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Difference between revisions of "DVI exam"

(→‎2022 221 problem 7)
(→‎2022 222 problem 7)
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==2022 222 problem 7==
 
==2022 222 problem 7==
 
[[File:MSU 2022 2 7.png|400px|right]]
 
[[File:MSU 2022 2 7.png|400px|right]]
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<cmath>r = 0.5, h = 3/2, KM = \frac {\sqrt{3}}{2},</cmath>
+
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>AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},</cmath>
+
 
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<cmath>AM + BM = AB \implies</cmath>
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<i><b>Solution 1</b></i>
 +
 
 +
Denote rhombus <math>ABCD, K = AC \cap BD, S</math> is the vertex of a pyramid <math>SK \perp ABC, I</math> is the center of the sphere, <math>IK = r = \frac {1}{2}, M \in AB, SM \perp AB, E</math> is the tangent point of <math>SM</math> and sphere, <math>\angle SMK = 60 ^\circ.</math>
 +
<cmath>IE = r, SK = SI + IK = \frac {3}{2}, KM = \frac {\sqrt{3}}{2}, SM = \sqrt{3}.</cmath>
 +
<cmath>AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},</cmath>  
 +
<cmath>AM + BM = AB = \sqrt{6}\implies</cmath>  
 
<cmath>\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}.</cmath>
 
<cmath>\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}.</cmath>
 +
<i><b>Solution 2</b></i>
 +
 +
The area of the rhombus <math>[ABCD]= AB^2 \cdot \sin 2\alpha.</math>
 +
 +
The area of the lateral surface is <math>[l]= 4 [SAB] = 2 \cdot AB \cdot SM.</math>
 +
<cmath>[ABCD] = [l] \cdot \cos 60 ^\circ =[l] \cdot \frac{1}{2} \implies AB \cdot SM = AB^2 \cdot \sin 2\alpha \implies</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>

Revision as of 06:15, 28 January 2024

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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