On February 21, the ceremony of presenting the Government Prizes in the field of education for 2018 took place at the House of the Government of the Russian Federation. The awards were presented to the laureates by the Deputy Chairman of the Government of the Russian Federation T.A. Golikov.

Among the laureates of the award are employees of the Laboratory for Working with Gifted Children. The award was given to teachers of the Russian national team at IPhO Vitaly Shevchenko and Alexander Kiselev, teachers of the Russian national team at IJSO Elena Mikhailovna Snigireva (chemistry) and Igor Kiselev (biology) and the head of the Russian team, MIPT vice-rector Artyom Anatolyevich Voronov.

The main achievements for which the team was awarded a government award are 5 gold medals for the Russian team at IPhO-2017 in Indonesia and 6 gold medals for the team at IJSO-2017 in Holland. Each student brought home gold!

Such a high result at the International Physics Olympiad was achieved by the Russian team for the first time. In the entire history of IPhO since 1967, neither the Russian team nor the USSR team has ever managed to win five gold medals before.

The complexity of the tasks of the Olympiad and the level of training of teams from other countries is constantly growing. However, the Russian team last years is in the top five teams in the world. In order to achieve high results, the teachers and the leadership of the national team are improving the system of preparation for the international in our country. Appeared training schools, where students study in detail the most difficult sections of the program. A database of experimental tasks is being actively created, performing which the guys are preparing for the experimental tour. Regular distance work is carried out, during the year of preparation, the guys receive about ten theoretical homework assignments. great attention is given to the qualitative translation of the conditions of the problems at the Olympiad itself. Training courses are being improved.

High results at international Olympiads are the result of the long work of a large number of teachers, employees and students of the Moscow Institute of Physics and Technology, personal teachers on the ground, and the hard work of the schoolchildren themselves. In addition to the above-mentioned laureates of the award, a huge contribution to the preparation of the national team was made by:

Fedor Tsybrov (creating tasks for qualification camps)

Alexey Noyan (experimental training of the national team, development of an experimental workshop)

Aleksey Alekseev (creating qualifying training tasks)

Arseniy Pikalov (preparation of theoretical materials and conducting seminars)

Ivan Erofeev (many years of work in all areas)

Alexander Artemiev (checking homework)

Nikita Semenin (creating qualifying training tasks)

Andrey Peskov (development and creation of experimental facilities)

Gleb Kuznetsov (experimental training of the national team)

8TH GRADE

SCHOOL STAGE TASKS

OF THE ALL-RUSSIAN OLYMPIAD FOR SCHOOLCHILDREN IN SOCIAL SCIENCE

FULL NAME. student ___________________________________________________________________________

Date of birth __________________________ Class ____,__ Date "_____" ______20__

Grade (max. 100 points) _________

Exercise 1. Choose the correct answer:

The Golden Rule of Morality says:

1) "An eye for an eye, a tooth for a tooth";

2) "Do not make yourself an idol";

3) “Treat people the way you want to be treated”;

4) "Honor thy father and thy mother."

Answer: ___

Task 2. Choose the correct answer:

The ability of a person to acquire and exercise rights and obligations by his actions is called: 1) legal capacity; 2) legal capacity; 3) emancipation; 4) socialization.

Answer: ___

(For the correct answer - 2 points)

Task 3. Choose the correct answer:

IN Russian Federation higher legal effect in the system of normative acts has

1) Decrees of the President of the Russian Federation 3) Criminal Code of the Russian Federation

2) The Constitution of the Russian Federation 4) Decrees of the Government of the Russian Federation

Answer: ___

(For the correct answer - 2 points)

Task 4. A scientist must correctly write concepts and terms. Fill in the correct letter(s) for the gaps.

1. Pr ... in ... legia - an advantage granted to someone.

2. D ... in ... den ... - income paid to shareholders.

3. T ... l ... rantn ... st - tolerance for other people's opinions.

Task 5. Fill in the gap in the row.

1. Genus, …….., nationality, nation.

2. Christianity, ………, Buddhism.

3. Production, distribution, ………, consumption.

Task 6. By what principle are the rows formed? Name the concept that is common to the terms below, uniting them.

1. Rule of law, separation of powers, guarantee of human rights and freedoms

2.Measure of value, means of accumulation, means of payment.

3. Custom, precedent, law.

1. ________________________________________________________

2.________________________________________________________

3.________________________________________________________

Task 7. Answer "yes" or "no":

1) Man is by nature a biosocial being.

2) Communication is understood only as the exchange of information.

3) Each person is individual.

4) In the Russian Federation the full amount a citizen receives rights and freedoms from the age of 14.

5) Every person is born as a person.

6) The Russian Parliament (Federal Assembly) consists of two chambers.

7) Society refers to self-developing systems.

8) If it is impossible to personally participate in the elections, it is allowed to issue a power of attorney to another person for the purpose of voting for the candidate specified in the power of attorney.

9) Progress historical development contradictory: both progressive and regressive changes can be found in it.

10) Individual, personality, individuality - concepts that are not identical.

4.1.
4.2.
4.3.
4.4.

For one correct answer - 2 points (Maximum score - 8).

KEYS TO THE OBJECTIVES

Exercise 1 ( For the correct answer - 2 points)

Task 2 ( For the correct answer - 2 points)

Task 3 ( For the correct answer - 2 points)

Task 4 ( 1 point for a correct letter. Maximum - 8 points)

1. Privilege. 2. Dividend. 3. Tolerance

Task 5 ( For each correct answer - 3 points. Maximum - 9 points)

1. Tribe. 2. Islam. 3. Exchange.

Task 6 ( For each correct answer - 4 points. Maximum - 12 points)

1. Signs of the rule of law

2. Functions of money

3. Sources of law.

Task 7 2 points for each correct answer. (Maximum per task - 20 points)

Municipal Stage Tasks All-Russian Olympiad schoolchildren in mathematics

Gorno-Altaisk, 2008

The municipal stage of the Olympiad is held on the basis of the Regulations on the All-Russian Olympiad for schoolchildren, approved by order of the Ministry of Education and Science of Russia dated 01.01.01 No. 000.

The stages of the Olympiad are held according to tasks compiled on the basis of general educational programs implemented at the levels of basic general and secondary (complete) general education.

Evaluation criteria

Tasks of mathematical Olympiads are creative, allow several different solutions. In addition, it is necessary to evaluate partial progress in problems (for example, analysis of an important case, proof of a lemma, finding an example, etc.). Finally, logical and arithmetic errors in the solutions are possible. The final scores for the task must take into account all of the above.

In accordance with the regulations for holding mathematical Olympiads for schoolchildren, each task is evaluated from 7 points.

Correspondence of the correctness of the solution and the points given is shown in the table.

	Correctness (falseness) of the decision
	Complete correct solution
	The right decision. There are some minor flaws that do not affect the overall solution.
	The decision is generally correct. However, the solution contains significant errors or missing cases that do not affect the logic of reasoning.
	One of the two (more complex) essential cases is correctly considered, or in a problem of the “estimate + example” type, the estimate is correctly obtained.
	Auxiliary statements are proved that help in solving the problem.
	Separate important cases in the absence of a solution (or in the case of an erroneous decision) are considered.
	Wrong decision, no progress.
	There is no solution.

It is important to note that any correct solution is worth 7 points. It is unacceptable to deduct points for the fact that the solution is too long, or for the fact that the student's solution differs from that given in methodological developments or from other decisions known to the jury.

At the same time, any arbitrarily long decision text that does not contain useful advances should be rated 0 points.

The procedure for holding the municipal stage of the Olympiad

The municipal stage of the Olympiad is held on the same day in November-December for students in grades 7-11. The recommended time for the Olympiad is 4 hours.

Topics for the tasks of the school and municipal stages of the Olympiad

Olympiad tasks for the school and municipal stages are compiled on the basis of mathematics programs for general educational institutions. It is also allowed to include tasks, the topics of which are included in the programs of school circles (electives).

The following are only those topics that are proposed to be used in the preparation of options for the tasks of the CURRENT academic year.

Magazines: Kvant, Mathematics at School

Books and teaching aids:

, Mathematical Olympiads of the Moscow Region. Ed. 2nd, rev. and additional – M.: Fizmatkniga, 200s.

, Mathematics. All-Russian Olympiads. Issue. 1. - M.: Enlightenment, 2008. - 192 p.

, Moscow Mathematical Olympiads. – M.: Enlightenment, 1986. – 303 p.

, Leningrad mathematical circles. - Kirov: Asa, 1994. - 272 p.

Collection of Olympiad problems in mathematics. - M.: MTSNMO, 2005. - 560 p.

Planimetry tasks . Ed. 5th rev. and additional - M.: MTSNMO, 2006. - 640 p.

, Kanel-, Moscow Mathematical Olympiads / Ed. . - M.: MTSNMO, 2006. - 456 p.

1. Instead of asterisks, put ten different numbers in the expression *+ ** + *** + **** = 3330 so that you get the correct equality.

2. The businessman Vasya took up trade. Every morning he
buys a commodity with some part of the money he has (perhaps with all the money he has). After dinner, he sells the purchased goods for twice as much as he bought. How should Vasya trade so that in 5 days he will have exactly rubles, if at first he had 1000 rubles.

3. Cut a 3 x 3 square into two parts and a 4 x 4 square into two parts so that the resulting four pieces can be folded into a square.

4. All natural numbers from 1 to 10 were written in a 2x5 table. After that, each of the sums of numbers in a row and in a column was calculated (in total, 7 sums were obtained). What is the largest number of these sums that can be prime numbers?

5. For a natural number N calculated the sums of all pairs of adjacent digits (for example, for N= The 35,207 sums are (8, 7, 2, 7)). Find the smallest N, for which among these sums there are all numbers from 1 to 9.

8 Class

1. Vasya raised a natural number A squared, wrote down the result on the board and erased the last 2005 digits. Could the last digit of the number left on the board be equal to one?

2. At the review of the troops of the Island of Liars and Knights (liars always lie, knights always tell the truth), the leader lined up all the soldiers. Each of the soldiers standing in the line said: "My neighbors in the line are liars." (The warriors standing at the ends of the line said: “My neighbor in the line is a liar.”) What largest number knights could be in the ranks if 2005 warriors came to the review?

3. The seller has an arrow scale for weighing sugar with two cups. Scales can show weight from 0 to 5 kg. In this case, sugar can only be placed on the left cup, and weights can be placed on any of the two cups. What is the smallest number of weights that a seller needs to have to weigh any amount of sugar from 0 to 25 kg? Explain the answer.

4. Find corners right triangle, if it is known that the point symmetrical to the vertex right angle with respect to the hypotenuse, lies on a straight line passing through the midpoints of the two sides of the triangle.

5. Cells of the 8x8 table are painted in three colors. It turned out that there is no three-cell corner in the table, all cells of which are of the same color (a three-cell corner is a figure obtained from a 2x2 square by deleting one cell). It also turned out that there is no three-celled corner in the table, all the cells of which are of three different colors. Prove that the number of cells of each color is even.

1. A set consisting of integers a, b, c, replaced with set a - 1, b + 1, c2. As a result, the resulting set coincided with the original. Find the numbers a, 6, c, if it is known that their sum is 2005.

2. Vasya took 11 consecutive natural numbers and multiplied them. Kolya took the same 11 numbers and added them up. Could the last two digits of Vasya's result coincide with the last two digits of Kolya's result?

3. Based on AC triangle ABC point taken D.
Prove that circles inscribed in triangles ABD And CBD, touch points cannot divide a segment BD into three equal parts.

4. Each of the points of the plane is colored in one of
three colors, all three colors being used. Is it true that for any such coloring it is possible to choose a circle on which there are points of all three colors?

5. A lame rook (a rook that can only move horizontally or only vertically exactly 1 square) went around the board 10 x 10 squares, visiting each square exactly once. In the first cell where the rook visited, we write the number 1, in the second - the number 2, in the third - 3, and so on up to 100. Could it be that the sum of the numbers written in two adjacent cells along the side is divisible by 4 ?

combinatorial tasks.

1. A set consisting of numbers a, b, c, replaced with a4 set - 2b2, b 4- 2c2, c4 - 2a2. As a result, the resulting set coincided with the original. Find the numbers a, b, c, if their sum is 3.

2. Each of the points of the plane is colored in one of
three colors, all three colors being used. Ver
but is it that with any such painting you can choose
a circle that has dots of all three colors?

3. Solve the equation in natural numbers

NOC (a; b) + gcd (a; b) = a b.(GCD - greatest common divisor, LCM - least common multiple).

4. Circle inscribed in a triangle ABC, concerns
parties AB And sun at points E And F respectively. points
M And N- bases of perpendiculars from points A and C to the line EF. Prove that if the sides of the triangle ABC form arithmetic progression and AC is the middle side, then ME + FN = EF.

5. Integers are placed in the cells of the 8x8 table.
It turned out that if you choose any three columns and any three rows of the table, then the sum of the nine numbers at their intersection will be equal to zero. Prove that all numbers in the table are equal to zero.

1. The sine and cosine of a certain angle turned out to be different roots square trinomial ax2 + bx + c. Prove that b2= a2 + 2ac.

2. For each of the 8 sections of a cube with an edge A, which are triangles with vertices in the midpoints of the edges of the cube, the point of intersection of the heights of the section is considered. Find the volume of a polyhedron with vertices at these 8 points.

3. Let y=k1 x + b1 , y = k2 x + b2 , y =k3 x + b3 - equations of three tangents to a parabola y=x2. Prove that if k3 = k1 + k2 , That b3 2 (b1 + b2 ).

4. Vasya called a natural number N. Then Peter
find the sum of the digits of a number N, then the sum of the digits
N+13N, then the sum of the digits N+2 13N, Then
the sum of the digits of a number N+ 3 13N etc. Could he
dy next time get more result
previous?

5. Is it possible to draw on the plane 2005 non-zero
vectors so that from any ten of them it is possible
choose three with zero sum?

PROBLEM SOLUTIONS

7th grade

1. For example, 5 + 40 + 367 + 2918 = 3330.

2. One of the options is the following. For the first four days, Vasya must buy goods with all the money he has. Then in four days he will have rubles (100) On the fifth day he must buy goods for 9,000 rubles. He will have 7,000 rubles left. After dinner, he will sell the goods for rubles, and he will have exactly rubles.

3. Answer. Two of the possible cutting examples are shown in Figures 1 and 2.

Rice. 1 +

Rice. 2

4 . Answer. 6.

If all 7 sums were prime numbers, then two sums of 5 numbers in particular would be prime. Each of these sums is greater than 5. If both of these sums were prime numbers greater than 5, then each of these sums would be odd (because only 2 is an even prime number). But if we add these sums, we get an even number. However, these two sums include all numbers from 1 to 10, and their sum is 55 - an odd number. Therefore, among the sums received, no more than 6 will be prime numbers. Figure 3 shows how to arrange the numbers in the table to get 6 simple sums (in our example, all sums of 2 numbers are 11, and. 1 + 2 + 3 + 7 + 6 = 19). Comment. For example without evaluation - 3 points.

Rice. 3

5. Answer.N=1

Number N at least ten digits, since there are 9 different sums. Therefore smallest number ten-digit, with each of the sums

1, ..., 9 must occur exactly once. Of two ten-digit numbers that begin with the same digit, the smaller one has the smaller first digit that differs. Therefore, the first digit of N is 1, the second is 0. The sum of 1 has already been met, so the smallest third digit is 2, and so on.

8 Class

1. Answer. Could.

Consider, for example, the number A = zero at the end of 1001). Then

A2 = 1 at the end of 2002 zero). If you erase the last 2005 digits, then the number 1 remains.

2. Answer. 1003.

Note that two warriors standing side by side could not be knights. Indeed, if they were both knights, they would both have told lies. Let's choose the warrior standing on the left and split the row of the remaining 2004 warriors into 1002 groups of two warriors standing side by side. Each such group has no more than one knight. That is, among the 2004 warriors under consideration, there are no more than 1002 knights. That is, there are no more than 1002 + 1 = 1003 knights in the line.

Consider the line: RLRLR ... RLRLR. There are exactly 1003 knights in such a line.

Comment. If only an answer is given, put 0 points, if only an example is given, - 2 points.

3. Answer. Two weights.

One weight is not enough for the seller, since a weight of at least 20 kg is required to weigh 25 kg of sugar. Having only such a weight, the seller will not be able to weigh, for example, 10 kg of sugar. Let us show that two weights are enough for the seller: one weighing 5 kg and one weighing 15 kg. Sugar weighing from 0 to 5 kg can be weighed without weights. To weigh from 5 to 10 kg of sugar, you need to put a weight of 5 kg on the right cup. To weigh 10 to 15 kg of sugar, place a 5 kg weight on the left cup and a 15 kg weight on the right cup. To weigh 15 to 20 kg of sugar, you need to put a 15 kg weight on the right cup. To weigh 20 to 25 kg of sugar, you need to put weights of 5 kg and 15 kg on the right cup.

4. Answer. 60°, 30°, 90°.

This problem provides a detailed solution. A straight line passing through the midpoints of the legs divides the height CH in half, so the desired point R MN, Where M And N- the midpoints of the leg and hypotenuse (Fig. 4), i.e. MN- middle line ABC.

Rice. 4

Then MN || sun=>P =BCH(as internal cross lying angles with parallel lines) => VSN =NPH (CHB = PHN = 90°

CH = PH - on the side and sharp corner) => HH =NH => CN= SW= A(in an isosceles triangle, the height is the bisector). But CN- median of a right triangle ABC, That's why CN = BN(clear if described near a triangle ABC circle) => BCN- equilateral, therefore, B - 60°.

5. Consider an arbitrary 2x2 square. It cannot contain cells of all three colors, since then it would be possible to find a three-celled corner, all cells of which are of three different colors. Also, in this 2x2 square, all cells cannot be of the same color, since then it would be possible to find a three-cell corner, all cells of which are of the same color. This means that there are only two colors of cells in this square. Note that in this square there cannot be 3 cells of the same color, since then it would be possible to find a three-cell corner, all cells of which are of the same color. That is, in this square there are 2 cells of two different colors.

Let us now divide the 8x8 table into 16 squares 2 x 2. Each of them either does not have cells of the first color, or two cells of the first color. That is, there are an even number of cells of the first color. Similarly, there are an even number of cells of the second and third colors.

Grade 9

1. Answer. 1003, 1002, 0.

Since the sets are the same, it follows that a + b + c = a -1 + b + 1 + c2. We get c = c2. That is, c \u003d 0 or c \u003d 1. Since c \u003d c2 , then a - 1 = b, b + 1 = a. This means that two cases are possible: the set b + 1, b, 0 and b + 1, b, 1. Since the sum of the numbers in the set is 2005, in the first case we get 2b + 1 = 2005, b = 1002 and set 1003, 1002, 0, in the second case we get 2 b + 2 = 2005, b = 1001, 5 is not an integer, i.e. the second case is impossible. Comment. If only the answer is given, then put 0 points.

2. Answer. Could.

Note that among 11 consecutive natural numbers, there are two that are divisible by 5, and there are two even numbers, so their product ends in two zeros. Note now that a + (a + 1) + (a + 2) + ... + (a + 10) = (a + 5) 11. If we take, for example, a = 95 (that is, Vasya chose the numbers 95, 96, ..., 105), then the sum will also end in two zeros.

3. Let E,F, TO,L, M, N- touch points (Fig. 5).
Let's pretend that DE = EF = Facebook= x. Then AK =
= AL = a, BL = BE= 2x, VM =bf= x,CM = CN = c,
DK = DE= x,DN = D.F. = 2 x=> A-B+ BC = a+ Zx + c =
= AC, which contradicts the triangle inequality.

Comment. It also proves the impossibility of equality bf = DE. In general, if for an inscribed triangle ABD circles E- point of contact and bf = DE, That F is the point at which the excircle AABD touches BD.

Rice. 5 A K D N C

4. Answer. Right.

A first color and dot IN l. If out of line l ABC, A, B and WITH). So outside the line l D) lies on a straight line l A And D, lI IN And D, l l

5. Answer. Couldn't.

Consider a chess coloring of a 10 x 10 board. Note that a lame rook moves from a white cell to a black one, and from a black cell to a white one. Let the rook start bypassing from the white square. Then 1 will be in a white cell, 2 - in a black one, 3 - in a white one, ..., 100 - in a black one. That is, odd numbers will be in white cells, and even numbers in black ones. But of the two adjacent cells on the side, one is black and the other is white. That is, the sum of the numbers written in these cells will always be odd, and will not be divisible by 4.

Comment. For “solutions”, in which only an example of some kind of bypass is considered, put 0 points.

Grade 10

1. Answer, a = b = c = - 1.

The fact that the sets coincide implies that their sums coincide. So, a4 2b2+ b 4 - 2c2 + c4 - 2a2 = a + b+ with =-3, (a+ (b2- 1) 2 + (c \u003d 0. From where a2 - 1 = b2 - 1 = c2 - 1 = 0, i.e. a = ±1, b = ±1, With= ± 1. Condition a + b+ with= -3 satisfy only a = b = c =- 1. It remains to verify that the found triple satisfies the conditions of the problem.

2. Answer. Right.

Let's assume that it is impossible to choose a circle that has points of all three colors. Pick a point A first color and dot IN second color and draw a line through them l. If out of line l there is a point C of the third color, then on the circle circumscribed about the triangle ABC, there are points of all three colors (for example, A, B and WITH). So outside the line l no dots of the third color. But since at least one point of the plane is colored in the third color, then this point (let's call it D) lies on a straight line l. If we now consider the points A And D, then one can similarly show that outside the line lI there are no dots of the second color. Having considered the points IN And D, it can be shown that outside the line l no dots of the first color. That is, outside the line l no colored dots. We got a contradiction with the condition. So, you can choose a circle on which there are points of all three colors.

3. Answer, a = b = 2.

Let gcd (a; b) = d. Then A= a1 d, b =b1 d, where gcd ( a1 ; b1 ) = 1. Then LCM (a; b)= a1 b1 d. From here a1 b1 d+ d = a1 db1 d, or a1 b1 + 1 = a1 b1 d. Where a1 b1 (d - 1) = 1. That is al = bl = 1 and d= 2, so a= b = 2.

Comment. Another solution could be obtained by using the equality LCM (a; b) GCD (a; b) = ab.

Comment. If only the answer is given, then put 0 points.

4. Let VR- the height of the isosceles triangle FBE (Fig. 6).

Then from the similarity of triangles AME ~ BPE it follows that https://pandia.ru/text/78/390/images/image028_3.gif" width="36 height=31" height="31">.