"problem_id":"f3a7e749-c3bf-4333-b01a-a4fce69c4e5a","collection_id":"942c14e8-6d80-4e81-bace-c21913394659","display_order":6,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"f3a7e749-c3bf-4333-b01a-a4fce69c4e5a","short_id":17597,"task_number":19,"topic_id":"7ee08b18-18e8-42f5-8ad4-c094688c771d","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":8,"content_text":"Найдите все значения параметров \$ a, b \$, при которых неравенство\n\\[ a^3 x^4 + 2ax^3 + b \\leqslant 2bx^2 + b^3 x + a \\]\nвыполняется для всех \$ x \$ из отрезка \$ [0, 1] \$.","answer_text":"\$a = b \\geqslant 0\$","answer":"\$a = b \\geqslant 0\$","solution_text":"Подставим \$ x = 0 \$ и \$ x = 1 \$. Получим \$ b \\leqslant a \$ и \$ a^3 + a \\leqslant b^3 + b \$. Из последнего следует, что \$ a \\leqslant b \$, ибо функция \$ f(x) = x^3 + x \$ возрастает. Стало быть, \$ a = b \$. Далее,\n\\[ a^3 x^4 + 2ax^3 + a \\leqslant 2ax^2 + a^3 x + a \\iff a^3 x^4 + 2ax^3 - 2ax^2 - a^3 x \\leqslant 0 \\iff ax\\left( a^2(x^3 - 1) + 2(x^2 - x) \\right) \\leqslant 0 \\iff ax(x - 1)\\left( a^2 x^2 + (a^2 + 2)x + a^2 \\right) \\leqslant 0. \\]\nПри \$ 0 < x < 1 \$ справедливо \$ x > 0 \$, \$ x - 1 < 0 \$, \$ a^2 x^2 + (a^2 + 2)x + a^2 > 0 \$. Стало быть, исходное неравенство выполняется для всех \$ x \$ из отрезка \$ [0, 1] \$ тогда и только тогда, когда \$ a = b \\geqslant 0 \$.","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_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"7ee08b18-18e8-42f5-8ad4-c094688c771d","name":"Задачи с параметрами","description":null},"tags":[{"id":"dd8fa188-ef7b-495c-b5b1-fc5dd69ca555","code":"tag-253","name":"Полные квадраты и факторизация"},{"id":"e7c29632-1fe7-4246-8016-d9ef8e73c2c8","code":"tag-255","name":"Разбор случаев"}]}},{"problem_id":"58622512-2856-46f8-92bc-18c47c6ae652","collection_id":"942c14e8-6d80-4e81-bace-c21913394659","display_order":7,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"58622512-2856-46f8-92bc-18c47c6ae652","short_id":17598,"task_number":20,"topic_id":"57632be7-7c51-406a-8a6f-58e0708032ea","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":6,"content_text":"Дана правильная треугольная пирамида \$ ABCS \$ с основанием \$ ABC \$ и вершиной \$ S \$. Плоскость \$ \\pi \$ перпендикулярна ребру \$ AS \$ и пересекает рёбра \$ AS \$, \$ BS \$ в точках \$ D \$, \$ E \$ соответственно. Известно, что \$ SD = AD \$ и \$ SE = 2BE \$. Найдите косинус угла между ребром \$ AS \$ и плоскостью основания \$ ABC \$.","answer_text":"\$1/\\sqrt{6}\$","answer":"\$1/\\sqrt{6}\$","solution_text":"Пусть \$ a \$ — длина ребра основания и \$ b \$ — длина бокового ребра. В прямоугольном треугольнике \$ SDE \$ имеем \$ SD = \\frac{1}{2}b \$ и \$ SE = \\frac{2}{3}b \$. Стало быть, \$ \\cos \\angle ASB = \\frac{3}{4} \$. Применяя теорему косинусов к треугольнику \$ ASB \$, получаем, что \$ a^2 = 2b^2 - 2b^2 \\cdot \\frac{3}{4} \$, откуда \$ b = a\\sqrt{2} \$. Пусть \$ O \$ — центр основания. Тогда в прямоугольном треугольнике \$ ASO \$ имеем \$ AS = b = a\\sqrt{2} \$ и \$ AO = a/\\sqrt{3} \$. Стало быть, \$ \\cos \\angle SAO = AO/AS = 1/\\sqrt{6} \$.","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_2022_ans.pdf","variant_key":"2022-v5","variant_label":"225","year":2022},"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":"57632be7-7c51-406a-8a6f-58e0708032ea","name":"Расстояния, углы, координаты и векторы","description":null},"tags":[{"id":"287f11b6-41e9-44fe-8b77-6768acbb9ffe","code":"tag-35","name":"Угол между прямой и плоскостью"},{"id":"c2f87176-38e1-4180-99e4-f61bcd4dbdb7","code":"tag-183","name":"Правильная треугольная пирамида"}]}}],"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. 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