Difference between revisions of "DVI exam"
(→2014 1 Problem 8) |
(→2014 1 Problem 8) |
||
Line 32: | Line 32: | ||
<i><b>Answer:<math> 3 \sqrt{2}, - 3 \sqrt {2}.</math></b></i> | <i><b>Answer:<math> 3 \sqrt{2}, - 3 \sqrt {2}.</math></b></i> | ||
+ | ==2015 1 Problem 7== | ||
+ | [[File:2015 7 distance.png|330px|right]] | ||
+ | A sphere is inscribed in a regular triangular prism with bases <math>ABCA'B'C'.</math> Find its radius if the distance between straight lines <math>AE</math> and <math>BD</math> is equal to <math>\sqrt{13},</math> where <math>E</math> and <math>D</math> are points lying on <math>A'B'</math> and <math>B'C'</math>, respectively, and <math>A'E : EB' = B'D : DC' = 1 : 2.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | The distance from the center of the sphere to the centers of the prism faces is equal to <math>R,</math> so <cmath>AA' = 2R, AB = 2 \sqrt{3} R.</cmath> | ||
+ | |||
+ | In order to find the distance <math>PQ</math> between the lines <math>\ell = AD</math> and <math>m = BE</math>, one can find the length of two perpendiculars <math>MM'</math> and <math>DE</math> to the line <math>m</math> that are perpendicular to each other. Then | ||
+ | <cmath>\frac {1}{PQ^2} = \frac{1}{DE^2} + \frac{1}{MM'^2}</cmath> | ||
+ | since, when viewed along a straight line <math>m</math>, the segment <math>PQ</math> is the altitude of a right triangle with legs <math>DE</math> and <math>MM'.</math> | ||
+ | |||
+ | The plane <math>\pi = BB'C'</math> containe the straight line <math>m.</math> The straight line <math>\ell</math> crossed <math>\pi</math> at the point <math>M \in BB'.</math> | ||
+ | <cmath>\frac {B'D}{DA'} = \frac {B'M}{AA'} =2 \implies B'M = 4R.</cmath> | ||
+ | In a right triangle <math>\triangle BKM</math> | ||
+ | <cmath>KM = BC = 2 \sqrt{3} R, BM = BB' + B'M = 6R, MM' \perp BE.</cmath> | ||
+ | <math>MM'</math> is the height falling on the hypotenuse, <math>\frac {1}{MM'^2} = \frac{1}{KM^2} + \frac{1}{BM^2}.</math> | ||
+ | |||
+ | Let <math>F</math> be the projection of <math>A</math> onto plane <math>\pi \implies F \in BC, BF = FC.</math> | ||
+ | |||
+ | Therefore <math>FM</math> is the projection of <math>\ell</math> onto plane <math>\pi, KM = 2 FB \implies m \cap MF</math> at the point <math>E.</math> | ||
+ | <cmath>\frac {B'D}{EB'} = 2 \implies DE \perp \pi \implies DE \perp m.</cmath> | ||
+ | <cmath>\frac {DE}{AF} = \frac {ME}{MF} = \frac {B'M}{MB} = \frac {2}{3} \implies DE = 2R.</cmath> | ||
+ | <cmath>\frac {1}{PQ^2} = \frac{1}{DE^2} + \frac{1}{MM'^2} = \frac{1}{DE^2} + \frac{1}{KM^2} + \frac{1}{BM^2} = \frac{1}{(2R)^2} + \frac{1}{(2\sqrt{3}R)^2} + \frac{1}{(6R)^2} = \frac{13}{(6R)^2}.</cmath> | ||
+ | <cmath>PO = \sqrt{13} = \frac{6R}{\sqrt{13}} \implies R = \frac {13}{6}.</cmath> | ||
+ | <i><b>Answer:<math>\frac {13}{6}.</math></b></i> | ||
==2016 1 Problem 8== | ==2016 1 Problem 8== |
Revision as of 14:15, 28 April 2024
DVI is an exam in mathematics at the Moscow State University named after M.V. Lomonosov. The first four problems have a standard level. Problem 5 is advanced level of geometry. Problem 6 is an advanced level equation or inequality. Problem 7 is advanced level of stereometry.
Below are the most difficult problems of this exam in recent years. The headings indicate the year when the problem was used, the variant option of the exam, and the number of the problem.
Contents
[hide]- 1 2014 1 Problem 6
- 2 2014 1 Problem 8
- 3 2015 1 Problem 7
- 4 2016 1 Problem 8
- 5 2020 201 problem 6
- 6 2020 202 problem 6
- 7 2020 203 problem 6
- 8 2020 204 problem 6
- 9 2020 205 problem 6
- 10 2020 206 problem 6
- 11 2021 215 problem 7
- 12 2022 221 problem 7
- 13 2022 222 problem 7
- 14 2022 222 problem 6
- 15 2022 224 problem 6
- 16 2023 231 problem 6
- 17 2023 231 EM problem 6
- 18 2023 232 problem 6
- 19 2023 233 problem 6
2014 1 Problem 6
Find all pares of real numbers satisfying the system of equations Solution
Denote Denote is the solution. Let If then if then therefore is the single root.
2014 1 Problem 8
Let
Find and
Solution where
Answer:
2015 1 Problem 7
A sphere is inscribed in a regular triangular prism with bases Find its radius if the distance between straight lines and is equal to where and are points lying on and , respectively, and
Solution
The distance from the center of the sphere to the centers of the prism faces is equal to so
In order to find the distance between the lines and , one can find the length of two perpendiculars and to the line that are perpendicular to each other. Then since, when viewed along a straight line , the segment is the altitude of a right triangle with legs and
The plane containe the straight line The straight line crossed at the point In a right triangle is the height falling on the hypotenuse,
Let be the projection of onto plane
Therefore is the projection of onto plane at the point Answer:
2016 1 Problem 8
Find the smallest value of the expression Solution Denote The shortest length of a broken line with fixed ends is equal to the distance between points and which is and is achieved if points and are collinear. Answer:
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:
2021 215 problem 7
The sphere touches all edges of the tetrahedron It is known that the products of the lengths of crossing edges are equal. It is also known that Find
Solution
The tangent segments from the common point to the sphere are equal.
Let us denote the segments from the vertex to the sphere by
Similarly, we define If then
If
The tetrahedron is a regular pyramid with a regular triangle with side at the base and side edges equal to
Answer: 3.
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:
2022 222 problem 6
Find all possible values of the product if it is known that and it is true
Solution
Let then for each equation is true, Let no solution.
Answer:
2022 224 problem 6
Find all triples of real numbers in the interval satisfying the system of equations
Solution
Denote Similarly,
Therefore Answer:
2023 231 problem 6
Let positive numbers be such that
Find the maximum value of
Solution Similarly Adding this equations, we get: If then
Answer:
Explanation for students
For the function under study it is required to find the majorizing function This function must be a linear combination of the given function and a constant,
At the supposed extremum point the functions and their derivatives must coincide
2023 231 EM problem 6
Find the maximum value and all argument values such that .
Solution because and signs of and are different, so Therefore
2023 232 problem 6
Let positive numbers be such that Find the maximum value of
Solution
It is clear that and Denote So If then
Answer:
2023 233 problem 6
Let positive numbers be such that
Find the maximum value of
Solution
Let Then
Equality is achieved if
Answer: