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Difference between revisions of "DVI exam"

(→‎2022 221 problem 7)
(→‎2022 221 problem 7)
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<cmath>\sin \angle MSN = \frac {1}{\sqrt{13}} = \frac {ID}{SN - IN}, IN = ID = \frac {AB}{1 + \sqrt{13}}.</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>
 
Answer: <math>\frac {12}{1 + \sqrt{13}}.</math>
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==2020 205 problem 6==
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[[File:2020 205 6.png|330px|right]]
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Let the quadrangular pyramid <math>ABCDS</math> with the base parallelogram <math>ABCD</math> be given.
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Point <math>E \in SB, \frac {SE}{EB} = 2.</math> Point <math>F \in SD, \frac {SF}{FD} = \frac {1}{2}.</math>
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Find the ratio in which the plane <math>AEF</math> divides the volume of the pyramid.
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<i><b>Solution</b></i>
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Let plane <math>AEF</math> cross edge <math>SC</math> at point <math>G.</math>
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We make the central projection from point <math>S</math>
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The images of points <math>A,E,F,G</math> are <math>A,B,D,C,</math> respectively.
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The image of <math>S</math> is the crosspoint of <math>AC</math> and <math>BD.</math>
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So lines <math>EF, SO,</math> and <math>AG</math> are crossed at point <math>H.</math>
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<cmath>\frac {2 OH}{SH} = \frac {BE}{SE} + \frac {DF}{SF} = \frac {AA}{SA}+ \frac {CG}{SG}.</cmath>
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<cmath>2 + \frac {1}{2} = 0 + \frac {CG}{SG} \implies \frac {CG}{SG} = \frac {5}{2}.</cmath>
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Let’s compare volumes of some tetrachedrons, denote the volume of <math>X</math> as <math>[X].</math>
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<cmath>\frac {[ABDS]}{[CBDS]} = \frac {[ABD]}{[CBD]} =1.</cmath>
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<cmath>\frac {[AEFS]}{[ABDS]} = \frac {[EFS]}{[BDS]} = \frac {SE \cdot SF}{SB \cdot SD} = \frac{2}{9}.</cmath>
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<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>
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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>
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Answer: 1 : 6.
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==2022 221 problem 7==
 
==2022 221 problem 7==

Revision as of 06:13, 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
2020 203 6 2.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 \[A_1 = A[x_1], A_2 = A[x_2], d = x_2 - x_1,\] \[x_0 = \frac{x_1 + x_2}{2}, A_0 = A[x_0] \implies\] \[V = \frac{d}{6} \cdot \left(A_1 + A_2 + 4 A_0 \right).\] 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 $P$ position. \[\frac {OE''}{ME} = \frac {\sqrt{3}}{2} \implies \frac {[E''F''G'']}{[EFG]} = 2 \left (\frac {OE''}{ME} \right)^2 = \frac {3}{2}.\] \[V = \frac{d}{6} \cdot ([EFG] + {[E'F'G']} + 4 [E''F''G'']) = d \cdot [EFG] \cdot \frac {4}{3} = \frac {1}{4} \cdot \frac {4}{3} = \frac {1}{3}.\]

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

2020 204 problem 6

2020 204 6.png

Let a regular triangular pyramid be given. The circumcenter of the sphere $O$ 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 $12.$

Solution

\[OP = ON \implies BP = BN, BS = 2 BP = 2 BN,\] \[AB = \sqrt{3}BN, \angle BSN = 30^\circ \implies\] \[SN = \frac {3}{2} SO = AB.\] \[NM = \frac {BN}{2} = \frac {AB}{2 \sqrt{3}} \implies\] \[\tan \angle MSN = \frac {1}{2\sqrt{3}} \implies\] \[\sin \angle MSN = \frac {1}{\sqrt{13}} = \frac {ID}{SN - IN}, IN = ID = \frac {AB}{1 + \sqrt{13}}.\] Answer: $\frac {12}{1 + \sqrt{13}}.$

2020 205 problem 6

2020 205 6.png

Let the quadrangular pyramid $ABCDS$ with the base parallelogram $ABCD$ be given.

Point $E \in SB, \frac {SE}{EB} = 2.$ Point $F \in SD, \frac {SF}{FD} = \frac {1}{2}.$

Find the ratio in which the plane $AEF$ divides the volume of the pyramid.

Solution

Let plane $AEF$ cross edge $SC$ at point $G.$ We make the central projection from point $S$ The images of points $A,E,F,G$ are $A,B,D,C,$ respectively. The image of $S$ is the crosspoint of $AC$ and $BD.$ So lines $EF, SO,$ and $AG$ are crossed at point $H.$ \[\frac {2 OH}{SH} = \frac {BE}{SE} + \frac {DF}{SF} = \frac {AA}{SA}+ \frac {CG}{SG}.\] \[2 + \frac {1}{2} = 0 + \frac {CG}{SG} \implies \frac {CG}{SG} = \frac {5}{2}.\] Let’s compare volumes of some tetrachedrons, denote the volume of $X$ as $[X].$ \[\frac {[ABDS]}{[CBDS]} = \frac {[ABD]}{[CBD]} =1.\] \[\frac {[AEFS]}{[ABDS]} = \frac {[EFS]}{[BDS]} = \frac {SE \cdot SF}{SB \cdot SD} = \frac{2}{9}.\] \[\frac {[GEFS]}{[CBDS]} = \frac {[EFS]}{[BDS]} \cdot \frac {GX}{CO} = \frac{2}{9} \cdot \frac {SG}{SC} = \frac{2}{9} \cdot \frac {2}{7} = \frac{4}{63}.\] \[\frac {[AEGFS]}{[ABDS]} = \frac {[AEFS]+[GEFS]}{[ABDS]} = \frac{2}{9} + \frac{4}{63} = \frac{2}{7} \implies \frac {[AEGFS]}{[ABCDS]} = \frac{1}{7}.\] Answer: 1 : 6.


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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