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5b:Tb90,Пусть $\pi_1$ и $\pi_2$ — две касательные плоскости к сфере, параллельные $\pi$, и пусть $T_1$ и $T_2$ — соответствующие точки касания. Тогда центр сферы $O$ совпадает с серединой отрезка $T_1T_2$. Пусть $T$ — точка пересечения $T_1T_2$ с плоскостью $\pi$. Пусть $\mathcal{B}_1$ — шар, находящийся между $\pi_1$ и $\pi$, а $\mathcal{B}_2$ — шар, находящийся между $\pi_2$ и $\pi$. Тогда диаметры $\mathcal{B}_1$ и $\mathcal{B}_2$ не могут превышать расстояния между $\pi_1$ и $\pi$ и между $\pi_2$ и $\pi$ соответственно. Причём равенство достигается ровно в одном случае — когда отрезки $TT_1$ и $TT_2$ суть соответственно диаметры $\mathcal{B}_1$ и $\mathcal{B}_2$. Следовательно, искомое отношение равно $TT_1/TT_2$. Не ограничивая общности, можно считать, что $TT_1\geqslant TT_2$. Тогда, если обозначить через $R$ радиус сферы и через $r$ — расстояние от точки $O$ до плоскости $\pi$, то \[\frac{TT_1}{TT_2}=\frac{R+r}{R-r}=\frac{R/r+1}{R/r-1}.\] Найдём $R/r$. Пусть $A$, $B$, $C$ — вершины параллелепипеда, через которые проходит плоскость $\pi$, и пусть $D$ — четвёртая вершина отсекаемого тетраэдра. Пусть $AD=a$, $BD=b$, $CD=c$. Поскольку плоскость $\pi$ делит диагональ параллелепипеда, исходящую из вершины $D$, в отношении $1:2$, расстояние от $O$ до $\pi$ в два раза меньше расстояния от $D$ до $\pi$. Стало быть, объём тетраэдра $ABCD$ равен одновременно $\frac{1}{3}\cdot 2r\cdot S(\triangle ABC)$ и $\frac{1}{6}abc$. Для площади $S(\triangle ABC)$ справедливо \[S(\triangle ABC)=\sqrt{S(\triangle ABD)^2+S(\triangle BCD)^2+S(\triangle ACD)^2}=\sqrt{\tfrac{1}{4}\bigl((ab)^2+(bc)^2+(ac)^2\bigr)},\] то есть \[r=\frac{abc}{4S(\triangle ABC)}=\frac{1}{2\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}.\] Учитывая, что $R=OD=\frac{1}{2}\sqrt{a^2+b^2+c^2}$, получаем \[R/r=\sqrt{a^2+b^2+c^2}\cdot\sqrt{\tfrac{1}{a^2}+\tfrac{1}{b^2}+\tfrac{1}{c^2}}=\sqrt{1+2+4}\cdot\sqrt{1+\tfrac{1}{2}+\tfrac{1}{4}}=7/2.\] Стало быть, искомое отношение равно \[\frac{TT_1}{TT_2}=\frac{7/2+1}{7/2-1}=9/5.\]41:[["$","script",null,{"id":"collection-schema","type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Course\",\"name\":\"ДВИ МГУ 2019 — вариант 1\",\"description\":\"Коллекция \\\"ДВИ МГУ 2019 — вариант 1\\\" для подготовки к экзамену. Содержит 8 задач. 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(математика)\",\"item\":\"https://sdamex.com/math/dvi-math\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Коллекции\",\"item\":\"https://sdamex.com/math/dvi-math/collections\"},{\"@type\":\"ListItem\",\"position\":4,\"name\":\"ДВИ МГУ 2019 — вариант 1\"}]}"}}],["$","nav",null,{"aria-label":"Хлебные крошки","className":"min-w-0 mb-4","children":["$","ol",null,{"className":"flex flex-wrap items-center gap-x-1 gap-y-0.5 text-xs text-muted-foreground","children":[["$","li","Математика-0",{"className":"flex min-w-0 items-center gap-x-1","children":[false,["$","$L58",null,{"prefetch":false,"href":"/math","className":"relative truncate transition-colors hover:text-foreground after:absolute after:-inset-y-2 after:inset-x-0","children":"Математика"}]]}],["$","li","ДВИ МГУ (математика)-1",{"className":"flex min-w-0 items-center gap-x-1","children":[["$","$L59",null,{"ref":"$undefined","iconNode":[["path",{"d":"m9 18 6-6-6-6","key":"mthhwq"}]],"className":"lucide-chevron-right h-3 w-3 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1","visible_problem_count":8,"year":2019},"created_at":"2026-06-10T10:30:00Z","updated_at":"2026-06-11T07:12:42.961126Z","problem_count":8,"visible_count":8,"total_count":8,"pdf_unlocked":false},"serverCollectionProblems":{"items":[{"problem_id":"d2fc1677-aa14-40e1-b360-aac6ca523e76","collection_id":"3cc201a9-d190-4201-a680-bd60465c72c7","display_order":1,"metadata":{},"created_at":"2026-06-11T07:12:42.961126Z","updated_at":"2026-06-11T07:12:42.961126Z","problem":{"id":"d2fc1677-aa14-40e1-b360-aac6ca523e76","short_id":18008,"task_number":14,"topic_id":"fbd591a9-072b-49ce-a91e-2ef4415a0351","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":3,"content_text":"Найдите наибольшее целое число, не превосходящее \$\\sqrt{2019\\cdot 2029-2016\\cdot 2032}\$.","answer_text":"\$6\$","answer":"\$6\$","solution_text":"Подкоренное выражение удобно упростить, заметив, что все четыре множителя группируются вокруг числа \$2024\$:\n\\[2019=2024-5,\\quad 2029=2024+5,\\quad 2016=2024-8,\\quad 2032=2024+8.\\]\nПрименяя формулу разности квадратов к каждому произведению, получаем\n\\[2019\\cdot 2029=(2024-5)(2024+5)=2024^2-5^2,\\qquad 2016\\cdot 2032=(2024-8)(2024+8)=2024^2-8^2.\\]\nПоэтому слагаемые с \$2024^2\$ взаимно уничтожаются:\n\\[2019\\cdot 2029-2016\\cdot 2032=\\bigl(2024^2-5^2\\bigr)-\\bigl(2024^2-8^2\\bigr)=8^2-5^2=64-25=39.\\]\nТаким образом,\n\\[\\sqrt{2019\\cdot 2029-2016\\cdot 2032}=\\sqrt{39}.\\]\nОценим этот корень. Так как \$6^2=36<39<49=7^2\$, то \$6<\\sqrt{39}<7\$. Следовательно, наибольшее целое число, не превосходящее \$\\sqrt{39}\$, равно \$6\$.","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2011-2019-v1","position_in_variant":1,"solution_origin":"sdamex-agents","solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2019_ans.pdf","variant":"июль 2019 года, вариант 1","variant_key":"2019-v1","year":2019},"created_at":"2026-06-11T07:12:42.961126Z","updated_at":"2026-08-31T18:36:47.303772Z","is_favorite":false,"is_solved":false,"is_reported":false,"topic":{"id":"fbd591a9-072b-49ce-a91e-2ef4415a0351","name":"Алгебраические преобразования","description":null},"tags":[{"id":"161ce81e-7848-4a50-990d-f97dfcbb50aa","code":"tag-22","name":"Преобразования выражений, включающих арифметические операции"},{"id":"31d4e4b7-a2c1-4cbd-85ff-1aa0dec4e314","code":"tag-45","name":"Сравнение чисел"},{"id":"da67531d-0011-489f-adab-84744c7aa977","code":"tag-208","name":"Преобразования выражений, включающих корни натуральной степени"}]}},{"problem_id":"39b307ac-3da0-41c0-b52e-f828d00310ff","collection_id":"3cc201a9-d190-4201-a680-bd60465c72c7","display_order":2,"metadata":{},"created_at":"2026-06-11T07:12:42.961126Z","updated_at":"2026-06-11T07:12:42.961126Z","problem":{"id":"39b307ac-3da0-41c0-b52e-f828d00310ff","short_id":18009,"task_number":14,"topic_id":"a4c5b1b9-850a-422b-b0b4-723ad67fab1c","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":2,"content_text":"Найдите \$a+b+c\$, если известно, что \$a+2b=3\$, \$b+2c=4\$, \$c+2a=5\$.","answer_text":"\$4\$","answer":"\$4\$","solution_text":"Сложим все три равенства почленно. В левой части каждая из переменных встречается ровно по три раза:\n\\[(a+2b)+(b+2c)+(c+2a)=3a+3b+3c=3(a+b+c).\\]\nВ правой части получаем \$3+4+5=12\$. Значит, \$3(a+b+c)=12\$, откуда\n\\[a+b+c=\\frac{12}{3}=4.\\]\nЭтого достаточно для ответа, но убедимся в корректности. Определитель системы отличен от нуля, поэтому она имеет единственное решение, а потому величина \$a+b+c\$ определена однозначно. Решив систему, находим \$a=\\dfrac{5}{3}\$, \$b=\\dfrac{2}{3}\$, \$c=\\dfrac{5}{3}\$; прямая подстановка даёт \$a+2b=\\dfrac{5}{3}+\\dfrac{4}{3}=3\$, \$b+2c=\\dfrac{2}{3}+\\dfrac{10}{3}=4\$, \$c+2a=\\dfrac{5}{3}+\\dfrac{10}{3}=5\$, и \$a+b+c=\\dfrac{5}{3}+\\dfrac{2}{3}+\\dfrac{5}{3}=4\$.","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2011-2019-v1","position_in_variant":2,"solution_origin":"sdamex-agents","solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2019_ans.pdf","variant":"июль 2019 года, вариант 1","variant_key":"2019-v1","year":2019},"created_at":"2026-06-11T07:12:42.961126Z","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":"4b7dfc01-ef7d-43d7-baec-c5f58da87aa6","code":"tag-75","name":"Системы уравнений"},{"id":"f8f70776-3b01-443a-9a7a-93f65d491a5f","code":"tag-239","name":"Линейные уравнения и неравенства"}]}},{"problem_id":"42ba835c-8f73-4f7c-816a-1f4e07d21dad","collection_id":"3cc201a9-d190-4201-a680-bd60465c72c7","display_order":3,"metadata":{},"created_at":"2026-06-11T07:12:42.961126Z","updated_at":"2026-06-11T07:12:42.961126Z","problem":{"id":"42ba835c-8f73-4f7c-816a-1f4e07d21dad","short_id":18010,"task_number":17,"topic_id":"56ee582a-3c52-4fd3-85c9-966671ec7454","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":4,"content_text":"Решите уравнение \$7\\sin x+2\\cos 2x=5\$.","answer_text":"\$x = \\pi/2+2k\\pi,\\ (-1)^k\\arcsin\\frac{3}{4}+k\\pi,\\ k\\in\\mathbb{Z}\$","answer":"\$x = \\pi/2+2k\\pi,\\ (-1)^k\\arcsin\\frac{3}{4}+k\\pi,\\ k\\in\\mathbb{Z}\$","solution_text":"\\[7\\sin x+2\\cos 2x=5 \\iff 7\\sin x+2-4\\sin^2 x=5 \\iff 4\\sin^2 x-7\\sin x+3=0 \\iff\\]\n\\[\\iff \\left[\\begin{array}{l}\\sin x=1\\\\ \\sin x=\\frac{3}{4}\\end{array}\\right. \\iff \\left[\\begin{array}{l}x=\\pi/2+2k\\pi,\\\\ x=(-1)^k\\arcsin\\frac{3}{4}+k\\pi,\\end{array}\\right.\\ k\\in\\mathbb{Z}.\\]","explanation":null,"score":10,"part":"2","metadata":{"dvi":true,"ingest_batch":"dvi-2011-2019-v1","position_in_variant":3,"solutions_url":"https://pk.math.msu.ru/wp-content/uploads/2025/06/dvi_2019_ans.pdf","variant":"июль 2019 года, вариант 1","variant_key":"2019-v1","year":2019},"created_at":"2026-06-11T07:12:42.961126Z","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":"728bc58c-1ce7-411d-8367-ebc8e9dfd104","code":"tag-106","name":"Квадратные уравнения"},{"id":"9564a83d-ccf3-4211-91ed-f6f568873b0e","code":"tag-131","name":"Тригонометрические уравнения"},{"id":"9dfedeed-365e-4cb8-a22b-e6ed50550eb1","code":"tag-139","name":"Тригонометрические уравнения, сводимые к целым на синус или косинус"},{"id":"ce7c2741-ac03-4ffe-afe8-d71552c8416c","code":"tag-196","name":"Синус и косинус двойного угла"}]}},{"problem_id":"5a7be6c6-8dc2-4141-8376-b74239a028e8","collection_id":"3cc201a9-d190-4201-a680-bd60465c72c7","display_order":4,"metadata":{},"created_at":"2026-06-11T07:12:42.961126Z","updated_at":"2026-06-11T07:12:42.961126Z","problem":{"id":"5a7be6c6-8dc2-4141-8376-b74239a028e8","short_id":18011,"task_number":16,"topic_id":"72c9fcf0-a99f-4128-b5bb-edf7f779b53c","exam_id":"a217734d-d4f5-4d66-ba1a-5248bb381f09","difficulty":6,"content_text":"Решите неравенство \$2^{\\log_2^2 x}+7x^{\\log_2 x}<16\$.","answer_text":"\\(\\frac{1}{2}=2 первичных баллов за геометрию (задания 15-19, 23-25); кодируется в скоринге пробника через geometry_tasks + geometry_min_points (0423)","geometry_tasks":[15,16,17,18,19,23,24,25],"hidden":false,"max_secondary_score":5,"mock_exam_enabled":true,"part1_tasks":19,"part2_tasks":[20,21,22,23,24,25],"score_conversion":[2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5],"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":2,"21":2,"22":2,"23":2,"24":2,"25":2},"task_count":25,"year":2026},"created_at":"2026-07-18T14:59:37.743898Z","updated_at":"2026-07-18T14:59:37.743898Z","problem_count":2181,"theory_count":104,"system_collection_count":89}]}]]

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