Difference between revisions of "DVI exam"
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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. | 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. | ||
− | ==2020 201 problem 6== | + | ==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> | ||
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Answer: <math>AG : GA' = 4 : 3.</math> | Answer: <math>AG : GA' = 4 : 3.</math> | ||
− | ==2020 202 problem 6== | + | ==2020 var 202 problem 6== |
[[File:2020 202 6.png|330px|right]] | [[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> | 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> | ||
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Answer: <math>\frac {11}{30}. </math> | Answer: <math>\frac {11}{30}. </math> | ||
− | + | ==2020 var 203 problem 6== | |
− | |||
− | ==2020 203 problem 6== | ||
[[File:2020 203 6 3.png|330px|right]] | [[File:2020 203 6 3.png|330px|right]] | ||
[[File:2020 203 6 2.png|330px|right]] | [[File:2020 203 6 2.png|330px|right]] | ||
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Answer: <math>\frac {1}{3}.</math> | Answer: <math>\frac {1}{3}.</math> | ||
− | ==2020 204 problem 6== | + | ==2020 var 204 problem 6== |
[[File:2020 204 6.png|300px|right]] | [[File:2020 204 6.png|300px|right]] | ||
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Answer: <math>\frac {12}{1 + \sqrt{13}}.</math> | Answer: <math>\frac {12}{1 + \sqrt{13}}.</math> | ||
− | ==2020 205 problem 6== | + | ==2020 var 205 problem 6== |
[[File:2020 205 6.png|330px|right]] | [[File:2020 205 6.png|330px|right]] | ||
Let the quadrangular pyramid <math>ABCDS</math> with the base parallelogram <math>ABCD</math> be given. | Let the quadrangular pyramid <math>ABCDS</math> with the base parallelogram <math>ABCD</math> be given. | ||
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<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> | <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. | Answer: 1 : 6. | ||
− | ==2020 206 problem 6== | + | |
+ | ==2020 var 206 problem 6== | ||
[[File:2020 206 6.png|330px|right]] | [[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> | 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> | ||
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Answer: <math>\frac {1}{\sqrt{3}}</math> | Answer: <math>\frac {1}{\sqrt{3}}</math> | ||
− | ==2021 215 problem 7== | + | ==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> | 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> | ||
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Answer: 3. | Answer: 3. | ||
− | ==2022 221 problem 7== | + | ==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]] | ||
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<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> | ||
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<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> | ||
− | ==2023 232 problem 6== | + | |
+ | ==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> | 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> | Find the maximum value of <math>\frac {a+ b}{c}.</math> | ||
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<i><b>Answer:<math>4.</math></b></i> | <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:
vladimir.shelomovskii@gmail.com, vvsss