Tasks of the municipal stage of the All-Russian Olympiad. Assignments for the municipal stage of the All-Russian Olympiad for schoolchildren in mathematics

Tasks of the municipal stage of the All-Russian Olympiad for 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 January 1, 2001 No. 000.

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

Evaluation criteria

Mathematical Olympiad tasks are creative and allow for several different solutions. In addition, it is necessary to evaluate partial progress in tasks (for example, analyzing an important case, proving a lemma, finding an example, etc.). Finally, logical and arithmetic errors in solutions are possible. The final score for the task must take into account all of the above.

In accordance with the regulations for holding mathematical Olympiads for schoolchildren, each problem is scored out of 7 points.

The correspondence between the correctness of the solution and the points awarded is shown in the table.

It is important to note that any correct solution is scored 7 points. It is unacceptable to deduct points because the solution is too long, or because the student’s solution differs from that given in the methodological developments or from other solutions known to the jury.

At the same time, any decision text, no matter how long, that does not contain useful advancements should be scored 0 points.

The procedure for holding the municipal stage of the Olympiad

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

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

Olympiad tasks at the school and municipal stages are compiled on the basis of mathematics programs for general education institutions. It is also allowed to include tasks whose topics are included in the programs of school clubs (electives).

Below are only those topics that are proposed to be used when compiling assignment options for the CURRENT academic year.

Magazines: “Quantum”, “Mathematics at school”

Books and teaching aids:

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

, Mathematics. All-Russian Olympiads. Vol. 1. – M.: Education, 2008. – 192 p.

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

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

Collection of Olympiad problems in mathematics. – M.: MTsNMO, 2005. – 560 p.

Planimetry problems . Ed. 5th revision and additional – M.: MTsNMO, 2006. – 640 p.

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

1. Instead of asterisks, replace the expression *+ ** + *** + **** = 3330 with ten different numbers so that the equation is correct.

2. Businessman Vasya started trading. Every morning he
buys goods with some part of the money he has (perhaps with all the money he has). After lunch, he sells the purchased goods for twice the price he bought. How should Vasya trade so that after 5 days he has exactly rubles, if at first he had 1000 rubles.

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

4. We wrote down all the natural numbers from 1 to 10 in a 2x5 table. After that, we calculated each of the sums of the numbers in a row and in a column (7 sums in total). 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= 35,207 amounts 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 the result on the board and erased the last 2005 digits. Could the last digit of the number remaining 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 warriors. Each of the warriors 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 is the largest number of knights that could be in the line if 2005 warriors came out to review?

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

4. Find the angles of a right triangle if it is known that the point symmetrical to the vertex of the right angle relative to the hypotenuse lies on the line passing through the midpoints of the two sides of the triangle.

5. The cells of the 8x8 table are painted in three colors. It turned out that the table does not have a three-cell corner, all the cells of which are the same color (a three-cell corner is a figure obtained from a 2x2 square by removing one cell). It also turned out that the table does not have a three-cell corner, all the cells of which are three different colors. Prove that the number of cells of each color is even.

1. Set consisting of integers a, b, c, replaced with set a - 1, b + 1, s2. As a result, the resulting set coincided with the original one. Find the numbers a, 6, c, if you know 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 one of
three colors, with all three colors 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) walked around a board of 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, etc. up to 100. Could it turn out that the sum of the numbers written in two adjacent cells on the side is divisible by 4 ?

Combinatorial problems.

1. A set consisting of numbers a, b, c, replaced with set a4 - 2b2, b 4- 2с2, с4 - 2а2. As a result, the resulting set coincided with the original one. Find the numbers a, b, c, if their sum is equal to - 3.

2. Each of the points of the plane is colored in one of
three colors, with all three colors used. Ver
but is it possible that with any such painting you can choose
a circle containing 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 dropped from points A and C to a straight line E.F.. Prove that if the sides of a triangle ABC form an arithmetic progression and AC is the middle side, then M.E. + FN = E.F..

5. The cells of an 8x8 table contain integers.
It turned out that if you select 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 of a square trinomial ax2 + bx + c. Prove that b2= a2 + 2ac.

2. For each of the 8 sections of a cube with an edge A, being triangles with vertices in the middle of the edges of the cube, the point of intersection of the section heights 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 named a natural number N. After which Petya
found the sum of the digits of a number N, then the sum of the digits of the number
N+13N, then the sum of the digits of the number N+2 13N, Then
sum of digits of a number N+ 3 13N etc. Could he each
next time get a better result
previous?

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

SOLUTIONS TO PROBLEMS

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 lunch, he will sell the goods in rubles, and he will have exactly rubles.

3. Answer. Two 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 in particular two sums of 5 numbers 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 (since 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 resulting sums, 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 an example without evaluation - 3 points.

Rice. 3

5. Answer.N=1

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

1, ..., 9 must appear exactly once. Of two ten-digit numbers that begin with the same digits, the one whose first differing digit is smaller is the smaller. Therefore, the first digit of N is 1, the second is 0. The sum of 1 has already been encountered, so the smallest third digit is 2, etc.

8 Class

1. Answer. She could.

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

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

2. Answer. 1003.

Note that the two warriors standing next to each other could not be knights. Indeed, if they were both knights, then they both told lies. Let's choose the warrior standing on the left and divide the row of the remaining 2004 warriors into 1002 groups of two warriors standing next to each other. There is no more than one knight in each such group. That is, among the 2004 warriors under consideration, there are no more than 1002 knights. That is, in total there are no more than 1002 + 1 = 1003 knights in the line.

Consider the line: RLRLR...RLRLR. In such a line there are exactly 1003 knights.

Comment. If only an answer is given, give 0 points; if only an example is given, give 2 points.

3. Answer. Two weights.

One weight will not be enough for the seller, since weighing 25 kg of sugar requires a weight weighing at least 20 kg. Having only such a weight, the seller will not be able to weigh, for example, 10 kg of sugar. Let us show that the seller only needs two weights: one weighing 5 kg and one weighing 15 kg. Sugar weighing from 0 to 5 kg can be weighed without weights. To weigh 5 to 10 kg of sugar, you need to place a 5 kg weight on the right cup. To weigh 10 to 15 kg of sugar, you need to 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 place a 15 kg weight on the right cup. To weigh 20 to 25 kg of sugar, you need to place 5 kg and 15 kg weights 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 middle of the leg and hypotenuse (Fig. 4), i.e. MN- middle line ABC.

Rice. 4

Then MN || Sun=>P =BCH(like internal crosswise angles with parallel lines) => VSN =N.P.H. (CHB = PHN = 90°,

CH = RN - along the side and acute angle) => VN =N.H. => CN= SV= A(in an isosceles triangle, the altitude is the bisector). But CN- median of a right triangle ABC, That's why CN = BN(obviously, if you describe it around 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-cell corner, all the cells of which are three different colors. Also, in this 2x2 square, all the cells cannot be the same color, since then it would be possible to find a three-cell corner, all the cells of which are the same color. This means that there are only two colored 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 the cells of which are 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 2 x 2 squares. Each of them either has no 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.

9th grade

1. Answer. 1003, 1002, 0.

From the fact that the sets coincide, the equality a + b + c = a -1 + b + 1 + c2 follows. We get c = c2. That is, c = 0 or c = 1. Since c = c2 , then a - 1 = b, b + 1 = a. This means that two cases are possible: 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 the 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 give 0 points.

2. Answer. They could.

Note that among 11 consecutive natural numbers, there are two divisible by 5, and there are two even numbers, so their product ends in two zeros. Let us now note that a + (a + 1) + (a + 2) + ... + (a + 10) = (a + 5) 11. If we take, for example, a = 95 (i.e. 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 = E.F. = FB= x. Then AK =
= AL = a, B.L. = BE= 2x, VM =B.F.= x,C.M. = CN = c,
DK = DE= x,DN = DF = 2 x=> AB + B.C. = a+ Zx + s =
= A.C., which contradicts the triangle inequality.

Comment. It also proves the impossibility of equality B.F. = DE. In general, if for inscribed in a triangle ABD circle E- point of contact and B.F. = DE, That F- 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 outside the 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. It couldn't.

Let's consider the chess coloring of a 10 x 10 board. Note that from a white square a lame rook moves into a black one, and from a black square into a white one. Let the rook begin its traversal from the white square. Then 1 will be in a white square, 2 - in a black one, 3 - in a white one, ..., 100 - in a black one. That is, white cells will contain odd numbers, and black cells will contain even numbers. But of the two adjacent cells, one is black and the other is white. That is, the sum of numbers written in these cells will always be odd and will not be divisible by 4.

Comment. For “solutions” that only consider an example of some kind of workaround, give 0 points.

Grade 10

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

Since the sets coincide, it follows that their sums coincide. So a4 - 2b2+ b 4 - 2с2 + с4 - 2а2 = а + b+ c =-3, (a+ (b2- 1)2 + (c= 0. Where from a2 - 1 = b2 - 1 = c2 - 1 = 0, i.e. a = ±1, b = ±1, With= ± 1. Condition a + b+ s= -3 satisfy only a = b = c =- 1. It remains to check that the found triple satisfies the conditions of the problem.

2. Answer. Right.

Let's assume that it is impossible to select a circle that contains points of all three colors. Let's choose a point A first color and dot IN second color and draw a straight line through them l. If outside the 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 there are no dots of a third color. But since at least one point of the plane is painted in a 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 similarly it can be shown that outside the line lI there are no dots of a second color. Having considered the points IN And D, it can be shown that outside the line l there are no dots of the first color. That is, outside the straight line l no colored dots. We received a contradiction with the condition. This means that you can choose a circle that has 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, which means 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 give 0 points.

4. Let VR- 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">.

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 Deputy Prime Minister of the Russian Federation T.A. Golikova.

Among the award winners are employees of the Laboratory for Working with Gifted Children. The award was received by 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, vice-rector of MIPT Artyom Anatolyevich Voronov.

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

This is the first time such a high result at the International Physics Olympiad has been achieved by the Russian team. In the entire history of the IPhO since 1967, neither the Russian nor the USSR national team had ever managed to win five gold medals.

The complexity of the Olympiad tasks and the level of training of teams from other countries is constantly growing. However, in recent years the Russian national team has been among the top five teams in the world. In order to achieve high results, the teachers and leadership of the national team are improving the system of preparation for international competitions in our country. Training schools have appeared where schoolchildren study in detail the most difficult sections of the program. A database of experimental tasks is being actively created, by completing which the children are preparing for the experimental tour. Regular distance work is carried out; during the year of preparation, children receive about ten theoretical homework assignments. Much attention is paid to high-quality translation of the conditions of the tasks 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, staff and students of MIPT, personal teachers on site, and the hard work of the schoolchildren themselves. In addition to the above-mentioned award winners, a huge contribution to the preparation of the national team was made by:

Fedor Tsybrov (creation of problems for qualification fees)

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

Alexey Alekseev (creation of qualification tasks)

Arseniy Pikalov (preparing theoretical materials and conducting seminars)

Ivan Erofeev (many years of work in all areas)

Alexander Artemyev (checking homework)

Nikita Semenin (creation of qualification tasks)

Andrey Peskov (development and creation of experimental installations)

Gleb Kuznetsov (experimental training of the national team)

8TH GRADE

SCHOOL TASKS

ALL-RUSSIAN OLYMPIAD FOR SCHOOLCHILDREN IN SOCIAL STUDIES

FULL NAME. student _____________________________________________________________________

Date of birth __________________________ Class ____,__ Date “_____” ______20__

Score (max. 100 points) _________

Exercise 1. Choose the correct answer:

The Golden Rule of Morality states:

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 your father and your mother.”

Answer: ___

Task 2. Choose the correct answer:

The ability of a person to acquire and exercise rights and obligations through 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 the Russian Federation, the highest legal force in the system of normative acts has

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

2) Constitution of the Russian Federation 4) Resolutions of the Government of the Russian Federation

Answer: ___

(For the correct answer - 2 points)

Task 4. A scientist must write concepts and terms correctly. Fill in the correct letter(s) in place of the blanks.

1. Pr…v…legia – an advantage granted to someone.

2. D...v...den... – income paid to shareholders.

3. T...l...t...ness - tolerance for other people's opinions.

Task 5. Fill in the blank in the row.

1. Clan, …….., nationality, nation.

2. Christianity, ………, Buddhism.

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

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

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

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

3. Custom, precedent, law.

1. ________________________________________________________

2.________________________________________________________

3.________________________________________________________

Task 7. Answer yes or no:

1) Man by nature is a biosocial being.

2) Communication refers only to the exchange of information.

3) Each person is individual.

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

5) Every person is born as an individual.

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

7) Society is a self-developing system.

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

9) The progress of historical development is contradictory: both progressive and regressive changes can be found in it.

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

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

KEYS TO ASSIGNMENTS

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 ( For a correctly indicated letter - 1 point. 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 a rule of law state

2. Functions of money

3. Sources of law.

Task 7 2 points for each correct answer. (Maximum for the task – 20 points)