Difference between revisions of "DVI exam"
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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== | ||
+ | [[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> | ||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let points <math>E,F,P, G, H, K</math> be the midpoints of <math>AB, AA', A'D', BB', B'C', B'A',</math> respectively. We need to prove that planes <math>GKH</math> and <math>EFP</math> are parallel, perpendicular to <math>B'D.</math> | ||
+ | Therefore, <math>B'D = \sqrt{3}.</math> | ||
+ | |||
+ | Point <math>O</math> is the midpoint <math>B'D \implies</math> | ||
+ | <cmath>B'O = \frac {\sqrt{3}}{2}, B'H = \frac {1}{2}, GH = \frac {\sqrt{2}}{2},</cmath> | ||
+ | <cmath>HQ = \frac{GH}{\sqrt{3}}, B'Q = \frac{\sqrt{3}}{6}, OQ = \frac {1}{\sqrt{3}} = IJ.</cmath> | ||
+ | For proof we can use one of the following methods: | ||
+ | |||
+ | 1. Vectors: <math>\vec {B'A'} = 2 \vec e_x, \vec {B'B} = 2 \vec e_y, \vec {B'C'} = 2 \vec e_z \implies </math> | ||
+ | <cmath>\vec {B'G} = \vec e_y, \vec {B'H} = \vec e_z, \vec {HG} = \vec e_y - \vec e_z, \vec {B'D} = \vec e_x + \vec e_y + \vec e_z.</cmath> | ||
+ | Scalar product <math>(\vec{B'D} \cdot \vec {HG}) = 0.</math> | ||
+ | Similarly, <cmath>\vec {B'E} = 2\vec e_y + \vec e_x, \vec {B'F} = 2\vec e_x+ \vec e_y, \vec {FE} = \vec e_y - \vec e_x.</cmath> | ||
+ | |||
+ | 2. <math>\angle B'OG = \angle B'OE = 90^\circ.</math> | ||
+ | |||
+ | 3. Rotating the cube around its axis <math>B'D</math> we find that the point <math>G</math> move to <math>H</math>, then to <math>K,</math> then to <math>G.</math> | ||
+ | |||
+ | Answer: <math>\frac {1}{\sqrt{3}}</math> | ||
==2022 221 problem 7== | ==2022 221 problem 7== |
Revision as of 06:30, 1 February 2024
Contents
[hide]2020 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 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 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 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 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 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:
2022 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 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: