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Difference between revisions of "DVI exam"

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Answer: <math>\frac {11}{30}. </math>
 
Answer: <math>\frac {11}{30}. </math>
  
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==2020 203 problem 6==
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[[File:2020 203 6 3.png|330px|right]]
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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>
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<i><b>Solution</b></i>
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Denote the vertices of polyhedron <math>E, F, G, E', F', G'.</math>
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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>
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This triangles lies in parallel planes, which are normal to cube diagonal <math>AC'.</math>
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The distance <math>d</math> between this planes is <cmath>\sqrt{3} - 2 \cdot \frac{\sqrt{3}}{6} = \frac {2}{\sqrt{3}}.</cmath>
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So the volume of the regular prism with base <math>\triangle EFG</math> and height <math>d</math> is
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<cmath>V_0 = \frac {\sqrt{3}}{8} \cdot \frac {2}{\sqrt{3}} = \frac {1}{4}.</cmath>
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Let the area <math>[A(x)]</math> be the quadratic function of <math>x.</math> Let
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<cmath>A_1 = A[x_1], A_2 = A[x_2], x_0 = \frac{x_1 + x_2}{2}, A_0 = A[x_0]  \implies</cmath>
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<cmath>V = (x_2 - x_1) \cdot \frac{A_1 + A_2 + 4 A_0}{6}.</cmath>
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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 position.
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<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>
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<cmath>V = (x_2 - x_1) \cdot \frac{{[EFG]} + {[E'F'G']} + 4 [E''F''G'']}{6} = (x_2 - x_1) \cdot [EFG] \cdot \frac {4}{3} = \frac {1}{4} \cdot \frac {4}{3} =  \frac {1}{3}.</cmath>
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Answer:  <math>\frac {1}{3}.</math>
  
  

Revision as of 04:20, 1 February 2024

2020 201 problem 6

2020 201 6.png

Let a triangular prism $ABCA'B'C'$ with a base $ABC$ be given, $D \in AB', E \in BC', F \in CA'.$ Find the ratio in which the plane $DEF$ divides the segment $AA',$ if $AD : DB' = 1 : 1,$ \[BE : EC' = 1 : 2, CF : FA' = 1 : 3.\]

Solution

Let $E',D',F'$ be the parallel projections of $D,E,F (DD' || AA' || EE' || FF')$ on the plane $ABC, H' = AE' \cap D'F', HH' || AA'.$ $\frac {BD'}{AD'} = \frac {BD}{AD} = 1, \frac {BE'}{CE'} = \frac {BE}{C'E} = \frac {1}{2} = k, \frac {CF'}{AF'} = \frac {CF}{A'F} = \frac {1}{3} = m.$

We use and get \[\frac {F'H'}{H'D'} = \frac {2}{k(m+1)} = 3 = \frac {FH}{HD}.\] \[\frac {E'H'}{AH'} = \frac {mk + 1}{k+1} = \frac {7}{9} = \frac {EH}{GH}.\] Let $DD' = x, FF'= y, DH = u, FH = v \implies HH' = \frac{yu + vx}{u+v }= \frac {7}{16}.$

Similarly $HH' = \frac{AG \cdot EH + EE' \cdot HG}{EH+HG} \implies AG = \frac {4}{7} \implies \frac {AG}{GA'} = \frac {4}{3}.$

Answer: $AG : GA' = 4 : 3.$

2020 202 problem 6

2020 202 6.png

Let a tetrahedron $ABCD$ be given, $AB = BC = CD = 5, CA = AD = DB = 6.$ Find the cosine of the angle $\varphi$ between the edges $BC$ and $AD.$

Solution

Let us describe a parallelepiped $AC'BD'B'DA'C$ around a given tetrahedron $ABCD.$

$AB = CD \implies AC'BD'$ and $B'DA'C$ are equal rectangles.

$AC = BD \implies AB'CD'$ and $C'DA'B$ are equal rectangles.

Denote $AC' = a, AD' = b, AB' = c \implies$ \[a^2 + b^2 = 5^2 = 25, a^2 + c^2 = 6^2 = 36.\] \[4AC'^2 = 4 a^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos \varphi,\] \[4AB'^2 = 4 c^2 = 5^2 + 6^2 + 2 \cdot 5 \cdot 6 \cos \varphi,\] \[4(c^2 - a^2) = 4(6^2 - 5^2) = 4 \cdot 5 \cdot 6 \cos \varphi \implies \cos \varphi = \frac {6^2 - 5^2}{5 \cdot 6} = \frac {11}{30}.\] Answer: $\frac {11}{30}.$


2020 203 problem 6

2020 203 6 3.png

Let a cube $ABCDA'B'C'D'$ with the base $ABCD$ and side edges $AA', BB', CC', DD', AB = 1$ be given. Find the volume of a polyhedron whose vertices are the midpoints of the edges $AB, AD, AA', CC', C'B', C'D'.$

Solution

Denote the vertices of polyhedron $E, F, G, E', F', G'.$ Triangles $\triangle EFG$ and $\triangle E'F'G'$ are equilateral triangles with sides $\frac {\sqrt{2}}{2}$ and areas $[EFG] = \frac {\sqrt{3}}{8}.$

This triangles lies in parallel planes, which are normal to cube diagonal $AC'.$ The distance $d$ between this planes is \[\sqrt{3} - 2 \cdot \frac{\sqrt{3}}{6} = \frac {2}{\sqrt{3}}.\] So the volume of the regular prism with base $\triangle EFG$ and height $d$ is \[V_0 = \frac {\sqrt{3}}{8} \cdot \frac {2}{\sqrt{3}} = \frac {1}{4}.\]

Let the area $[A(x)]$ be the quadratic function of $x.$ Let \[A_1 = A[x_1], A_2 = A[x_2], x_0 = \frac{x_1 + x_2}{2}, A_0 = A[x_0] \implies\] \[V = (x_2 - x_1) \cdot \frac{A_1 + A_2 + 4 A_0}{6}.\] Suppose, we move point $P$ along axis $AC'$ and cross the solid by plane contains $P$ and normal to axis. Distance from $P$ to each crosspoint this plane with the edge change proportionally position $P$ along axes, so the area is quadratic function from position. \[\frac {OE''}{ME} = \frac {\sqrt{3}}{2} \implies \frac {[E''F''G'']}{[EFG]} = 2 \left (\frac {OE''}{ME} \right)^2 = \frac {3}{2}.\] \[V = (x_2 - x_1) \cdot \frac{{[EFG]} + {[E'F'G']} + 4 [E''F''G'']}{6} = (x_2 - x_1) \cdot [EFG] \cdot \frac {4}{3} = \frac {1}{4} \cdot \frac {4}{3} = \frac {1}{3}.\]

Answer: $\frac {1}{3}.$


2022 221 problem 7

MSU 2022 7.png
MSU 2022 7a.png

The volume of a triangular prism $ABCA'B'C'$ with base $ABC$ and side edges $AA', BB', CC'$ is equal to $72.$ Find the volume of the tetrahedron $DEFG,$ where $D$ is the centroid of the face $ABC'A', E$ is the point of intersection of the medians of $\triangle A'B'C', F$ is the midpoint of the edge $AC$ and $G$ is the midpoint of the edge $BC.$

Solution

Let us consider the uniform triangular prism $ABCA'B'C'.$ Let $M$ be the midpoint of $AB, M'$ be the midpoint of $A'B', K$ be the midpoint of $CM, L$ be the midpoint of $C'M', 2 FG = AB.$

The area $[KED]$ of $\triangle KED$ in the sum with the areas of triangles $[KEL], [EDM'], [KDM]$ is half the area of rectangle $CC'M'M,$ so \[\frac {[KED]}{[CC'M'M]} = \frac {1}{2} - \frac {1}{12}- \frac {1}{12}-\frac {1}{8} = \frac {5}{24}.\] \[FG \perp ED.\] Denote the distance between these lines $h.$ The volume of the tetrahedron is $U = \frac {ED \cdot h \cdot FG}{6}.$ \[\frac {ED \cdot h}{2} = \frac {5}{24} \cdot CC' \cdot CM \implies U =\frac{5}{24 \cdot 3} \cdot CC' \cdot CM \cdot FG.\] The volume of the prism is $V = \frac{CM \cdot AB}{2} \cdot CC' =CC' \cdot CM \cdot FG = 72.$ \[\frac {U}{V} = \frac {\frac {5}{72} CC' \cdot CM \cdot FG}{CC' \cdot CM \cdot FG} = \frac {5}{72} \implies U = 5.\]

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 222 problem 7

MSU 2022 2 7.png

A sphere of diameter $1$ is inscribed in a pyramid at the base of which lies a rhombus with an acute angle $2\alpha$ and side $\sqrt{6}.$ Find the angle $2\alpha$ if it is known that all lateral faces of the pyramid are inclined to plane of its base at an angle of $60^\circ.$

Solution 1

Denote rhombus $ABCD, K = AC \cap BD, S$ is the vertex of a pyramid $SK \perp ABC, I$ is the center of the sphere, $IK = r = \frac {1}{2}, M \in AB, SM \perp AB, E$ is the tangent point of $SM$ and sphere, $\angle SMK = 60 ^\circ.$ \[IE = r, SK = SI + IK = \frac {3}{2}, KM = \frac {\sqrt{3}}{2}, SM = \sqrt{3}.\] \[AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},\] \[AM + BM = AB = \sqrt{6}\implies\] \[\tan \alpha + \frac {1}{\tan \alpha} = 2 \sqrt {2} \implies \tan \alpha = \sqrt {2} - 1 \implies \tan 2 \alpha = 1 \implies 2 \alpha = \frac {\pi}{4}.\] Solution 2

The area of the rhombus $[ABCD]= AB^2 \cdot \sin 2\alpha.$

The area of the lateral surface is $[l]= 4 [SAB] = 2 \cdot AB \cdot SM.$ \[[ABCD] = [l] \cdot \cos 60 ^\circ =[l] \cdot \frac{1}{2} \implies AB \cdot SM = AB^2 \cdot \sin 2\alpha \implies\] \[\sin {2 \alpha} = \frac {SM}{AB} = \frac {\sqrt{3}}{\sqrt{6}} = \frac {1}{\sqrt{2}}.\] Answer:$\frac {\pi}{4}.$

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