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5b:T59e,Положим $ t = \log_x \left( x^2 + \frac{3}{2} \right) $. Тогда, поскольку \[ t \leqslant \frac{4}{t} \iff \frac{(t-2)(t+2)}{t} \leqslant 0 \iff \left[ \begin{array}{l} 0 < t \leqslant 2 \\ t \leqslant -2 \end{array} \right. \iff \left[ \begin{array}{l} \frac{1}{t} \geqslant \frac{1}{2} \\ t \leqslant -2 \end{array} \right., \] получаем \[ \log_x \left( x^2 + \tfrac{3}{2} \right) \leqslant 4 \log_{x^2 + \frac{3}{2}}(x) \iff \left[ \begin{array}{l} \log_{x^2 + \frac{3}{2}}(x) \geqslant \frac{1}{2} \\ \log_x \left( x^2 + \tfrac{3}{2} \right) \leqslant -2 \end{array} \right. \iff \begin{cases} x > 0 \\ \left[ \begin{array}{l} \log_{x^2 + \frac{3}{2}}(x^2) \geqslant 1 \\ \log_{x^{-2}}\left( x^2 + \tfrac{3}{2} \right) \geqslant 1 \end{array} \right. \end{cases} \iff \begin{cases} x > 0 \\ \left[ \begin{array}{l} \dfrac{x^2 - \left( x^2 + \frac{3}{2} \right)}{\left( x^2 + \frac{3}{2} \right) - 1} \geqslant 0 \\ \dfrac{\left( x^2 + \frac{3}{2} \right) - x^{-2}}{x^{-2} - 1} \geqslant 0 \end{array} \right. \end{cases} \iff \begin{cases} x > 0 \\ \left[ \begin{array}{l} x \in \varnothing \\ \dfrac{x^4 + \frac{3}{2}x^2 - 1}{x^2 - 1} \leqslant 0 \end{array} \right. \end{cases} \iff \begin{cases} x > 0 \\ \dfrac{(x^2 + 2)\left( x^2 - \frac{1}{2} \right)}{x^2 - 1} \leqslant 0 \end{cases} \iff \begin{cases} x > 0 \\ \frac{1}{2} \leqslant x^2 < 1 \end{cases} \iff \frac{1}{\sqrt{2}} \leqslant x < 1. \]5c:T740,Будем считать, что $ 0 < a < 1 $, $ 0 < x < \pi/4 $, $ 0 < y < \pi/4 $. Тогда значения всех тангенсов определены и никакие знаменатели не обращаются в нуль. Домножая на произведения знаменателей и учитывая, что $ \operatorname{tg}(x+y) = \dfrac{\operatorname{tg} x + \operatorname{tg} y}{1 - \operatorname{tg} x \operatorname{tg} y} $, получаем \[ \operatorname{tg}^2\left( a(x+y) \right) + \operatorname{tg}\left( a(x+y) \right)\operatorname{tg}(x+y)\left( \operatorname{tg} x \operatorname{tg} y - 1 \right) + \operatorname{tg} x \operatorname{tg} y = 0 \iff \operatorname{tg}^2\left( a(x+y) \right) - \left( \operatorname{tg} x + \operatorname{tg} y \right)\operatorname{tg}\left( a(x+y) \right) + \operatorname{tg} x \operatorname{tg} y = 0 \iff \left( \operatorname{tg}\left( a(x+y) \right) - \operatorname{tg} x \right)\left( \operatorname{tg}\left( a(x+y) \right) - \operatorname{tg} y \right) = 0 \iff \left[ \begin{array}{l} \operatorname{tg}\left( a(x+y) \right) = \operatorname{tg} x \\ \operatorname{tg}\left( a(x+y) \right) = \operatorname{tg} y \end{array} \right. \iff \left[ \begin{array}{l} a(x+y) = x \\ a(x+y) = y \end{array} \right. \iff \left[ \begin{array}{l} y = \frac{1-a}{a}x \\ y = \frac{a}{1-a}x \end{array} \right. \] При любом $ a \in (0, 1) $, $ a \ne \frac{1}{2} $, любому $ x $, удовлетворяющему $ 0 < x < \min\left( \frac{1-a}{a}, \frac{a}{1-a} \right) \cdot \frac{\pi}{4} $, будут соответствовать два значения $ y $ в интервале $ 0 < y < \pi/4 $. Если же $ a = \frac{1}{2} $, получаем соответствие $ y = x $, которое взаимно однозначно.5d:T54e,Положим $ h = AA' $. Введём прямоугольную систему координат с центром в $ A $ и осями $ AB $, $ AD $ и $ AA' $. Тогда \[ \vec{AE} = \left( \sqrt{2},\ \frac{\sqrt{2}}{2},\ \frac{h}{2} \right), \qquad \vec{D'F} = \vec{AF} - \vec{AD'} = \left( \frac{\sqrt{2}}{2},\ \frac{\sqrt{2}}{2},\ 0 \right) - (0, \sqrt{2}, h) = \left( \frac{\sqrt{2}}{2},\ -\frac{\sqrt{2}}{2},\ -h \right). \] Ортогональность этих векторов означает, что $ 1 - \frac{1}{2} - \frac{h^2}{2} = 0 $, откуда $ h = 1 $. Стало быть, середины отрезков $ AE $ и $ D'F $ имеют координаты \[ \frac{1}{2}\vec{AE} = \left( \frac{\sqrt{2}}{2},\ \frac{\sqrt{2}}{4},\ \frac{1}{4} \right), \qquad \frac{1}{2}\left( \vec{AF} + \vec{AD'} \right) = \frac{1}{2}\left( \frac{\sqrt{2}}{2},\ \frac{3\sqrt{2}}{2},\ 1 \right) = \left( \frac{\sqrt{2}}{4},\ \frac{3\sqrt{2}}{4},\ \frac{1}{2} \right). \] Квадрат расстояния же между этими точками равен \[ \left( \frac{\sqrt{2}}{4} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{1}{4} \right)^2 = \frac{1}{8} + \frac{1}{2} + \frac{1}{16} = \frac{11}{16}. \] Стало быть, само расстояние равно $ \frac{\sqrt{11}}{4} $.41:[["$","script",null,{"id":"collection-schema","type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Course\",\"name\":\"ДВИ МГУ 2022 — вариант 226\",\"description\":\"Коллекция \\\"ДВИ МГУ 2022 — вариант 226\\\" для подготовки к экзамену. Содержит 7 задач. 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Известно, что \$ a_2 + a_4 + a_6 + \\ldots + a_{30} = 45 \$ и что \$ a_3 + a_6 + a_9 + \\ldots + a_{30} = 100 \$. Найдите разность этой прогрессии.","answer_text":"\$14\$","answer":"\$14\$","solution_text":"Пусть \$ d \$ — разность прогрессии. Тогда, во-первых, \$ 45 = 15(a_2 + a_{30})/2 = 15(2a_1 + 30d)/2 = 15(a_1 + 15d) \$, откуда \$ a_1 + 15d = 3 \$, и, во-вторых, \$ 100 = 10(a_3 + a_{30})/2 = 10(2a_1 + 31d)/2 = 5(2a_1 + 31d) \$, откуда \$ 2a_1 + 31d = 20 \$. Стало быть, \$ (2a_1 + 31d) - 2(a_1 + 15d) = 20 - 2 \\cdot 3 \$, то есть \$ d = 14 \$.","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-v6","variant_label":"226","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":"ec133604-ce66-4c15-84f7-9262bb200a7a","code":"tag-227","name":"Задачи на прогрессии"}]}},{"problem_id":"c59ea120-e7b9-4a12-955f-121fb6faac7b","collection_id":"5ca2e948-e214-437e-b157-38d51de05a43","display_order":3,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"c59ea120-e7b9-4a12-955f-121fb6faac7b","short_id":17601,"task_number":17,"topic_id":"56ee582a-3c52-4fd3-85c9-966671ec7454","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":5,"content_text":"Решите уравнение \$ \\operatorname{tg} x = 4\\sin x - \\sqrt{3} \$.","answer_text":"\$x = \\pi/3 + 2k\\pi,\\ 2\\pi/9 + \\frac{2}{3}k\\pi,\\ k \\in \\mathbb{Z}\$","answer":"\$x = \\pi/3 + 2k\\pi,\\ 2\\pi/9 + \\frac{2}{3}k\\pi,\\ k \\in \\mathbb{Z}\$","solution_text":"Заметим, что \$ \\sin x \$ и \$ \\cos x \$ одновременно в нуль не обращаются. Стало быть,\n\\[ \\operatorname{tg} x = 4\\sin x - \\sqrt{3} \\iff \\sin x = 4\\sin x \\cos x - \\sqrt{3}\\cos x \\iff \\sin 2x = \\frac{1}{2}\\sin x + \\frac{\\sqrt{3}}{2}\\cos x \\iff \\sin 2x = \\sin\\left( x + \\frac{\\pi}{3} \\right) \\iff \\left[ \\begin{array}{l} 2x = x + \\frac{\\pi}{3} + 2k\\pi \\\\ 2x = -x + \\frac{2\\pi}{3} + 2k\\pi \\end{array} \\right.,\\ k \\in \\mathbb{Z} \\iff \\left[ \\begin{array}{l} x = \\frac{\\pi}{3} + 2k\\pi \\\\ x = \\frac{2\\pi}{9} + \\frac{2}{3}k\\pi \\end{array} \\right.,\\ 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-v6","variant_label":"226","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":"9564a83d-ccf3-4211-91ed-f6f568873b0e","code":"tag-131","name":"Тригонометрические уравнения"},{"id":"a617e4cd-3e3c-4ea5-a251-0d7a26e59838","code":"tag-154","name":"Равенство тригонометрических функций"},{"id":"d3ec0e69-d667-41d1-ae0a-a79b81adcdd9","code":"tag-200","name":"Синус, косинус и тангенс суммы и разности двух углов"}]}},{"problem_id":"0a384079-114d-4f5c-8298-a4563d18ff18","collection_id":"5ca2e948-e214-437e-b157-38d51de05a43","display_order":4,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"0a384079-114d-4f5c-8298-a4563d18ff18","short_id":17602,"task_number":16,"topic_id":"72c9fcf0-a99f-4128-b5bb-edf7f779b53c","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":6,"content_text":"Решите неравенство \$ \\log_x \\left( x^2 + \\dfrac{3}{2} \\right) \\leqslant 4 \\log_{x^2 + \\frac{3}{2}}(x) \$.","answer_text":"\$x \\in \\left[ \\frac{1}{\\sqrt{2}}, 1 \\right)\$","answer":"\$x \\in \\left[ \\frac{1}{\\sqrt{2}}, 1 \\right)\$","solution_text":"$5b","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-v6","variant_label":"226","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":"6352f8cc-638a-4f87-beeb-1608b2daf4f4","code":"tag-249","name":"Замена переменной"},{"id":"6dd5bf32-95ce-48b6-924b-b4ffe052f767","code":"tag-250","name":"Логарифм с переменным основанием"}]}},{"problem_id":"7bdaf072-d6f6-4822-b87d-5b2d93d2507f","collection_id":"5ca2e948-e214-437e-b157-38d51de05a43","display_order":5,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"7bdaf072-d6f6-4822-b87d-5b2d93d2507f","short_id":17603,"task_number":18,"topic_id":"1148d06d-7ecc-4ba9-b2d9-947ac91b9c9d","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":7,"content_text":"Окружность, проходящая через вершину \$ A \$ треугольника \$ ABC \$, касается его стороны \$ BC \$ в точке \$ D \$ и пересекает стороны \$ AC \$ и \$ AB \$ в точках \$ E \$ и \$ F \$ соответственно. Известно, что \$ AF = 3BF \$, \$ BD = CD \$, \$ AE = 2CE \$ и что \$ ED = \\sqrt{10} \$. Найдите \$ BC \$.","answer_text":"\$4\\sqrt{3}\$","answer":"\$4\\sqrt{3}\$","solution_text":"Положим \$ x = BF \$, \$ y = BD \$. Тогда \$ AB = 4x \$, \$ BC = 2y \$. По теореме о касательной и секущей \$ y^2 = 4x^2 \$, то есть \$ BC = 2y = 4x = AB \$.\n\nОбозначим через \$ L \$ точку пересечения прямых \$ AB \$ и \$ ED \$. Обозначим также через \$ M \$ и \$ N \$ середины сторон \$ AC \$ и \$ AB \$ соответственно. Треугольники \$ AEL \$ и \$ MED \$ подобны. При этом \$ ME = AE - AM = AE - \\frac{1}{2}AC = AE - \\frac{1}{2} \\cdot \\frac{3}{2}AE = \\frac{1}{4}AE \$. Стало быть, \$ ED = \\frac{1}{4}EL \$, откуда \$ DL = 3ED = 3\\sqrt{10} \$. Кроме того, из того же подобия получаем, что \$ AL = 4MD = 2AB = 8x \$. Применяя ещё раз теорему о касательной и секущей, получаем \$ 8x \\cdot 5x = 4\\sqrt{10} \\cdot 3\\sqrt{10} \$, то есть \$ x = \\sqrt{3} \$. Стало быть, \$ BC = AB = 4x = 4\\sqrt{3} \$.","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-v6","variant_label":"226","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":"91f89d3c-fa9c-4ac7-8db7-bedf8feba097","code":"tag-258","name":"Степень точки"},{"id":"a7109bee-7556-446b-98b9-32d957694fec","code":"tag-155","name":"Подобие"}]}},{"problem_id":"92f6a12a-131d-4cda-a00b-9032aca26706","collection_id":"5ca2e948-e214-437e-b157-38d51de05a43","display_order":6,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"92f6a12a-131d-4cda-a00b-9032aca26706","short_id":17604,"task_number":19,"topic_id":"7ee08b18-18e8-42f5-8ad4-c094688c771d","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":9,"content_text":"Найдите все значения параметра \$ a \$ из интервала \$ (0, 1) \$, при которых для каждого \$ x \$ из интервала \$ (0, \\pi/4) \$ существует не более одного значения \$ y \$ в интервале \$ (0, \\pi/4) \$, такого что\n\\[ \\frac{\\operatorname{tg} x \\operatorname{tg} y}{\\operatorname{tg}\\left( a(x+y) \\right)} = \\frac{\\operatorname{tg}(x+y) - \\operatorname{tg}\\left( a(x+y) \\right)}{1 + \\operatorname{tg}(x+y)\\operatorname{tg}\\left( a(x+y) \\right)}. \\]","answer_text":"\$a = 1/2\$","answer":"\$a = 1/2\$","solution_text":"$5c","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-v6","variant_label":"226","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":"63383330-9060-4445-8f93-c53c52a03571","code":"tag-89","name":"Уравнения с параметром"},{"id":"dcec6e03-173e-4ea7-9b6a-cb935a3bdda5","code":"tag-212","name":"Тригонометрические формулы суммы или разности аргументов"},{"id":"dd8fa188-ef7b-495c-b5b1-fc5dd69ca555","code":"tag-253","name":"Полные квадраты и факторизация"}]}},{"problem_id":"a63a21bc-1fbf-4620-bfaf-c58e2cfe9884","collection_id":"5ca2e948-e214-437e-b157-38d51de05a43","display_order":7,"metadata":{},"created_at":"2026-06-10T20:49:51.310551Z","updated_at":"2026-06-10T20:49:51.310551Z","problem":{"id":"a63a21bc-1fbf-4620-bfaf-c58e2cfe9884","short_id":17605,"task_number":20,"topic_id":"57632be7-7c51-406a-8a6f-58e0708032ea","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":6,"content_text":"Основание \$ ABCD \$ прямоугольного параллелепипеда \$ ABCDA'B'C'D' \$ с боковыми рёбрами \$ AA' \$, \$ BB' \$, \$ CC' \$, \$ DD' \$ является квадратом со стороной \$ \\sqrt{2} \$. Известно, что \$ AE \\perp D'F \$, где \$ E \$ — центр грани \$ BCC'B' \$, \$ F \$ — центр квадрата \$ ABCD \$. 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