Difference between revisions of "DVI exam"
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− | ==2020 201 problem 6== | + | DVI is an exam in mathematics at the Moscow State University named after M.V. Lomonosov. The first four problems have a standard level. |
+ | Problem 5 is advanced level of geometry. | ||
+ | Problem 6 is an advanced level equation or inequality. | ||
+ | Problem 7 is advanced level of stereometry. | ||
+ | |||
+ | Below are the most difficult problems of this exam in recent years. The headings indicate the year when the problem was used, the variant option of the exam, and the number of the problem. | ||
+ | |||
+ | ==2011 Problem 8== | ||
+ | Solve the system of equations | ||
+ | <cmath>\left\{ | ||
+ | <i><b>Standard Solution</b></i> | ||
+ | <cmath>\left\{ | ||
+ | Denote | ||
+ | <cmath>\left\{ | ||
+ | We get <cmath>\left\{ | ||
+ | First equation define inner points of the circle with radius <math>1</math> and the circle. | ||
+ | The distance from the straight line to the origin of the coordinate system <math>d</math> is | ||
+ | <cmath>\frac {1}{d^2} = \frac {1}{3^2} + \frac {(2 \sqrt{2})^2}{3^2} = 1 \implies d = 1,</cmath> | ||
+ | so the system of the equations define the only tangent point of the circle and the line. | ||
+ | <cmath>u = \frac {2 \sqrt{2}}{3}, v = \frac {1}{3} \implies x = \frac {5}{9}, y = \frac {1}{9}.</cmath> | ||
+ | <i><b>Short Solution</b></i> | ||
+ | <cmath>9(2 x^2 + 4xy + 11 y^2) \le 9 \le (4x + 7y)^2 \implies 2(x - 5y)^2 \le 0 \implies x = 5y \implies</cmath> | ||
+ | <cmath>81 y^2 \le 1, 27 y \ge 3 \implies y = \frac {1}{9}, x = \frac {5}{9}.</cmath> | ||
+ | |||
+ | ==2012 Problem 8== | ||
+ | [[File:2012 7.png|330px|right]] | ||
+ | Let the tetrahedron <math>SABC, AC = BC = 5, AB = 6, AS = BS = 7, CS = 4</math> be given. | ||
+ | |||
+ | A right circular cylinder is located so that the circle of its upper base touches each of the faces which contains vertex <math>S.</math> | ||
+ | |||
+ | The circle of the lower base lies in the <math>ABC</math> plane and touches straight lines <math>AC</math> and <math>BC.</math> | ||
+ | |||
+ | Find the height <math>h</math> of the cylinder. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>M</math> the midpoint <math>AB.</math> Plane <math>SCM</math> is the bisector plane of segment <math>AB, SCM \perp ABC.</math> | ||
+ | <math>CM = \sqrt{AC^2 - \frac {AB^2}{4}}= 4 = SC.</math> | ||
+ | |||
+ | The inradius of <math>\triangle ABC</math> equal to <math>\frac {3}{2},</math> distance from incenter <math>I</math> to vertex <math>C</math> is <math>IC = 4 - \frac {3}{2} = \frac {5}{2}.</math> | ||
+ | |||
+ | Denote <math>D</math> the foot from <math>S</math> to <math>\overline{MC} \implies SD = \sqrt{15}, CD = 1.</math> | ||
+ | |||
+ | Denote <math>KK'L</math> the crosssection of <math>SABC</math> by plane of the upper base of cylinder, <math>O'</math> is the incenter <math>\triangle KK'L, F'</math> is the point of tangency incircle of <math>\triangle KK'L</math> and <math>KK'.</math> | ||
+ | |||
+ | Denote <math>F</math> and <math>O</math> the foots from <math>F'</math> and <math>O'</math> to <math>\overline{MC}.</math> Denote the radius <math>OF = r = O'F'.</math> | ||
+ | |||
+ | The circle of the lower base inscribed in angle equal to <math>\angle ACB,</math> so | ||
+ | <cmath>\frac {CO}{FO} = \frac{5}{3} \implies CO = \frac{5r}{3}, CF = \frac{2r}{3}.</cmath> | ||
+ | <cmath>\triangle MF'F \sim \triangle MSD \implies \frac {MF}{MD} = \frac {F'F}{SD}.</cmath> | ||
+ | Projection from the point <math>S</math> maps <math>Q'</math> onto <math>I \implies</math> | ||
+ | <cmath>\triangle IO'O \sim \triangle ISD \implies \frac {IO}{ID} = \frac {O'O}{SD}.</cmath> | ||
+ | <math>h = O'O = F'F \implies \frac {O'O}{SD} = \frac {F'F}{SD} =\frac {MF}{MD} = \frac {IO}{ID},</math> | ||
+ | <cmath>\frac {4 + \frac {2r}{3} } {4+1} = \frac {\frac{5}{2} + \frac {5r}{3} } { \frac {5}{2}+ 1} \implies r = \frac {1}{4} \implies h = \frac {5 \sqrt{15}} {6}.</cmath> | ||
+ | |||
+ | <i><b>Answer: <math>\frac {5 \sqrt{15}} {6}.</math></b></i> | ||
+ | |||
+ | ==2014 1 Problem 6== | ||
+ | Find all pares of real numbers <math>(x,y)</math> satisfying the system of equations | ||
+ | <cmath>\left\{ | ||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>t= y^{\frac{2}{3}} \implies y = t^{\frac{3}{2}}.</math> | ||
+ | <cmath>\left\{ | ||
+ | Denote <math>u = \frac {x}{4}, v = \frac {t}{4}.</math> | ||
+ | <cmath>\left\{ | ||
+ | <math> u = v = 1, x = 4, y = 8</math> is the solution. Let | ||
+ | <cmath>F(u) = u^{\frac{3}{2}} + (2 - u)^{\frac{3}{2}} \implies F'(u) = 1.5(\sqrt{u} - \sqrt{2 - u}).</cmath> | ||
+ | If <math>u > 1</math> then <math>F'(u) > 0,</math> if <math>u < 1</math> then <math>F'(u) < 0,</math> therefore <math>u = 1</math> is the single root. | ||
+ | |||
+ | ==2014 1 Problem 8== | ||
+ | Let <math>f(x,y) = y + \sqrt{6 - 6x^2 - 14 y^2 - 18 xy}, g(x,y) = y - \sqrt{6 - 6x^2 - 14 y^2 - 18 xy}.</math> | ||
+ | |||
+ | Find <math>max_x max_y (f(x,y), g(x,y))</math> and <math>min_x min_y (f(x,y), g(x,y)).</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | <cmath>6 - 6x^2 - 14 y^2 - 18 xy = 6 - 6 \left(x^2 + 3xy + \frac {9}{4} y^2\right) + \frac {6 \cdot 9}{4} y^2 - 14 y^2 = 6 - 6 \cdot \left(x + \frac {3y}{2} \right)^2 - \frac {1}{2} y^2,</cmath> | ||
+ | <cmath>6 - 6 \cdot (x + \frac {3y}{2})^2 - \frac {1}{2} y^2 \le 6 - \frac {1}{2} y^2.</cmath> | ||
+ | <math>f(x,y) \ge g(x,y) \implies max_x max_y (f(x,y), g(x,y)) = max_x max_y f(x,y) = max_y (y + \sqrt{6 - \frac {1}{2} y^2}),</math> where <math>y > 0.</math> | ||
+ | <cmath>\frac {u+v+w}{3} \le \sqrt{\frac {u^2 + u^2 + w^2}{3}} \implies</cmath> | ||
+ | <cmath>y + \sqrt{6 - \frac {1}{2} y^2} = \frac {y}{2}+\frac {y}{2}+\sqrt{6 - \frac {1}{2} y^2} \le \sqrt {3} \cdot \sqrt{6 - \frac {1}{2} y^2 + \frac {1}{4} y^2 + \frac {1}{4} y^2} = \sqrt {3 \cdot 6} = 3 \sqrt{2}.</cmath> | ||
+ | <cmath>f(-x,-y) = -g(x,y) \implies min_x min_y (f(x,y), g(x,y)) = min_x min_y (g(x,y)) = - max_x max_y f(x,y) = - 3 \sqrt{2}.</cmath> | ||
+ | |||
+ | <i><b>Answer:<math> 3 \sqrt{2}, - 3 \sqrt {2}.</math></b></i> | ||
+ | ==2015 1 Problem 7== | ||
+ | [[File:2015 7 distance.png|330px|right]] | ||
+ | A sphere is inscribed in a regular triangular prism with bases <math>ABCA'B'C'.</math> Find its radius if the distance between straight lines <math>AE</math> and <math>BD</math> is equal to <math>\sqrt{13},</math> where <math>E</math> and <math>D</math> are points lying on <math>A'B'</math> and <math>B'C'</math>, respectively, and <math>A'E : EB' = B'D : DC' = 1 : 2.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | The distance from the center of the sphere to the centers of the prism faces is equal to <math>R,</math> so <cmath>AA' = 2R, AB = 2 \sqrt{3} R.</cmath> | ||
+ | |||
+ | In order to find the distance <math>PQ</math> between the lines <math>\ell = AD</math> and <math>m = BE</math>, one can find the length of two perpendiculars <math>MM'</math> and <math>DE</math> to the line <math>m</math> that are perpendicular to each other. Then | ||
+ | <cmath>\frac {1}{PQ^2} = \frac{1}{DE^2} + \frac{1}{MM'^2}</cmath> | ||
+ | since, when viewed along a straight line <math>m</math>, the segment <math>PQ</math> is the altitude of a right triangle with legs <math>DE</math> and <math>MM'.</math> | ||
+ | |||
+ | The plane <math>\pi = BB'C'</math> containe the straight line <math>m.</math> The straight line <math>\ell</math> crossed <math>\pi</math> at the point <math>M \in BB'.</math> | ||
+ | <cmath>\frac {B'D}{DA'} = \frac {B'M}{AA'} =2 \implies B'M = 4R.</cmath> | ||
+ | In a right triangle <math>\triangle BKM</math> | ||
+ | <cmath>KM = BC = 2 \sqrt{3} R, BM = BB' + B'M = 6R, MM' \perp BE.</cmath> | ||
+ | <math>MM'</math> is the height falling on the hypotenuse, <math>\frac {1}{MM'^2} = \frac{1}{KM^2} + \frac{1}{BM^2}.</math> | ||
+ | |||
+ | Let <math>F</math> be the projection of <math>A</math> onto plane <math>\pi \implies F \in BC, BF = FC.</math> | ||
+ | |||
+ | Therefore <math>FM</math> is the projection of <math>\ell</math> onto plane <math>\pi, KM = 2 FB \implies m \cap MF</math> at the point <math>E.</math> | ||
+ | <cmath>\frac {B'D}{EB'} = 2 \implies DE \perp \pi \implies DE \perp m.</cmath> | ||
+ | <cmath>\frac {DE}{AF} = \frac {ME}{MF} = \frac {B'M}{MB} = \frac {2}{3} \implies DE = 2R.</cmath> | ||
+ | <cmath>\frac {1}{PQ^2} = \frac{1}{DE^2} + \frac{1}{MM'^2} = \frac{1}{DE^2} + \frac{1}{KM^2} + \frac{1}{BM^2} = \frac{1}{(2R)^2} + \frac{1}{(2\sqrt{3}R)^2} + \frac{1}{(6R)^2} = \frac{13}{(6R)^2}.</cmath> | ||
+ | <cmath>PO = \sqrt{13} = \frac{6R}{\sqrt{13}} \implies R = \frac {13}{6}.</cmath> | ||
+ | <i><b>Answer:<math>\frac {13}{6}.</math></b></i> | ||
+ | |||
+ | ==2016 2 Problem 7== | ||
+ | [[File:2016 7.png|330px|right]] | ||
+ | [[File:2016 7 top.png|330px|right]] | ||
+ | [[File:2016 7 side.png|330px|right]] | ||
+ | Let the base of the regular pyramid with vertex <math>S</math> be the hexagon <math>ABCDEF</math> with side <math>5.</math> The plane <math>\pi</math> is parallel to the edge <math>AB</math>, perpendicular to the plane <math>SDE</math> and intersects the edge <math>BC</math> at point <math>K,</math> so that <math>\frac {BK}{KC} = \frac {3}{2}.</math> The lines along which <math>\pi</math> intersects the <math>BCS</math> plane and the base plane are perpendicular. | ||
+ | |||
+ | Find the area of the triangle cut off by the plane <math>\pi</math> from the face <math>CDS.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>K' = \pi \cap AF, L = \pi \cap SC, L' = \pi \cap SF, M = \pi \cap SD, M' = \pi \cap SE,</math> | ||
+ | |||
+ | <math>G,O,H,N</math> are the midpoints of <math>KK',CF,DE, MM',</math> respectively. | ||
+ | |||
+ | Plane <math>SGH</math> is the plane symmetry of pyramid, <math>SGH \perp KK' \implies SGH \perp \pi.</math> | ||
+ | |||
+ | By condition <math>GN = \pi \cap SGN \perp SH,</math> so exist point <math>Q = SO \cap GN, Q \in LL'.</math> | ||
+ | |||
+ | <math>KL</math> is the line along which <math>\pi</math> intersects the <math>BCS</math> plane, <math>KK'</math> is the line along which <math>\pi</math> intersects the base plane, so <math>\angle LKK' = 90^\circ \implies LK || GN.</math> | ||
+ | |||
+ | We use the top wiew and get | ||
+ | <cmath>\frac {SL}{LC} = \frac{KB + CB} {CK}= \frac{3+5} {2} = 4. \frac{GO}{OH} = \frac{CK}{CD} = \frac{2}{5}.</cmath> | ||
+ | <cmath>AB = 5 \implies OH = \frac {5 \sqrt{3}}{2} \implies GO = \sqrt{3}.</cmath> | ||
+ | <cmath>Q \in LL' \implies \frac {SQ}{QO} = 4.</cmath> | ||
+ | Denote <math>h = SO, \alpha = \angle SHO = \angle GQO</math> and use the side wiew. | ||
+ | |||
+ | <cmath>\tan \alpha = \frac {SO}{OH} = \frac {GO}{QH} \implies OH \cdot GO = \frac {15}{2} = SO \cdot QO = \frac {h^2}{5} \implies</cmath> | ||
+ | <cmath>h = 5 \sqrt {\frac {3}{2}} \implies \tan \alpha = \sqrt{2} \implies \cos \alpha = \frac {1}{\sqrt{3}}.</cmath> | ||
+ | Triangle <math>\triangle OCD</math> is the regular triangle with side <math>5</math>, so | ||
+ | <cmath>[OCD] = \frac {25 \sqrt{3}}{4} \implies [SCD] = \frac{OCD}{\cos \alpha} = \frac {75}{4}.</cmath> | ||
+ | <math>SH = \sqrt {SO^2 + OH^2} = \frac {15}{2}, NH = GH \cos \alpha = \frac {7}{2} \implies \frac {SN}{SH} =\frac{8}{15}= \frac{SM}{SD}.</math> | ||
+ | <cmath>[SML] = [SCD] \cdot \frac {SL}{SC} \cdot \frac {SM}{SD} = \frac {75}{4} \cdot \frac{4}{5} \cdot \frac{8}{15} = 8.</cmath> | ||
+ | <i><b>Answer: 8.</b></i> | ||
+ | |||
+ | ==2016 2 Problem 8== | ||
+ | Find the smallest value of the expression | ||
+ | <cmath>f=\sqrt{13 + \log^2_a \cos \frac {x}{a} + \log_a \cos^4 \frac {x}{a}}+\sqrt{97 + \log^2_a \sin \frac {x}{a} - \log_a \sin^8 \frac {x}{a}} + \sqrt{20 + \log^2_a \tan \frac {x}{a} + \log_a \tan^4 \frac {x}{a}}.</cmath> | ||
+ | <i><b>Solution</b></i> | ||
+ | <cmath>13 + \log^2_a \cos \frac {x}{a} + \log_a \cos^4 \frac {x}{a} = 9 + \log^2_a \cos \frac {x}{a} + 4\log_a \cos \frac {x}{a} + 4 = 3^2 + (\log_a \cos \frac {x}{a} + 2)^2,</cmath> | ||
+ | <cmath>97 + \log^2_a \sin \frac {x}{a} - \log_a \sin^8 \frac {x}{a} = 9^2 + (4 - \log_a \sin \frac {x}{a} )^2,</cmath> | ||
+ | <cmath>20 + \log^2_a \tan \frac {x}{a} + \log_a \tan^4 \frac {x}{a} = 16 + (\log_a \tan \frac {x}{a} + 2)^2 = 4^2 + (\log_a \sin \frac {x}{a} - \log_a \cos \frac {x}{a}+ 2)^2.</cmath> | ||
+ | Denote <math>\vec {AB} = (3, \log_a \cos \frac {x}{a} + 2), \vec {BC} = (9, 4 - \log_a \sin \frac {x}{a}), \vec {CD} = (4, \log_a \sin \frac {x}{a} - \log_a \cos \frac {x}{a}+ 2).</math> | ||
+ | <cmath>f = |\vec {AB}| + |\vec {BC}| + |\vec {CD}|.</cmath> | ||
+ | <cmath>\vec {AB} + \vec {BC} + \vec {CD} = (3 +9 + 4, \log_a \cos \frac {x}{a} + 2 +4 - \log_a \sin \frac {x}{a} + \log_a \sin \frac {x}{a} - \log_a \cos \frac {x}{a}+ 2) = (16,8).</cmath> | ||
+ | The shortest length of a broken line <math>ABCD</math> with fixed ends is equal to the distance between points <math>A</math> and <math>D,</math> which is <math>|AD|</math> and is achieved if points <math>A,B,C,</math> and <math>D</math> are collinear. | ||
+ | <cmath>|\vec AD| = \sqrt {16^2 + 8^2} = 8 \sqrt{5}, \vec {AB} = (3, 1.5),\vec {BC} = (9, 4.5), \vec {CD} (4,2).</cmath> | ||
+ | <cmath>\log_a \tan \frac {x}{a} + 2 = 2 \implies \log_a \tan \frac {x}{a} = 0 \implies \tan \frac {x}{a} = 1.</cmath> | ||
+ | <cmath>\sin \frac {x}{a} >0 \implies \sin \frac {x}{a} = \frac {1}{\sqrt{2}}, \cos \frac {x}{a} = \frac {1}{\sqrt{2}}.</cmath> | ||
+ | <cmath>4 - \log_a \sin \frac {x}{a} = 4.5 \implies \log_a \frac {1}{\sqrt{2}} = - \frac{1}{2} \implies a = 2.</cmath> | ||
+ | <cmath>\sin \frac {x}{2} = \cos \frac {x}{2} = \frac {1}{\sqrt{2}} \implies \frac {x}{2} = \frac {\pi}{4} + 2 k \pi \implies x = \frac {\pi}{2} + 4k \pi.</cmath> | ||
+ | <i><b>Answer:<math> min(f) = 8 \sqrt{5}, a = 2, x = \frac {\pi}{2} + 4k \pi.</math></b></i> | ||
+ | |||
+ | ==2020 var 201 problem 6== | ||
[[File:2020 201 6.png|330px|right]] | [[File:2020 201 6.png|330px|right]] | ||
Let a triangular prism <math>ABCA'B'C'</math> with a base <math>ABC</math> be given, <math>D \in AB', E \in BC', F \in CA'.</math> Find the ratio in which the plane <math>DEF</math> divides the segment <math>AA',</math> if <math>AD : DB' = 1 : 1,</math> <cmath>BE : EC' = 1 : 2, CF : FA' = 1 : 3.</cmath> | Let a triangular prism <math>ABCA'B'C'</math> with a base <math>ABC</math> be given, <math>D \in AB', E \in BC', F \in CA'.</math> Find the ratio in which the plane <math>DEF</math> divides the segment <math>AA',</math> if <math>AD : DB' = 1 : 1,</math> <cmath>BE : EC' = 1 : 2, CF : FA' = 1 : 3.</cmath> | ||
Line 17: | Line 180: | ||
Answer: <math>AG : GA' = 4 : 3.</math> | Answer: <math>AG : GA' = 4 : 3.</math> | ||
− | ==2022 221 problem 7== | + | ==2020 var 202 problem 6== |
+ | [[File:2020 202 6.png|330px|right]] | ||
+ | Let a tetrahedron <math>ABCD</math> be given, <math>AB = BC = CD = 5, CA = AD = DB = 6.</math> Find the cosine of the angle <math>\varphi</math> between the edges <math>BC</math> and <math>AD.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let us describe a parallelepiped <math>AC'BD'B'DA'C</math> around a given tetrahedron <math>ABCD.</math> | ||
+ | |||
+ | <math>AB = CD \implies AC'BD'</math> and <math>B'DA'C</math> are equal rectangles. | ||
+ | |||
+ | <math>AC = BD \implies AB'CD'</math> and <math>C'DA'B</math> are equal rectangles. | ||
+ | |||
+ | Denote <math>AC' = a, AD' = b, AB' = c \implies</math> | ||
+ | <cmath>a^2 + b^2 = 5^2 = 25, a^2 + c^2 = 6^2 = 36.</cmath> | ||
+ | <cmath>4AC'^2 = 4 a^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos \varphi,</cmath> | ||
+ | <cmath>4AB'^2 = 4 c^2 = 5^2 + 6^2 + 2 \cdot 5 \cdot 6 \cos \varphi,</cmath> | ||
+ | <cmath>4(c^2 - a^2) = 4(6^2 - 5^2) = 4 \cdot 5 \cdot 6 \cos \varphi \implies \cos \varphi = \frac {6^2 - 5^2}{5 \cdot 6} = \frac {11}{30}.</cmath> | ||
+ | Answer: <math>\frac {11}{30}. </math> | ||
+ | |||
+ | ==2020 var 203 problem 6== | ||
+ | [[File:2020 203 6 3.png|330px|right]] | ||
+ | [[File:2020 203 6 2.png|330px|right]] | ||
+ | Let a cube <math>ABCDA'B'C'D'</math> with the base <math>ABCD</math> and side edges <math>AA', BB', CC', DD', AB = 1</math> be given. Find the volume of a polyhedron whose vertices are the midpoints of the edges <math>AB, AD, AA', CC', C'B', C'D'.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote the vertices of polyhedron <math>E, F, G, E', F', G'.</math> | ||
+ | Triangles <math>\triangle EFG</math> and <math>\triangle E'F'G'</math> are equilateral triangles with sides <math>\frac {\sqrt{2}}{2}</math> and areas <math>[EFG] = \frac {\sqrt{3}}{8}.</math> | ||
+ | |||
+ | This triangles lies in parallel planes, which are normal to cube diagonal <math>AC'.</math> | ||
+ | The distance <math>d</math> between this planes is <cmath>\sqrt{3} - 2 \cdot \frac{\sqrt{3}}{6} = \frac {2}{\sqrt{3}}.</cmath> | ||
+ | So the volume of the regular prism with base <math>\triangle EFG</math> and height <math>d</math> is | ||
+ | <cmath>V_0 = \frac {\sqrt{3}}{8} \cdot \frac {2}{\sqrt{3}} = \frac {1}{4}.</cmath> | ||
+ | |||
+ | Let the area <math>[A(x)]</math> be the quadratic function of <math>x.</math> Let | ||
+ | <cmath>A_1 = A[x_1], A_2 = A[x_2], d = x_2 - x_1,</cmath> <cmath>x_0 = \frac{x_1 + x_2}{2}, A_0 = A[x_0] \implies</cmath> | ||
+ | <cmath>V = \frac{d}{6} \cdot \left(A_1 + A_2 + 4 A_0 \right).</cmath> | ||
+ | Suppose, we move point <math>P</math> along axis <math>AC'</math> and cross the solid by plane contains <math>P</math> and normal to axis. Distance from <math>P</math> to each crosspoint this plane with the edge change proportionally position <math>P</math> along axes, so the area is quadratic function from <math>P</math> position. | ||
+ | <cmath>\frac {OE''}{ME} = \frac {\sqrt{3}}{2} \implies \frac {[E''F''G'']}{[EFG]} = 2 \left (\frac {OE''}{ME} \right)^2 = \frac {3}{2}.</cmath> | ||
+ | <cmath>V = \frac{d}{6} \cdot ([EFG] + {[E'F'G']} + 4 [E''F''G'']) = d \cdot [EFG] \cdot \frac {4}{3} = \frac {1}{4} \cdot \frac {4}{3} = \frac {1}{3}.</cmath> | ||
+ | |||
+ | Answer: <math>\frac {1}{3}.</math> | ||
+ | |||
+ | ==2020 var 204 problem 6== | ||
+ | [[File:2020 204 6.png|300px|right]] | ||
+ | |||
+ | Let a regular triangular pyramid be given. The circumcenter of the sphere <math>O</math> is equidistant from the edge and from the plane of the base of the pyramid. Find the radius of the sphere inscribed in this pyramid if the length of the edge of its base is <math>12.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <cmath>OP = ON \implies BP = BN, BS = 2 BP = 2 BN,</cmath> | ||
+ | <cmath>AB = \sqrt{3}BN, \angle BSN = 30^\circ \implies</cmath> | ||
+ | <cmath>SN = \frac {3}{2} SO = AB.</cmath> | ||
+ | <cmath>NM = \frac {BN}{2} = \frac {AB}{2 \sqrt{3}} \implies</cmath> | ||
+ | <cmath>\tan \angle MSN = \frac {1}{2\sqrt{3}} \implies</cmath> | ||
+ | <cmath>\sin \angle MSN = \frac {1}{\sqrt{13}} = \frac {ID}{SN - IN}, IN = ID = \frac {AB}{1 + \sqrt{13}}.</cmath> | ||
+ | Answer: <math>\frac {12}{1 + \sqrt{13}}.</math> | ||
+ | |||
+ | ==2020 var 205 problem 6== | ||
+ | [[File:2020 205 6.png|330px|right]] | ||
+ | Let the quadrangular pyramid <math>ABCDS</math> with the base parallelogram <math>ABCD</math> be given. | ||
+ | |||
+ | Point <math>E \in SB, \frac {SE}{EB} = 2.</math> Point <math>F \in SD, \frac {SF}{FD} = \frac {1}{2}.</math> | ||
+ | |||
+ | Find the ratio in which the plane <math>AEF</math> divides the volume of the pyramid. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let plane <math>AEF</math> cross edge <math>SC</math> at point <math>G.</math> | ||
+ | We make the central projection from point <math>S</math> | ||
+ | The images of points <math>A,E,F,G</math> are <math>A,B,D,C,</math> respectively. | ||
+ | The image of <math>S</math> is the crosspoint of <math>AC</math> and <math>BD.</math> | ||
+ | So lines <math>EF, SO,</math> and <math>AG</math> are crossed at point <math>H.</math> | ||
+ | <cmath>\frac {2 OH}{SH} = \frac {BE}{SE} + \frac {DF}{SF} = \frac {AA}{SA}+ \frac {CG}{SG}.</cmath> | ||
+ | <cmath>2 + \frac {1}{2} = 0 + \frac {CG}{SG} \implies \frac {CG}{SG} = \frac {5}{2}.</cmath> | ||
+ | Let’s compare volumes of some tetrachedrons, denote the volume of <math>X</math> as <math>[X].</math> | ||
+ | <cmath>\frac {[ABDS]}{[CBDS]} = \frac {[ABD]}{[CBD]} =1.</cmath> | ||
+ | <cmath>\frac {[AEFS]}{[ABDS]} = \frac {[EFS]}{[BDS]} = \frac {SE \cdot SF}{SB \cdot SD} = \frac{2}{9}.</cmath> | ||
+ | <cmath>\frac {[GEFS]}{[CBDS]} = \frac {[EFS]}{[BDS]} \cdot \frac {GX}{CO} = \frac{2}{9} \cdot \frac {SG}{SC} = \frac{2}{9} \cdot \frac {2}{7} = \frac{4}{63}.</cmath> | ||
+ | <cmath>\frac {[AEGFS]}{[ABDS]} = \frac {[AEFS]+[GEFS]}{[ABDS]} = \frac{2}{9} + \frac{4}{63} = \frac{2}{7} \implies \frac {[AEGFS]}{[ABCDS]} = \frac{1}{7}.</cmath> | ||
+ | Answer: 1 : 6. | ||
+ | |||
+ | ==2020 var 206 problem 6== | ||
+ | [[File:2020 206 6.png|330px|right]] | ||
+ | Given a cube <math>ABCDA'B'C'D'</math> with the base <math>ABCD</math> and side edges <math>AA', BB', CC', DD' =1.</math> Find the distance between the line passing through the midpoints of the edges <math>AB</math> and <math>AA'</math> and the line passing through the midpoints of the edges <math>BB'</math> and <math>B'C'.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let points <math>E,F,P, G, H, K</math> be the midpoints of <math>AB, AA', A'D', BB', B'C', B'A',</math> respectively. We need to prove that planes <math>GKH</math> and <math>EFP</math> are parallel, perpendicular to <math>B'D.</math> | ||
+ | Therefore, <math>B'D = \sqrt{3}.</math> | ||
+ | |||
+ | Point <math>O</math> is the midpoint <math>B'D \implies</math> | ||
+ | <cmath>B'O = \frac {\sqrt{3}}{2}, B'H = \frac {1}{2}, GH = \frac {\sqrt{2}}{2},</cmath> | ||
+ | <cmath>HQ = \frac{GH}{\sqrt{3}}, B'Q = \frac{\sqrt{3}}{6}, OQ = \frac {1}{\sqrt{3}} = IJ.</cmath> | ||
+ | For proof we can use one of the following methods: | ||
+ | |||
+ | 1. Vectors: <math>\vec {B'A'} = 2 \vec e_x, \vec {B'B} = 2 \vec e_y, \vec {B'C'} = 2 \vec e_z \implies </math> | ||
+ | <cmath>\vec {B'G} = \vec e_y, \vec {B'H} = \vec e_z, \vec {HG} = \vec e_y - \vec e_z, \vec {B'D} = \vec e_x + \vec e_y + \vec e_z.</cmath> | ||
+ | Scalar product <math>(\vec{B'D} \cdot \vec {HG}) = 0.</math> | ||
+ | Similarly, <cmath>\vec {B'E} = 2\vec e_y + \vec e_x, \vec {B'F} = 2\vec e_x+ \vec e_y, \vec {FE} = \vec e_y - \vec e_x.</cmath> | ||
+ | |||
+ | 2. <math>\angle B'OG = \angle B'OE = 90^\circ.</math> | ||
+ | |||
+ | 3. Rotating the cube around its axis <math>B'D</math> we find that the point <math>G</math> move to <math>H</math>, then to <math>K,</math> then to <math>G.</math> | ||
+ | |||
+ | Answer: <math>\frac {1}{\sqrt{3}}</math> | ||
+ | |||
+ | ==2021 var 215 problem 7== | ||
+ | The sphere touches all edges of the tetrahedron <math>ABCD.</math> It is known that the products of the lengths of crossing edges are equal. It is also known that <math>AB = 3, BC = 1.</math> Find <math>AC.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | The tangent segments from the common point to the sphere are equal. | ||
+ | |||
+ | Let us denote the segments from the vertex <math>A</math> to the sphere by <math>a.</math> | ||
+ | |||
+ | Similarly, we define <math>b, c, d.</math> | ||
+ | <cmath>AB = a + b = 3, BC = b + c = 1, a - c = (a+b) - (b+c) = 3 - 1 = 2.</cmath> | ||
+ | <cmath>AB \cdot CD = AD \cdot BC \implies 3(c+d) = 1(a + d) \implies a = 3c + 2d</cmath> | ||
+ | <cmath>a = c + 2 \implies c + d = 1 \implies b = d.</cmath> | ||
+ | <cmath>AD = AB = 3, AD \cdot BC = 3 \cdot 1 = 3 = (a+c)(b+ d) = (3 - b + 1 - b) \cdot 2b.</cmath> | ||
+ | If <math>b = \frac {3}{2}</math> then <math>c < 0.</math> | ||
+ | |||
+ | If <math>b = \frac {1}{2} = d = c, a = \frac {5}{2}, AC = 3.</math> | ||
+ | |||
+ | The tetrahedron <math>ABCD</math> is a regular pyramid with a regular triangle with side <math>1</math> at the base and side edges equal to <math>3.</math> | ||
+ | |||
+ | Answer: 3. | ||
+ | |||
+ | ==2022 var 221 problem 7== | ||
[[File:MSU 2022 7.png|330px|right]] | [[File:MSU 2022 7.png|330px|right]] | ||
[[File:MSU 2022 7a.png|330px|right]] | [[File:MSU 2022 7a.png|330px|right]] | ||
Line 39: | Line 331: | ||
<i><b>Answer: 5.</b></i> | <i><b>Answer: 5.</b></i> | ||
− | ==2022 222 problem 7== | + | ==2022 var 222 problem 7== |
[[File:MSU 2022 2 7.png|400px|right]] | [[File:MSU 2022 2 7.png|400px|right]] | ||
A sphere of diameter <math>1</math> is inscribed in a pyramid at the base of which lies a rhombus with an acute angle <math>2\alpha</math> and side <math>\sqrt{6}.</math> Find the angle <math>2\alpha</math> if it is known that all lateral faces of the pyramid are inclined to plane of its base at an angle of <math>60^\circ.</math> | A sphere of diameter <math>1</math> is inscribed in a pyramid at the base of which lies a rhombus with an acute angle <math>2\alpha</math> and side <math>\sqrt{6}.</math> Find the angle <math>2\alpha</math> if it is known that all lateral faces of the pyramid are inclined to plane of its base at an angle of <math>60^\circ.</math> | ||
Line 58: | Line 350: | ||
<cmath>\sin {2 \alpha} = \frac {SM}{AB} = \frac {\sqrt{3}}{\sqrt{6}} = \frac {1}{\sqrt{2}}.</cmath> | <cmath>\sin {2 \alpha} = \frac {SM}{AB} = \frac {\sqrt{3}}{\sqrt{6}} = \frac {1}{\sqrt{2}}.</cmath> | ||
<i><b>Answer:<math> \frac {\pi}{4}.</math></b></i> | <i><b>Answer:<math> \frac {\pi}{4}.</math></b></i> | ||
+ | |||
+ | ==2022 var 222 problem 6== | ||
+ | Find all possible values of the product <math>xy</math> if it is known that <math>x,y \in \left [ 0, \frac{\pi}{2} \right)</math> and it is true | ||
+ | <cmath>\frac{1 - \sin(x - y)}{1 - \cos(x - y)}= \frac{1 - \sin(x + y)}{1 - \cos(x + y)}.</cmath> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>y = 0,</math> then for each <math>x</math> equation is true, <math>xy = 0.</math> Let <math>y > 0.</math> | ||
+ | <cmath>1 - \sin (x - y) - \cos (x + y) + \sin (x – y) \cos (x + y) = 1 - \sin (x + y) - \cos (x - y) + \sin (x + y) \cos (x - y).</cmath> | ||
+ | <cmath>\sin (x + y) - \sin (x - y) + \cos (x - y) - \cos (x + y) = \sin ((x + y) - (x - y)),</cmath> | ||
+ | <cmath>2 \cos x \sin y + 2 \sin x \sin y = 2 \cos y \sin y,</cmath> | ||
+ | <cmath> \cos x + \sin x = \cos y .</cmath> | ||
+ | <math>\cos y < 0, x \in \left [ 0, \frac{\pi}{2} \right) \implies \cos x + \sin x \ge 1,</math> no solution. | ||
+ | |||
+ | <i><b>Answer:<math>0.</math></b></i> | ||
+ | |||
+ | ==2022 var 224 problem 6== | ||
+ | Find all triples of real numbers <math>(x,y,z)</math> in the interval <math>\left ( 0; \frac {\pi}{2} \right)</math> satisfying the system of equations | ||
+ | <cmath>\begin{equation} \left\{ \begin{aligned} | ||
+ | \sin x &= \sin y - \sin z \cos (x+z) ,\ | ||
+ | \cos x &= \cos z + \cos y \cos (x+y) . | ||
+ | \end{aligned} \right.\end{equation}</cmath> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>u = x + z \implies</math> | ||
+ | <cmath>x = u - z, \sin (u - z) = \sin u \cos z - \sin z \cos u = \sin x = \sin y - \sin z \cos (x+z) \implies</cmath> | ||
+ | <cmath>\sin y = \cos z \sin (x+z).</cmath> | ||
+ | Similarly, <cmath>\cos z = \sin y \sin (x+y).</cmath> | ||
+ | |||
+ | <cmath>\begin{equation} \left\{ \begin{aligned} | ||
+ | \sin (x+z) = \frac {\sin y}{\cos z},\ | ||
+ | \sin (x+y) = \frac {\cos z}{\sin y}. | ||
+ | \end{aligned} \right.\end{equation}</cmath> | ||
+ | Therefore | ||
+ | <cmath>x+y = x+z = \frac{ \pi}{2} \implies y = z,</cmath> | ||
+ | <cmath>\sin y = \cos z \implies x = y = z = \frac {\pi}{4}.</cmath> | ||
+ | <i><b>Answer:<math>\left (\frac {\pi}{4},\frac {\pi}{4},\frac {\pi}{4} \right ).</math></b></i> | ||
+ | |||
+ | ==2023 var 231 problem 6== | ||
+ | Let positive numbers <math>a,b,c</math> be such that <math>\frac {1}{a+1} + \frac {1}{b+1} +\frac {1}{c+1} = 1. </math> | ||
+ | |||
+ | Find the maximum value of <math>\frac {a}{2 + a^2} + \frac {b}{2 + b^2} + \frac {c}{2 + c^2}.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | <cmath>\frac {a}{a^2 +2} \le \frac {1}{6} +\frac {1}{2(1+a)} \Leftrightarrow 6a + 6a^2 \le a^3 + a^2 +2a + 2 + 3a ^2 + 6 \Leftrightarrow</cmath> | ||
+ | <cmath>a^3 - 2 a^2 - 4a + 8 \ge 0 \Leftrightarrow (a -2)^2 (a+2) \ge 0.</cmath> | ||
+ | Similarly | ||
+ | <cmath>\frac {b}{b^2 +2} \le \frac {1}{6} +\frac {1}{2(1+b)}, \frac {c}{c^2 +2} \le \frac {1}{6} +\frac {1}{2(1+c)}.</cmath> | ||
+ | Adding this equations, we get: | ||
+ | <cmath>\frac {a}{2 + a^2} + \frac {b}{2 + b^2} + \frac {c}{2 + c^2} \le \frac {1}{2} \left (1 + \frac {1}{a+1} + \frac {1}{b+1} +\frac {1}{c+1} \right ) = 1.</cmath> | ||
+ | If <math>a = b = c = 2</math> then <math>\frac {a}{2 + a^2} + \frac {b}{2 + b^2} + \frac {c}{2 + c^2} = 1.</math> | ||
+ | |||
+ | <i><b>Answer:<math>1.</math></b></i> | ||
+ | |||
+ | <i><b>Explanation for students</b></i> | ||
+ | |||
+ | For the function under study <math>F(x) = \frac {x}{x^2 +2}</math> it is required to find the majorizing function <math>G(x) \ge F(x).</math> This function must be a linear combination of the given function <math>g(x) = \frac {1}{x+1}</math> and a constant, <math>G(x) = k + m \cdot g(x).</math> | ||
+ | |||
+ | At the supposed extremum point <math>x_0 = 2</math> the functions and their derivatives must coincide <math>G(x_0) = F(x_0), G'(x_0) = F'(x_0).</math> | ||
+ | <cmath>F(x_0) = \frac {x_0}{x_0^2 +2} = \frac {1}{3} = G(x_0) = k + m \cdot \frac {1}{x_0+1} = k + \frac {m}{3} \implies 1 = 3k + m.</cmath> | ||
+ | <cmath>F'(x_0) = \frac {2 - x_0^2}{(x_0^2 + 2)^2} = -\frac{1}{18} = G'(x_0) = m \cdot g'(x_0) = - \frac {m}{(1+ x_0)^2} = -\frac{m}{9} \implies m = \frac{1}{2}, k = \frac {1}{6}.</cmath> | ||
+ | |||
+ | ==2023 var 231 EM problem 6== | ||
+ | <cmath>F(x) = log_{\frac{5}{2}} (2 + \cos x) \cdot log_{\frac{5}{2}} (3 - \cos x).</cmath> | ||
+ | Find the maximum value <math>F_m = max (F(x))</math> and all argument values <math>x_0</math> such that <math>F_m = F(x_0)</math>. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | <cmath>a + t = 2 + \cos x; a - t = 3 - \cos x \implies a = \frac {5}{2}, t = \cos x - \frac {1}{2}.</cmath> | ||
+ | <cmath>u = \frac {t}{a} = \frac {\cos x - \frac {1}{2}}{a}\implies </cmath> | ||
+ | <cmath>F(u) = log_a {a(1 + u)} \cdot log_a {a(1 - u)} = (1 + log_a (1 + u)) \cdot (1 + log_a (1 - u)) = 1 + log_a (1- u^2) + log_a (1+ u) \cdot log_a (1 - u) \le 1,</cmath> | ||
+ | because <cmath>1- u^2 \le 1 \implies log_a (1- u^2) \le 0</cmath> | ||
+ | and signs of <math>log_a (1 + u)</math> and <math>log_a (1 - u)</math> are different, so <math>log_a (1 + u) \cdot log_a (1 - u) \le 0.</math> | ||
+ | Therefore <cmath>F_m = 1, \cos x_0 = \frac {1}{2}, x_0 = \pm \frac{\pi}{3} + 2 k \pi.</cmath> | ||
+ | |||
+ | ==2023 var 232 problem 6== | ||
+ | Let positive numbers <math>a,b,c</math> be such that <cmath>\left ( a+b+c \right) \cdot \left ( \frac {1}{a} + \frac {1}{b} +\frac {1}{c} \right ) = 10.</cmath> | ||
+ | Find the maximum value of <math>\frac {a+ b}{c}.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | <cmath>\left ( a+b+c \right) \cdot \left ( \frac {1}{a} + \frac {1}{b} +\frac {1}{c} \right ) = 10 \implies</cmath> | ||
+ | <cmath>\frac {b}{a} + \frac {a}{b} + \frac {b}{c} + \frac {a}{c} + \frac {c}{a} + \frac {c}{b} = 7.</cmath> | ||
+ | |||
+ | <cmath>\frac {a+b}{c} + \frac {c(a+b)}{ab} + \frac {b}{a} +\frac {a}{b} = 7.</cmath> | ||
+ | <cmath>\frac {a+b}{c} + \frac {c}{a+b}\cdot \frac {(a+b)^2}{ab} + \frac {b}{a} +\frac {a}{b} = 7.</cmath> | ||
+ | It is clear that <math>\frac {b}{a} +\frac {a}{b} \ge 2</math> and <math>\frac {(a+b)^2}{ab} \ge 4,</math> | ||
+ | Denote <math>X = \frac {a+b}{c}.</math> | ||
+ | So <cmath>7 = \frac {a+b}{c} + \frac {c}{a+b}\cdot \frac {(a+b)^2}{ab} + \frac {b}{a} +\frac {a}{b} \ge \frac {a+b}{c} + \frac {4c}{a+b}+ 2 = X + \frac {4}{X} + 2,</cmath> | ||
+ | <cmath>X + \frac {4}{X} \le 5 \implies (X - 4)(X - 1) \le 0 \implies X \le 4.</cmath> | ||
+ | If <math>a = b = 2, c = 1</math> then <math>\frac {a+b}{c} = 4.</math> | ||
+ | |||
+ | <i><b>Answer:<math>4.</math></b></i> | ||
+ | |||
+ | ==2023 var 233 problem 6== | ||
+ | Let positive numbers <math>a,b,c</math> be such that <math>a^2 + b^2 + c^2 = 1.</math> | ||
+ | |||
+ | Find the maximum value of <math>a b + b c \sqrt{3}.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>\vec X = \{a; c \}, \vec Y = \{1; \sqrt{3} \}</math> Then | ||
+ | <cmath>a + c \sqrt{3} = \vec X \cdot \vec Y \le | \vec X| \cdot |\vec Y| = \sqrt{a^2 + c^2} \cdot \sqrt{1 + 3 } = 2 \sqrt{a^2 + c^2}.</cmath> | ||
+ | |||
+ | <cmath>2 u v \le u^2 + v^2 \implies 2 b \cdot \sqrt{a^2 + c^2} \le b^2 + (a^2 + c^2) = 1.</cmath> | ||
+ | |||
+ | Equality is achieved if <cmath>\frac {c}{a} = \frac {\sqrt{3}}{1}, b^2 = a^2 + c^2 \implies a = \frac{1}{2 \sqrt{2}}, b = \frac{1}{ \sqrt{2}}, c = \frac{\sqrt{3}}{2 \sqrt{2}}.</cmath> | ||
+ | |||
+ | <i><b>Answer: <math> 1.</math></b></i> | ||
+ | |||
+ | ==2024 Problem 18 (EGE)== | ||
+ | [[File:2024 18 EGE.gif|330px|right]] | ||
+ | |||
+ | Find those values of the parameter a for which the system of equations has exactly one solution: | ||
+ | <cmath>\left\{ | ||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | 1. Special case <math>a = 0 \implies x = \frac{3}{2}, y = 0</math> exactly one solution. | ||
+ | |||
+ | 2. <math>|y| \ge 0 \implies \frac {3 - 2x}{x} \ge 0 \implies x \in \left [ 0; \frac{3}{2} \right ].</math> | ||
+ | |||
+ | 3. We solve the first equation with respect <math>y</math> and get <math>y = \pm \left( \frac{3}{x} - 2 \right ).</math> | ||
+ | |||
+ | This solution is shown in the diagram by red curve. | ||
+ | |||
+ | We solve the second equation with respect <math>y</math> and get | ||
+ | <cmath>y = \frac {3 - 2x}{2a} - \frac{1}{2} = -\frac {x}{a} + \frac{3 - a}{2a}.</cmath> | ||
+ | This solution is shown in the diagram by segments which connect point <math>\left ( \frac{3}{2}, -\frac{1}{2} \right)</math> with axis <math>x = 0.</math> | ||
+ | |||
+ | Each solution of the system is shown by the point of crosspoint red curve with segment. | ||
+ | |||
+ | If <math>a = \frac {1}{3}</math> then segment (colored by blue) is tangent to red curve (discriminant is zero), so we have two solutions (1,1) and <math>\approx(1.4,-0.4).</math> | ||
+ | |||
+ | If <math>a \in (0, \frac{1}{3})</math> we get three solutions (colored by yellow). | ||
+ | |||
+ | In other cases the system has exactly one solution. | ||
+ | |||
+ | <i><b>Answer: </b></i> <math>(- \infty,0 ]\cup (\frac{1}{3}, \infty ).</math> | ||
+ | ==2024 Test problem 7== | ||
+ | Find all values of the parameter a for which there is at least one solution to the inequality <cmath>\frac {1}{x} + \frac{2}{a} \le \frac {3}{a - x} - \frac{1}{x+a}</cmath> on the interval <math>x \in [2,3]</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <math>\frac {1}{x} + \frac{2}{a} \le \frac {3}{a - x} - \frac{1}{x+a} \leftrightarrow F(x,a) \ge 0,</math> where <math>F(x,a) = (a + 2x) \cdot\left( \frac{2}{a^2-x^2}-\frac{1}{ax} \right) = \frac {2(x+a/2)(x+a_1) (x-a_2)}{ax(x+a)(a - x)},</math> where <math>a_1 = a(\sqrt{2}+1), a_2= a(\sqrt{2}-1).</math> | ||
+ | |||
+ | The equation <math>F(x,a) = 0</math> has solutions <math>x= -\frac{a}{2}, x = -a_1,</math> and <math>x = a_2.</math> | ||
+ | <cmath>F(2,a) = \frac {(4+a)(2+a_1) (2-a_2)}{2a(2+a)(a -2)}.</cmath> | ||
+ | <math>F(2,a) \ge 0</math> if <math>a \in [-4, -2) \cup [2-2\sqrt{2},0) \cup (2, 2 \sqrt{2}+2],</math> so given inequality has the solution <math>x=2</math> for these <math>a.</math> | ||
+ | <cmath>F(3,a) = \frac {(6+a)(3+a_1) (3-a_2)}{3a(3+a)(a -3)}.</cmath> | ||
+ | <math>F(3,a) \ge 0</math> if <math>a \in [-6, -3) \cup [3-3\sqrt{2},0) \cup (3, 3 \sqrt{2}+3],</math> so given inequality has the solution <math>x=3</math> for these <math>a.</math> | ||
+ | |||
+ | <math>a \in (-\infty, -6), x \in [2,3] \implies F(x,a) < 0,</math> no solution of the given inequality. | ||
+ | |||
+ | <math>F(x,-2) < 0</math> no solution of the inequality if <math>a = -2.</math> | ||
+ | |||
+ | <math>a \in (-2, 3(1 - \sqrt{2})).</math> If <math>x \in [2,3] F(x,a) < 0 \implies</math> no solution of the inequality. | ||
+ | |||
+ | <math>a \in (0, 2). </math> If <math>x \in [2,3] F(x,a) < 0 \implies</math> no solution of the given inequality. | ||
+ | |||
+ | <math>a \in (3 + 3\sqrt{2},\infty).</math> If <math>x \in [2,3] F(a,x) < 0 \implies</math> no solution of the given inequality. | ||
+ | |||
+ | ==2024 var 241 Problem 2== | ||
+ | The natural numbers <math>a_1,...a_n</math> form a strictly increasing arithmetic progression. Find all possible values of <math>n</math> if it is known that <math>n</math> is odd, <math>n > 1</math> and <math>a_1 + a_2+...+a_n = 2024.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <math>n</math> is odd, so <math>a_1 + a_2+...+a_n = n a_\frac{n+1}{2} = 2024 = 8 \cdot 11 \cdot 23.</math> | ||
+ | |||
+ | Let <math>n = 11 \implies a_6 = 8 \cdot 23 = 184,</math> the common difference may be <math>1,</math> increasing arithmetic progression exist. | ||
+ | |||
+ | Let <math>n = 23 \implies a_{12} = 8 \cdot 11 = 88,</math> the common difference may be <math>1,</math> increasing arithmetic progression exist. | ||
+ | |||
+ | Let <math>n = 11 \cdot 23 = 253 \implies a_{127} = 8 \implies a_{100} < 0</math> can not be the natural number. | ||
+ | |||
+ | Answer: <math>11,23.</math> | ||
+ | |||
+ | ==2024 var 242 Problem 7== | ||
+ | [[File:2024 1 problem 7.png|390px|right]] | ||
+ | The base of the pyramid is the trapezoid <math>ABCD, AD||BC, AD = 2BC.</math> | ||
+ | |||
+ | A sphere of radius <math>1</math> touches the plane of the base of the pyramid and the planes of its lateral faces <math>ADS</math> and <math>BCS</math> at points <math>P,Q,</math> and <math>T,</math> respectively. | ||
+ | |||
+ | Find the ratio in which the volume of the pyramid is divided by the plane <math>ADT,</math> if the face <math>ADS</math> is perpendicular to the plane <math>ABD</math> and the height of the pyramid is <math>4.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | A sketch of the given pyramid is shown in the diagram. The planes <math>SAD</math> and <math>SBC</math> intersect along the straight line <math>SK||AD||BC,</math> that is, the planes <math>ABCD, SBC, SAD</math> form the lateral surface of a prism into which a sphere with center at point <math>I</math> is inscribed. | ||
+ | |||
+ | The plane <math>\pi</math> containing the point <math>I</math> and perpendicular to <math>AD</math> contains points <math>P, Q,</math> and <math>T.</math> Plane <math>\pi</math> intersects parallel lines <math>AD, BC,</math> and <math>SK</math> at points <math>N,L,</math> and <math>K,</math> respectively. | ||
+ | |||
+ | Let <math>EF</math> be the line parallel to <math>AD, E \in BS, F \in CS.</math> The plane <math>AETFD</math> cuts off the pyramid <math>SAEFD</math> with volume <math>v</math> from the pyramid <math>SABCD</math> with volume <math>V.</math> | ||
+ | |||
+ | <math>KN \perp AD</math> and equal to the distance from <math>S</math> to <math>AD, KN = 4, LN \perp AD, LN \perp KN</math> and equal to the distance between <math>BC</math> and <math>AD.</math> | ||
+ | <cmath>V = \frac {1}{3} KN \cdot KL \cdot \frac { AD + BC}{2} = BC \cdot \frac {KN \cdot KL}{2} = BC \cdot [KNL].</cmath> | ||
+ | Consider a right triangle <math>KLN ([KLN]</math> is the area of <math>\triangle KLN)</math> into which a circle <math>PQT</math> with radius <math>r = 1</math> is inscribed. | ||
+ | <cmath>QN = PN = r, KQ = KN - NQ = 3 = KT, x = TL = PL \implies</cmath> | ||
+ | <cmath>(x+1)^2 + 4^2 = (x+3)^2 \implies x = 2 \implies \frac {EF}{BC} = \frac {KT}{KL} = \frac {3}{5}.</cmath> | ||
+ | We are looking for <math>v.</math> Let <math>h</math> be the distance from <math>S</math> to the plane <math>AETFD.</math> | ||
+ | <cmath>v = \frac {1}{3} h \cdot NT \cdot \frac {AD + EF}{2} = BC \cdot (2 + \frac{3}{5}) \cdot \frac {h \cdot NT}{2} = BC \cdot \frac{13}{15} \cdot [KNT].</cmath> | ||
+ | <cmath>\frac {[KNT]} {[KNL]} = \frac {KT}{KL} = \frac {3}{5} \implies \frac {v}{V} = \frac{13}{15} \cdot \frac{3}{5} = \frac{13}{25} \implies \frac {v}{V-v} = \frac{13}{12}.</cmath> | ||
+ | Answer: <math>13 : 12.</math> | ||
+ | |||
+ | ==2024 var 243 Problem 6== | ||
+ | Solve the system of equations in the positive <math>x, y, z:</math> | ||
+ | <cmath>\left\{ | ||
+ | |||
+ | <i><b>Solution (after Natalia Zakharova)</b></i> | ||
+ | <cmath>x^4 + x^2\cdot y^2 + y^4 = (x^2 + xy + y^2) \cdot (x^2 - xy + y^2),</cmath> | ||
+ | <cmath>y^4 + y^2 \cdot z^2 + z^4 = (y^2 + yz + z^2) \cdot (y^2 - yz + z^2),</cmath> | ||
+ | <cmath>z^4 + z^2\cdot x^2 + x^4 = (x^2 + xz + z^2) \cdot (x^2 - xz + z^2) \implies</cmath> | ||
+ | <cmath>\frac {(x^4 + x^2\cdot y^2 + y^4) (y^4 + y^2 \cdot z^2 + z^4)(z^4 + z^2\cdot x^2 + x^4)}{(x^2 + xy + y^2) (y^2 + yz + z^2)(z^2 + zx + x^2)} = (x^2 - xy + y^2) (y^2 - yz + z^2)(z^2 - zx + x^2) = (xyz)^2.</cmath> | ||
+ | <cmath>x^2 - xy + y^2 \ge 2xy - xy = xy, y^2 - yz + z^2 \ge yz, z^2 - zx + x^2 \ge zx \implies</cmath> | ||
+ | <cmath>(xyz)^2 = (x^2 - xy + y^2) (y^2 - yz + z^2)(z^2 - zx + x^2) \ge xy \cdot yz \cdot zx = (xyz)^2 \implies x = y = z \implies x = \frac{1}{3}.</cmath> | ||
+ | Answer: <math>x = y = z = \frac{1}{3}.</math> | ||
+ | |||
+ | ==2024 var 244 Problem 7== | ||
+ | [[File:2024 244 problem 7.png|320px|right]] | ||
+ | [[File:2024 244 problem 7a.png|320px|right]] | ||
+ | Let <math>ABCDA'B'C'D'</math> be the cube, <math>AB = 1</math>. Let <math>K \in A'B', L \in B'B,</math> | ||
+ | <math>M \in BC, N \in CD, Q \in DD', P \in A'D', \angle A'AK = \angle LAK,</math> | ||
+ | <cmath>\angle BAM = \angle MAN, \angle DAQ = \angle PAQ,</cmath> | ||
+ | <cmath>A'K + LB = BM + ND = DQ + PA' = \frac {5}{4}.</cmath> | ||
+ | |||
+ | Find the ratio in which the plane <math>KMQ</math> divides the volume of the cube. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | 1. Let <math>F</math> lie on the ray <math>B'A', A'F = BL.</math> | ||
+ | <cmath>AB = AA', \angle ABL = \angle AA'F \implies \triangle AA'F = \triangle ABL \implies AL = AF.</cmath> | ||
+ | <cmath>\angle KAF = \angle A'AF + \angle KAA' = \angle BAL + \angle KAA' =</cmath> | ||
+ | <cmath>= 90^\circ - \angle KAL = 90^\circ - \angle KAA' = \angle AKF \implies</cmath> | ||
+ | <cmath>AF = KF = KA' + BL = \frac {5}{4} = AL.</cmath> | ||
+ | So <math>BL = \sqrt{AL^2 – AB^2} = \frac {3}{4} \implies KA' = KB' = \frac {1}{2}.</math> | ||
+ | |||
+ | Similarly, <math>M</math> is the midpoint <math>BC, Q</math> is the midpoint <math>DD'.</math> | ||
+ | |||
+ | 2. <math>KQ = KM = QM, AQ = AK = AM = C'M = C'K = C'Q =\frac {\sqrt{5}}{2} \implies </math> | ||
+ | |||
+ | regular pyramids are equal <math>AKMQ = C'KMQ.</math> So <math>O</math> (midpoint <math>AC'</math>) lies in plane <math>KMQ.</math> | ||
+ | |||
+ | Let <math>M'</math> be the midpoint <math>A'D' \implies M'</math> symmetric to <math>M</math> with respect <math>O,</math> so <math>M' \in KMQ.</math> | ||
+ | |||
+ | Similarly <math>K' \in KMQ, Q' \in KMQ,</math> where <math>K'</math> midpoint <math>CD, Q'</math> the midpoint <math>BB'.</math> | ||
+ | |||
+ | For each point on the edges of the solid forming a part of the cube cut off by a plane <math>KMQ</math> from the side of vertex <math>A,</math> one can find a point symmetrical relative to the center of the cube <math>O</math> on the edges of the solid forming another part of the cube. | ||
+ | |||
+ | It means that these parts are congruent and the plane <math>KMQ</math> divides the cube in half. | ||
+ | |||
+ | Answer: <math>1:1.</math> | ||
+ | ==2024 var 245 Problem 6== | ||
+ | Let <math>a,b,c,</math> and <math>d</math> be the positive real numbers such that <math>a+b+c+d = 1.</math> Find the minimal value of <math> \frac {a^2}{1-a} + \frac {b^2}{1-b}+ \frac {c^2}{1-c} + \frac {d^2}{1-d}.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <cmath> \frac {a^2}{1-a} + a = \frac {a}{1-a} = \frac{a}{b+c+d} = \frac{1}{\frac{b}{a}+\frac {c}{a}+\frac{d}{a}} \le \frac{1}{9} \left( \frac{a}{b}+\frac {a}{c}+\frac{a}{d}\right).</cmath> | ||
+ | For the last transform we use unequality between the harmonic mean and the arithmetic mean for three numbers. | ||
+ | Therefore | ||
+ | <cmath> \frac {a^2}{1-a} + \frac {b^2}{1-b}+ \frac {c^2}{1-c} + \frac {d^2}{1-d} \le \frac{1}{9} \left( \frac{a}{b}+\frac{b}{a}+\frac {a}{c}+\frac{c}{a}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{b}{d}+\frac{d}{b}+\frac{c}{d}+\frac{d}{c}\right) - a - b - c - d \le \frac{12}{9} - 1 = \frac{1}{3}.</cmath> | ||
+ | Equality we get if <math>a = b = c = d = \frac {1}{4}.</math> | ||
+ | |||
+ | Answer: <math>\frac{1}{3}.</math> | ||
+ | ==2024 var 246 Problem 5== | ||
+ | [[File:2024 var 246 5.png|390px|right]] | ||
+ | Let <math>\triangle ABC</math> be given, <math>\angle BAC > 90^\circ.</math> Point <math>D</math> is located on side <math>BC</math> so that <math>AC = CD,</math> the circle <math>\odot ACD</math> touches <math>AB</math> at point <math>A.</math> | ||
+ | |||
+ | Point <math>E</math> is located on ray <math>AD</math> so that <math>CE = EA = AB.</math> | ||
+ | |||
+ | Find the ratio of <math>BC : AB.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <cmath>\frac {\overset{\Large\frown} {AD}}{2} = \angle BAD = \angle ACB.</cmath> | ||
+ | Triangles <math>\triangle CAD</math> and <math>\triangle EAC</math> are isosceles with a common angle <math>\angle DAC \implies</math> | ||
+ | <cmath>\angle AEC = \angle ACD = \angle BAD \implies AB||CE.</cmath> | ||
+ | <math>AB = CE, AB||CE \implies ABEC</math> is the parallelogram <math>\implies BC = 2 BD.</math> | ||
+ | <cmath>\triangle ABD \sim \triangle CBA \implies \frac {AB}{BC} = \frac {BD}{AB} \implies 2 AB^2 = BC^2 \implies \frac {BC}{AB} = \sqrt{2}.</cmath> | ||
+ | Answer: <math>\sqrt{2}.</math> | ||
+ | ==2024 var 246 Problem 6== | ||
+ | |||
+ | <math>f(x) = x^4 – 12x^3 +ax^2+bx +81,</math> where <math>a</math> and <math>b</math> are real numbers. | ||
+ | |||
+ | Find <math>f(5)</math> if <math>f(x) = (x - c_1)(x - c_2)(x - c_3)(x – c_4)</math> where | ||
+ | |||
+ | 1) <math>c_i, i = 1..4</math> are real numbers, | ||
+ | |||
+ | 2) <math>c_i, i = 1..4</math> are positive numbers. | ||
+ | |||
+ | <i><b>Solution (after Natalia Zakharova)</b></i> | ||
+ | |||
+ | 1) <math>f(x) = (x^2 - t^2)(x^2 - 12x - \frac{9^2}{t^2})</math> has four real roots | ||
+ | <cmath>c_1 = t, c_2 = -t, c_3 = 6 + \sqrt{36 + 81/t^2}, c_4 = 6 - \sqrt{36 + 81/t^2}</cmath> | ||
+ | if <math>t \ne 0.</math> | ||
+ | <cmath>f(5) = (t^2 - 25)(35 + \frac{9^2}{t^2}) = 35t^2 - \frac{2025}{t^2} - 794.</cmath> | ||
+ | <math>F(5)</math> can take any real values. There are positive and negative roots <math>c_1 \cdot c_2 < 0.</math> | ||
+ | |||
+ | 2) <math>c_1 + c_2 + c_3 + c_4 = 12, c_1 \cdot c_2 \cdot c_3 \cdot c_4 = 81,</math> | ||
+ | <cmath>\frac {c_1 + c_2 + c_3 + c_4}{4} = 3 = \sqrt[4]{c_1 \cdot c_2 \cdot c_3 \cdot c_4} \implies c_1 = c_2 = c_3 = c_4 = 3 \implies f(x) = (x-3)^4, f(5) = 2^4 = 16.</cmath> | ||
+ | Answer: <math>1)</math> any real number, <math>2) 16.</math> | ||
+ | |||
+ | ==2024 var 246 Problem 7== | ||
+ | [[File:2024 var 246 7.png|300px|right]] | ||
+ | The distance from the midpoint <math>M</math> of the height <math>SO</math> of a regular quadrangular pyramid <math>SABCD</math> to the lateral face is <math>MF' = \sqrt{2}</math> and to the lateral edge is <math>ME' = \sqrt{3}.</math> | ||
+ | |||
+ | Find the volume of the pyramid. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>K</math> be the midpoint <math>AB, OF \perp SK, OE \perp AS \implies</math> | ||
+ | <cmath>EO = 2 E'M = 2 \sqrt{3}, FO = 2 F'M = 2 \sqrt{3}.</cmath> | ||
+ | Denote <math>AB = 2a, SO = h \implies KO = a, AO = a \sqrt{2}.</math> | ||
+ | |||
+ | Let us express the heights of right triangles through their legs: | ||
+ | <cmath>\frac {1}{EO^2} = \frac {1}{AO^2} + \frac {1}{SO^2} \implies \frac {1}{12} = \frac {1}{2a^2} + \frac {1}{h^2}.</cmath> | ||
+ | <cmath>\frac {1}{FO^2} = \frac {1}{KO^2} + \frac {1}{SO^2} \implies \frac {1}{8} = \frac {1}{a^2} + \frac {1}{h^2}.</cmath> | ||
+ | <cmath>\frac {1}{2a^2} = \frac {1}{8} - \frac {1}{12} = \frac {1}{24} \implies a^2 = 12.</cmath> | ||
+ | <cmath>\frac {1}{h^2} = \frac {1}{8} - \frac {1}{12} = \frac {1}{24} \implies h = 2 \sqrt{6} \implies V = \frac {h \cdot 4a^2}{3} = 32 \sqrt{6}.</cmath> | ||
+ | Answer: <math>32 \sqrt{6}.</math> | ||
+ | |||
+ | ==2024 var 247 Problem 6== | ||
+ | [[File:2024 7 problem 6.png|300px|right]] | ||
+ | Real numbers <math>a, b,</math> and <math>c</math> satisfy the system of equations | ||
+ | <cmath>\left\{ | ||
+ | Find the largest possible value of <math>c.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | In coordinates <math>a, b,</math> and <math>c</math> the first equation defines the plane <math>ABC,</math> the second - a sphere with the center at the origin. They are shown in the diagram. | ||
+ | |||
+ | The solution of the given system (if it exists) is a circle symmetrical with respect to the plane <math>a = b.</math> This plane intersects the plane of the first equation along the line <math>CED</math> on which the points of maximum (E) and minimum (D) of the values of <math>c</math> are located. | ||
+ | |||
+ | At these points the system takes the form | ||
+ | <cmath>\left\{ | ||
+ | <cmath>(4 - c)^2 = 4a^2 = 2(8 - c^2) \implies c^2 - 8c = - 2 c^2.</cmath> | ||
+ | These system has two solutions <math>c_1 = 0, c_2 = \frac {8}{3},</math> so solution of the given system exist. | ||
+ | |||
+ | Answer: <math>\frac {8}{3}.</math> | ||
+ | |||
+ | ==2024 var 247 Problem 7== | ||
+ | [[File:2024 247 problem 7.png|300px|right]] | ||
+ | Let the cube <math>ABCDA'B'C'D'</math> be given. Let points <math>K \in A'B', L \in BC, M \in CD,</math> and <math>N \in A'D'</math> be given, <math>A'K = BL, A'N = DM.</math> | ||
+ | |||
+ | Let the plane <math>A'BD</math> cross the plane <math>ALN</math> by the line <math>\ell,</math> and cross the plane <math>AKM</math> by line <math>\ell'.</math> | ||
+ | |||
+ | Find the angle between <math>\ell</math> and <math>\ell'.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Denote <math>F = AL \cap BD, E = AN \cap A'D \implies \ell = EF,</math> | ||
+ | |||
+ | <math>E' = AM \cap BD, F' = A'B \cap AK \implies \ell' = E'F'.</math> | ||
+ | <cmath>A'K = BL, \angle BA'B' = \angle KA'F' = \angle CBD = \angle LBF, \angle AKA' = \angle F'KA' =</cmath> | ||
+ | <cmath>= \angle ALB = \angle FLB \implies \triangle A'KF' = \triangle BLF \implies A'F' = BF.</cmath> | ||
+ | Similarly, <math>A'E = DE'.</math> | ||
+ | <math>A'B = BD = DA' \implies A'BD</math> is the regular triangle. | ||
+ | |||
+ | Denote <math>r</math> the rotation of the plane <math>A'BD</math> around the center of <math>\triangle A'BD</math> by an angle of <math>60^\circ</math> which maps point <math>A'</math> into point <math>B.</math> | ||
+ | |||
+ | The transforming <math>r</math> maps point <math>F'</math> into point <math>F,</math> point <math>E'</math> into point <math>E,</math> that is, line <math>\ell' = E'F'</math> maps into line <math>\ell.</math> | ||
+ | |||
+ | The angle between these lines is <math>60^\circ.</math> | ||
+ | |||
+ | Answer: <math>60^\circ.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 07:59, 30 July 2024
DVI is an exam in mathematics at the Moscow State University named after M.V. Lomonosov. The first four problems have a standard level. Problem 5 is advanced level of geometry. Problem 6 is an advanced level equation or inequality. Problem 7 is advanced level of stereometry.
Below are the most difficult problems of this exam in recent years. The headings indicate the year when the problem was used, the variant option of the exam, and the number of the problem.
Contents
[hide]- 1 2011 Problem 8
- 2 2012 Problem 8
- 3 2014 1 Problem 6
- 4 2014 1 Problem 8
- 5 2015 1 Problem 7
- 6 2016 2 Problem 7
- 7 2016 2 Problem 8
- 8 2020 var 201 problem 6
- 9 2020 var 202 problem 6
- 10 2020 var 203 problem 6
- 11 2020 var 204 problem 6
- 12 2020 var 205 problem 6
- 13 2020 var 206 problem 6
- 14 2021 var 215 problem 7
- 15 2022 var 221 problem 7
- 16 2022 var 222 problem 7
- 17 2022 var 222 problem 6
- 18 2022 var 224 problem 6
- 19 2023 var 231 problem 6
- 20 2023 var 231 EM problem 6
- 21 2023 var 232 problem 6
- 22 2023 var 233 problem 6
- 23 2024 Problem 18 (EGE)
- 24 2024 Test problem 7
- 25 2024 var 241 Problem 2
- 26 2024 var 242 Problem 7
- 27 2024 var 243 Problem 6
- 28 2024 var 244 Problem 7
- 29 2024 var 245 Problem 6
- 30 2024 var 246 Problem 5
- 31 2024 var 246 Problem 6
- 32 2024 var 246 Problem 7
- 33 2024 var 247 Problem 6
- 34 2024 var 247 Problem 7
2011 Problem 8
Solve the system of equations Standard Solution Denote We get First equation define inner points of the circle with radius and the circle. The distance from the straight line to the origin of the coordinate system is so the system of the equations define the only tangent point of the circle and the line. Short Solution
2012 Problem 8
Let the tetrahedron be given.
A right circular cylinder is located so that the circle of its upper base touches each of the faces which contains vertex
The circle of the lower base lies in the plane and touches straight lines and
Find the height of the cylinder.
Solution
Denote the midpoint Plane is the bisector plane of segment
The inradius of equal to distance from incenter to vertex is
Denote the foot from to
Denote the crosssection of by plane of the upper base of cylinder, is the incenter is the point of tangency incircle of and
Denote and the foots from and to Denote the radius
The circle of the lower base inscribed in angle equal to so Projection from the point maps onto
Answer:
2014 1 Problem 6
Find all pares of real numbers satisfying the system of equations Solution
Denote Denote is the solution. Let If then if then therefore is the single root.
2014 1 Problem 8
Let
Find and
Solution where
Answer:
2015 1 Problem 7
A sphere is inscribed in a regular triangular prism with bases Find its radius if the distance between straight lines and is equal to where and are points lying on and , respectively, and
Solution
The distance from the center of the sphere to the centers of the prism faces is equal to so
In order to find the distance between the lines and , one can find the length of two perpendiculars and to the line that are perpendicular to each other. Then since, when viewed along a straight line , the segment is the altitude of a right triangle with legs and
The plane containe the straight line The straight line crossed at the point In a right triangle is the height falling on the hypotenuse,
Let be the projection of onto plane
Therefore is the projection of onto plane at the point Answer:
2016 2 Problem 7
Let the base of the regular pyramid with vertex be the hexagon with side The plane is parallel to the edge , perpendicular to the plane and intersects the edge at point so that The lines along which intersects the plane and the base plane are perpendicular.
Find the area of the triangle cut off by the plane from the face
Solution
Denote
are the midpoints of respectively.
Plane is the plane symmetry of pyramid,
By condition so exist point
is the line along which intersects the plane, is the line along which intersects the base plane, so
We use the top wiew and get Denote and use the side wiew.
Triangle is the regular triangle with side , so Answer: 8.
2016 2 Problem 8
Find the smallest value of the expression Solution Denote The shortest length of a broken line with fixed ends is equal to the distance between points and which is and is achieved if points and are collinear. Answer:
2020 var 201 problem 6
Let a triangular prism with a base be given, Find the ratio in which the plane divides the segment if
Solution
Let be the parallel projections of on the plane
We use and get Let
Similarly
Answer:
2020 var 202 problem 6
Let a tetrahedron be given, Find the cosine of the angle between the edges and
Solution
Let us describe a parallelepiped around a given tetrahedron
and are equal rectangles.
and are equal rectangles.
Denote Answer:
2020 var 203 problem 6
Let a cube with the base and side edges be given. Find the volume of a polyhedron whose vertices are the midpoints of the edges
Solution
Denote the vertices of polyhedron Triangles and are equilateral triangles with sides and areas
This triangles lies in parallel planes, which are normal to cube diagonal The distance between this planes is So the volume of the regular prism with base and height is
Let the area be the quadratic function of Let Suppose, we move point along axis and cross the solid by plane contains and normal to axis. Distance from to each crosspoint this plane with the edge change proportionally position along axes, so the area is quadratic function from position.
Answer:
2020 var 204 problem 6
Let a regular triangular pyramid be given. The circumcenter of the sphere is equidistant from the edge and from the plane of the base of the pyramid. Find the radius of the sphere inscribed in this pyramid if the length of the edge of its base is
Solution
Answer:
2020 var 205 problem 6
Let the quadrangular pyramid with the base parallelogram be given.
Point Point
Find the ratio in which the plane divides the volume of the pyramid.
Solution
Let plane cross edge at point We make the central projection from point The images of points are respectively. The image of is the crosspoint of and So lines and are crossed at point Let’s compare volumes of some tetrachedrons, denote the volume of as Answer: 1 : 6.
2020 var 206 problem 6
Given a cube with the base and side edges Find the distance between the line passing through the midpoints of the edges and and the line passing through the midpoints of the edges and
Solution
Let points be the midpoints of respectively. We need to prove that planes and are parallel, perpendicular to Therefore,
Point is the midpoint For proof we can use one of the following methods:
1. Vectors: Scalar product Similarly,
2.
3. Rotating the cube around its axis we find that the point move to , then to then to
Answer:
2021 var 215 problem 7
The sphere touches all edges of the tetrahedron It is known that the products of the lengths of crossing edges are equal. It is also known that Find
Solution
The tangent segments from the common point to the sphere are equal.
Let us denote the segments from the vertex to the sphere by
Similarly, we define If then
If
The tetrahedron is a regular pyramid with a regular triangle with side at the base and side edges equal to
Answer: 3.
2022 var 221 problem 7
The volume of a triangular prism with base and side edges is equal to Find the volume of the tetrahedron where is the centroid of the face is the point of intersection of the medians of is the midpoint of the edge and is the midpoint of the edge
Solution
Let us consider the uniform triangular prism Let be the midpoint of be the midpoint of be the midpoint of be the midpoint of
The area of in the sum with the areas of triangles is half the area of rectangle so Denote the distance between these lines The volume of the tetrahedron is The volume of the prism is
An arbitrary prism is obtained from a regular one as a result of an affine transformation.
All points on the tetrahedron are defined affinely, which means that the volume ratio will be preserved.
Answer: 5.
2022 var 222 problem 7
A sphere of diameter is inscribed in a pyramid at the base of which lies a rhombus with an acute angle and side Find the angle if it is known that all lateral faces of the pyramid are inclined to plane of its base at an angle of
Solution 1
Denote rhombus is the vertex of a pyramid is the center of the sphere, is the tangent point of and sphere, Solution 2
The area of the rhombus
The area of the lateral surface is Answer:
2022 var 222 problem 6
Find all possible values of the product if it is known that and it is true
Solution
Let then for each equation is true, Let no solution.
Answer:
2022 var 224 problem 6
Find all triples of real numbers in the interval satisfying the system of equations
Solution
Denote Similarly,
Therefore Answer:
2023 var 231 problem 6
Let positive numbers be such that
Find the maximum value of
Solution Similarly Adding this equations, we get: If then
Answer:
Explanation for students
For the function under study it is required to find the majorizing function This function must be a linear combination of the given function and a constant,
At the supposed extremum point the functions and their derivatives must coincide
2023 var 231 EM problem 6
Find the maximum value and all argument values such that .
Solution because and signs of and are different, so Therefore
2023 var 232 problem 6
Let positive numbers be such that Find the maximum value of
Solution
It is clear that and Denote So If then
Answer:
2023 var 233 problem 6
Let positive numbers be such that
Find the maximum value of
Solution
Let Then
Equality is achieved if
Answer:
2024 Problem 18 (EGE)
Find those values of the parameter a for which the system of equations has exactly one solution: Solution
1. Special case exactly one solution.
2.
3. We solve the first equation with respect and get
This solution is shown in the diagram by red curve.
We solve the second equation with respect and get This solution is shown in the diagram by segments which connect point with axis
Each solution of the system is shown by the point of crosspoint red curve with segment.
If then segment (colored by blue) is tangent to red curve (discriminant is zero), so we have two solutions (1,1) and
If we get three solutions (colored by yellow).
In other cases the system has exactly one solution.
Answer:
2024 Test problem 7
Find all values of the parameter a for which there is at least one solution to the inequality on the interval
Solution
where where
The equation has solutions and if so given inequality has the solution for these if so given inequality has the solution for these
no solution of the given inequality.
no solution of the inequality if
If no solution of the inequality.
If no solution of the given inequality.
If no solution of the given inequality.
2024 var 241 Problem 2
The natural numbers form a strictly increasing arithmetic progression. Find all possible values of if it is known that is odd, and
Solution
is odd, so
Let the common difference may be increasing arithmetic progression exist.
Let the common difference may be increasing arithmetic progression exist.
Let can not be the natural number.
Answer:
2024 var 242 Problem 7
The base of the pyramid is the trapezoid
A sphere of radius touches the plane of the base of the pyramid and the planes of its lateral faces and at points and respectively.
Find the ratio in which the volume of the pyramid is divided by the plane if the face is perpendicular to the plane and the height of the pyramid is
Solution
A sketch of the given pyramid is shown in the diagram. The planes and intersect along the straight line that is, the planes form the lateral surface of a prism into which a sphere with center at point is inscribed.
The plane containing the point and perpendicular to contains points and Plane intersects parallel lines and at points and respectively.
Let be the line parallel to The plane cuts off the pyramid with volume from the pyramid with volume
and equal to the distance from to and equal to the distance between and Consider a right triangle is the area of into which a circle with radius is inscribed. We are looking for Let be the distance from to the plane Answer:
2024 var 243 Problem 6
Solve the system of equations in the positive
Solution (after Natalia Zakharova) Answer:
2024 var 244 Problem 7
Let be the cube, . Let
Find the ratio in which the plane divides the volume of the cube.
Solution
1. Let lie on the ray So
Similarly, is the midpoint is the midpoint
2.
regular pyramids are equal So (midpoint ) lies in plane
Let be the midpoint symmetric to with respect so
Similarly where midpoint the midpoint
For each point on the edges of the solid forming a part of the cube cut off by a plane from the side of vertex one can find a point symmetrical relative to the center of the cube on the edges of the solid forming another part of the cube.
It means that these parts are congruent and the plane divides the cube in half.
Answer:
2024 var 245 Problem 6
Let and be the positive real numbers such that Find the minimal value of
Solution
For the last transform we use unequality between the harmonic mean and the arithmetic mean for three numbers. Therefore Equality we get if
Answer:
2024 var 246 Problem 5
Let be given, Point is located on side so that the circle touches at point
Point is located on ray so that
Find the ratio of
Solution
Triangles and are isosceles with a common angle is the parallelogram Answer:
2024 var 246 Problem 6
where and are real numbers.
Find if where
1) are real numbers,
2) are positive numbers.
Solution (after Natalia Zakharova)
1) has four real roots if can take any real values. There are positive and negative roots
2) Answer: any real number,
2024 var 246 Problem 7
The distance from the midpoint of the height of a regular quadrangular pyramid to the lateral face is and to the lateral edge is
Find the volume of the pyramid.
Solution
Let be the midpoint Denote
Let us express the heights of right triangles through their legs: Answer:
2024 var 247 Problem 6
Real numbers and satisfy the system of equations Find the largest possible value of
Solution
In coordinates and the first equation defines the plane the second - a sphere with the center at the origin. They are shown in the diagram.
The solution of the given system (if it exists) is a circle symmetrical with respect to the plane This plane intersects the plane of the first equation along the line on which the points of maximum (E) and minimum (D) of the values of are located.
At these points the system takes the form These system has two solutions so solution of the given system exist.
Answer:
2024 var 247 Problem 7
Let the cube be given. Let points and be given,
Let the plane cross the plane by the line and cross the plane by line
Find the angle between and
Solution
Denote
Similarly, is the regular triangle.
Denote the rotation of the plane around the center of by an angle of which maps point into point
The transforming maps point into point point into point that is, line maps into line
The angle between these lines is
Answer:
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