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Difference between revisions of "DVI exam"

(→‎2022 222 problem 7)
(→‎2022 221 problem 7)
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==2020 201 problem 6==
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[[File:2020 201 6.png|330px|right]]
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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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<i><b>Solution</b></i>
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Let <math>E',D',F'</math> be the parallel projections of <math>D,E,F (DD' || AA' || EE' || FF')</math> on the plane <math>ABC, H' = AE' \cap D'F', HH' || AA'.</math>
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<math>\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.</math>
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We use  and get
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<cmath>\frac {F'H'}{H'D'} = \frac {2}{k(m+1)} = 3 = \frac {FH}{HD}.</cmath>
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<cmath>\frac {E'H'}{AH'} = \frac {mk + 1}{k+1} = \frac {7}{9} = \frac {EH}{GH}.</cmath>
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Let <math>DD' = x, FF'= y, DH = u, FH = v \implies HH' = \frac{yu + vx}{u+v }= \frac {7}{16}.</math>
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Similarly <math>HH' = \frac{GA \cdot EH + EE' \cdot HG}{EH+HG} </math> <math>\implies AA'= \frac {4}{7} \implies \frac {AG}{GA'} = \frac {4}{3}.</math>
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Answer: <math>AG : GA' = 4 : 3.</math>
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==2022 221 problem 7==
 
==2022 221 problem 7==
 
[[File:MSU 2022 7.png|330px|right]]
 
[[File:MSU 2022 7.png|330px|right]]

Revision as of 03:35, 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 $HH' = \frac{GA \cdot EH + EE' \cdot HG}{EH+HG}$ $\implies AA'= \frac {4}{7} \implies \frac {AG}{GA'} = \frac {4}{3}.$

Answer: $AG : GA' = 4 : 3.$

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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