School stage of the All-Russian Schoolchildren Olympics.
About boy Marata
This is a story that I heard from the neighbor of my tueshka. I will write on the face of a neighbor - the former teacher who is already retired.
When this story happened, I already worked for 27 years in a teacher and did a lot of different kids, even though the school was and a little designed for nearby villages.
There was August, we, teachers, prepared for the new school year, constituted plans, discussed the clock.
As I remember - it happened on August 25th. I went out of the teacher and went towards the Cabinet of Mathematics, I was class teacher Grade 3, this cabinet was assigned to us. The cleaner gave me the key (there was no wrap at school, for him there was a cleaner) and I had a stack of plans and benefits to prepare for the upcoming classes in silence. Opening the door, I'm just dumbfounded: I was sitting at my student Marat, the window was openly - the floor was recently painted. Failing to Shalun that he climbs through the windows, and not entering the door, I asked if he was ready for the upcoming study, which he received an unexpected answer: he would not come to school.
Why? I asked.
He said that parents go far, now they will not live here, he came to say goodbye, because I have a favorite teacher.
Wanted to him and family of a good way, I went out to pour myself a cup of coffee. Returning, the boy was no longer saw.
It has come on September 1. On the school playground was noisy and my class came in a good fighting of the Spirit, only the girl dinar was gloomy clouds. I knew that they were friends with Marat from the first class and thought that she was sad because of his departure. But the director of the school approached and called me aside.
- Have you already heard about the tragedy? She asked - in your class did not become a student.
My heart shoved me, because I sincerely loved these kids.
"Marat" said the director of the school - they drove with their parents on the mountain road and the brakes refused in an old car. Found the whole family at the foot of the mountain, everyone was dead.
For a minute I closed in my eyes, I turned away so that the children did not see my tears. I remembered how Marat came a couple of days ago to say goodbye and thought that then I saw him for the last time.
After the lessons, I called the dinar to myself and asked to talk to me so that the child would become easier. The girl burst into the streams of words broke out of a children's soul. But here I ran a nervous chill to me: Marat's family immediately began as summer vacation, I went to visit grandparents. In early June, they broke. Families guys lived next door and firmly friendly and family dinars called Marat's relatives.
I understood in a stupor: the boy came to say goodbye to me! And apparently thoughts, I voiced in a rumor, to what Dinara answered me calmly: he also came to say goodbye to me. Came to the playground and presented this pebble - his beloved. And the girl extended me a beautiful smooth pebble, which I often watched on the desk at Marat. And then - continued Dinara - in the evening they called the post office and called my father. He brought terrible news.
So, it would seem, an unprecedented case, happened to us, with ordinary people.
So I finished my story Elderly neighbor of my tuyushka.
That night I slept terrible. I thought all the time, as such a thing could be, on the one hand, it was simply not believed that, on the other, the elderly teacher is unlikely to tell the fairy tales, the more such, because of which it can be crazy. Woman she is calm, intelligent, nothing to tell to tell the unprecedencies, she said the truth.
- The gardener wants to plant six gooseberry bushes, so that at a distance of 2m from each of them rose exactly three bushes of the gooseberry. Can he do it?
Answer: Yes. For example, if on two sides of the ABCD square to construct the correct AEB and DCF triangles, then for each point the condition will be performed, since de \u003d ec
Criteria:
There is a faithful example without justifying the equality / inequality of the parties - 4 points;
There is a faithful example with a full justification - 7 points;
Only the answer - 0 points
Solution: Spertu multipliers t, m, and. Then the expression takes the form. Fraction accepts the greatest value for the smallest denominator and the largest numerator. Consequently e \u003d 1, and numbers and, k, and equal to 9.8,7 numbers. Numbers m, a, t May be arbitrary.
Criteria:
There is only an example with a correct answer - 7 points.
There is only an example - 4 points.
- Lisers with family and nine tails live in the magical kingdom. Those who have 7 tails are always lying, and those who have 9 tails, always speak the truth. One day, three foxes brought a conversation among themselves.
Redhead fox: "We have 27 tails together."
Gray Fox: "It really is so!"
White Fox: "Stupidity, Redhead says nonsense!"
How many tails was every fox? (Justify the answer.)
Solution: If the redhead told the truth, then all three would have 9 tails. But then the White would say the truth, and this is incorrect. Then redhead lies, and gray, respectively, too. Then White says the truth.
Answer: Redhead had 7 tails, in gray - 7, White - 9.
Criteria:
- The boy Marat can rise from the first floor to the fifth floor, and Dasha's girl during the same time he managed to run only to the fourth. Dasha is twice as fast as it rises, and Marat descends at the same speed as Dasha. The children decided to compete and coming from the first floor to 25, starting at the same time. Marat, reaching 25 floors, began to descend to meet the loser Dasha. How much time will take place from the beginning of the competition until the meeting?
Solution: For a minute, Marat rises to the 4th floor upwards, and Dasha - 3 floors up. For the same moment both can go down on 6 floors down. In order to defeat the marat to overcome 24 floors. After 6 minutes, Marat reaches the finish, and Dasha rises only on 18 floors (up to 19). Now the distance between them is 6 floors, and the rate of approximation 3 + 6 \u003d 9 floors per minute. To meet them will need 40 seconds.
Answer: 6 minutes and 40 seconds
Criteria:
Only the answer, without explanation - 1 point;
The solution with a full justification is 7 points.
- In the ABC triangle, all parties are equal to 2017 cm. Points M, N, P, K are located as shown in the figure. It is known that CK + PC \u003d MA + AN \u003d 2017 See Find the angle of Kon.
Solution: Note that CK + PC \u003d AP + PC and MA + AN \u003d MA + MC. Then CK \u003d AP and AN \u003d MC. Consequently, the triangles APN and MKC are equal. ∠anp \u003d ∠cmk and ∠apn + ∠anp \u003d 120o. Then ∠mpo + ∠pmo \u003d 120o. ∠kon \u003d ∠Pom \u003d 60o.
Answer: ∠Kon \u003d 60o
Criteria:
The solution with a full justification is 7 points.
- A natural number is called a palindrome if it does not change when it is written in the reverse order (for example, 626 - Palindrome, and 2017 - no). Imagine the number of 2017 as the sum of two palindromes.
Solution: for example, 1331 + 686 \u003d 2017.
Criteria:
The presence of any faithful example is 7 points.
- Airat and Dina together weigh 84 kg, Dina and Tanya - 76 kg, Tanya and Sasha - 77 kg, Sasha and Masha - 67 kg, Masha and Airat - 64 kg. Who is heavier all and how much does he weigh?
Solution: A + D \u003d 84, D + T \u003d 76, T + C \u003d 77, C + M \u003d 67, M + A \u003d 64. Mix all equations and obtain 2 (a + d + t + s + m) \u003d 368. Then a + d + t + c + m \u003d 184. Using the second and fourth equality from the condition, we obtain A + 76 + 67 \u003d 184. Consequently a \u003d 41, d \u003d 43, T \u003d 33, C \u003d 44, m \u003d 23.
Answer: Heavy - Sasha. Sasha weighs 44 kg.
Criteria:
Only the answer, without explanation, without specifying weight - 0 points;
Only the answer, without explanation, indicating the weight - 3 points;
The solution with a full justification is 7 points.
- Damir drew a square 5 hours 5 on the airtal sheet and paints each minute in one cell. Lesha considers the number of previously bordering (on the side) of previously painted cells and records this number on the board. Prove that when all the cells are painted, the amount of numbers on the board will be equal to 40.Things: we note that Lesha considers the number of borders of this cell, for which both neighboring cells are painted. Performing your operations, Lached every border considers one and only once. Then, the sum of all numbers is equal to the number of boundary segments, namely 2 * 4 * 5 \u003d 40.
- Locate the area of \u200b\u200bthe painted part of the parallelogram, if the area of \u200b\u200bthe large parallelogram is equal to 40 (the vertices of all parallelograms except the largest in the middle of the respective parties)?
Solution: In the ABCD parallelogram, we will spend the segments of EG and FH. They are parallel to the sides. Then 4 smaller parallelogram are formed. In each of them, the diagonal divides parallelograms into two equal parts. Consequently, the total area of \u200b\u200bthe "angular" triangles AEH, EBF, FCG, GDH is equal to the EFGH parallelogram area.
The task is given that all quadrangles are parallelograms. It is not necessary to prove it! Then the area of \u200b\u200bthe "angular" triangles of the largest parallelogram is equal to 20. The second - 10, in the third - 5. Subscribe from the area of \u200b\u200bthe entire parallelogram of the area of \u200b\u200b"angular" triangles of the first and third parallelograms. 40-20-5 \u003d 25.
Criteria:
Only the answer, without explanation - 1 point;
The solution with a full justification is 7 points.
- Instead of skips, insert such numbers to express
Became the identity.
Solution: Let the numbers be missed
Substitute to the equation. We get ,. Substitute we get
Then. Substitute we get then.
Criteria:
Only the answer, without explanation - 4 points;
The solution with a full justification is 7 points.
School stage of the All-Russian Olympiad of schoolchildren in mathematics
- Is 72017 + 72018 + 72019 divided into 19?
Decision: .
Answer: Yes.
Criteria:
Only the answer, without explanation - 0 points;
The solution with a full justification is 7 points.
- In the ABCD rectangle on the side of the CD, the middle m was noted on the side of the AD - the middle of N. CN segments and Am intersect at the point K. How many times the square of the AKCB quadrangle is more than the MDNK quadrangle area?
Solution: ED - Median Triangle ACD. It is known that the medians of the triangle divide it to six areometric. Then area of \u200b\u200bthe triangles AEK, CEK, CMK, DMK, DKN, ANK are equal. And the area of \u200b\u200bthe ACD triangle is equal to ABC Square. Then the attitude 
.
Answer: 4 times.
Criteria:
Only the answer, without explanation - 1 point;
The solution with a full justification is 7 points.
Solution: Let's see the mind 
. Transform
. Then the schedule will take the form
Criteria:
Only the right schedule, without explanation - 4 points;
The solution with a full justification is 7 points.
- In the village of Hobbits, everyone either always tells the truth, or always lies. The wizard invited several of the hobbits to himself and asked each of them about each of the others, the "belly" of one or "liar". A total of 54 responses of "Pravdolub" and 56 liar answers were received. How many times could the wizard hear the truth?
Solution: If n is invited by the Hobbits, it is given n (n - 1) \u003d 54 + 56 \u003d 110 of the responses, from where n \u003d 11. Let from these 11 hobbits T of belly and (11 - t) liars.
The answer "Liar" can only give a liar about the belt and the belt about the liar, such phrases were 2t (11 - t) \u003d 56, from where t \u003d 4 or t \u003d 7. If the truthly, four, they gave 4 ⋅ 10 \u003d 40 truthful answers . If the beluters are seven, they gave 7 ⋅ 10 \u003d 70 truthful answers.
Comment. Pay attention to the fact that it follows from the condition that half of the answers "Liar" are truthful. But it is not immediately clear what the proportion of the truthful responses is "Pravdolub".
Criteria:
Complete solution - 7 points.
The correctly found both cases (how many truths and liars), but
the number of truthful answers is incorrectly calculated - 4 points.
There are 2 situations described in the task. If correctly disassembled only
one, then put 3 points.
Both responses are given without an explanation - 1 point.
Only one of the answers is given - 0 points. The answer: 40 or 70
- The merchant jewelry has 61 weight gains 1g, 2g, ..., 61g. He put them in a row so that the weight of each, starting with the second, is a divider of the sum of the scales of all previous giri. The first weight of weighs 61g, the second - 1g. Find the weight of the Third Giri.
Decision. The sum of all numbers except the latter is divided into the last number,
so, the sum of all numbers is also divided into the last number. The sum of all numbers
from 1 to 61 equal to 31 ⋅ 61. So the last number is 1, 31 or 61. Since 1 and
61 stand in the first and second places, last number - 31. Third number -
the divider of the number 61 + 1 \u003d 62, that is, it is equal to 1, 2 or 31. We know that numbers 1
and 31 are not located in third place, so there is a number 2 in third place.
Comment. Give an example how are the numbers on the other cards
(or prove its existence) is not required.
Criteria:
Complete correct solution - 7 points.
It is argued that on the third card - number 2 or number 19, but
there are no other progress - 1 point.
School stage of the All-Russian Olympiad of schoolchildren in mathematics
- Find some pair of natural numbers a and b, boulen 1, satisfying the equation A13 · B31 \u003d 62017.
Decision. It is enough to bring one example.
Since, suitable a \u003d.
Boy Marat
(History about one immigration in the USA)
Moscow did not believe in tears ... as always.
Marat turned 20 years old, but it did not stop him from "chatting without a case," as his parents were expressed.
Since childhood, Marat wanted to dance, and he wanted to dance ballroom dancing, and it was completely bad in the eyes of others.
It was terrible, and "as the one, we are-so-giving birth !!!". But he had talents. Marat consisted of flexible durable muscles, and he did not grow fat.
Parents moved to Moscow to earn money from one of the countries of the neighboring countries, so Marat from childhood was granted to himself, although he went to school.
After school, he went to the dance circle at the Metro "Airport", a circle is paid, but he was taken without money, for talents. There, he studied Tango for many years, Salsa, and of course his favorite ballroom dancing.
In the circle, 25 girls and 2 boys studied. When Marata was fourteen a circle won the regional competition when he was sanging sixteen a circle won the urban. At sixteen years, thank God, the school ended, and Marat could devote himself to dancing. By that time, the parents were despair to influence him so that he chose himself a decent profession, such as plumbing, like dad.
Marat began to maintain dance lessons. He was not taken to the army, since he remained a citizen of another state, or for some other reasons that I am unknown. And his many friends were taken away, which added to marat even more popular among girls in his circle of communication.
When he was eighteen years old, Marat's circle took the second All-Russian place. And the second because the Parliament of Marat Natasha fell ill with a runny nose, and she had to replace Katya. But then Marat learned that the couple took something to someone had given someone for it.
A few weeks later, the circle was closed, Marat did not understand, too, why - someone needed this place in the basement of the house at the "Airport".
And besides, Natasha turned his leg, and in general "it was necessary and think about the profession," Natasha Parents said and determined it in Miit - Moscow Institute of Transport Engineers. (The Institute, which I know very well, since he has grown, due to certain circumstances, many Millionaire-Millionaires of Israel and the United States, and absolutely not led to a lot of fame and billions of dollars in American science). Very long, many students of this institute have arranged permanent residence in the United States.
But to the girl Natasha Kotelnikova, it did not have anything to do.
And then Marat thought, and what to do in life. You can become a locks-sanitary room, and easy to earn 30 thousand rubles per month.
But Marat decided to leave in the US, and there to dance. In those times, immigration and in the United States dreamed of many talented, unusual and capable people in the country. But not everyone was easily done for various reasons, which I will write about.
Then he turned to me.
We began to communicate with him on Skype and plan his move. In private dance lessons, he earned the desired amount for several years.
We have chosen such a way to arrive at permanent residence in the US who are in the people and the specialists in their anema were called "refugee". What perhaps most light way in our time immigrate in the USA.
But it was necessary to make refugee history. Marat suggested say that he he is gay and he oppressed him.
Marat: "Well, so, I can say that I gay. As a dancer I come out, and I can have all the geian manneuriums with ease. And then I spent my whole life with the girls and they already fed up. "
I: "So can you be gay?"
Marat: "No, no, I'm not gay ..."
I'm sure?"
Marat, after some pause: "Yes, not, definitely not gay!"
I: "Then we will not be able to say that you are gay, and besides, to prove that you are gay, you will need to provide photos of your partner,
where are you kissing, and photos where your things lie in one closet.
Do you have such a partner? "
(Really there are such requirements when people are trying so sidel immigrate in the USA)
Marat: "Not, there is no such partner ..."
In the end, Marat came to me to America. We used the other, the real story of his life, which was enough for legal registration of his permanent residence in the United States.
In it real life In the same place with his head, there were enough circumstances, which is truly, without any speculations, they gave him such an opportunity. He just did not even guessed to communicate with me.
Recently, Marat received his believed "refugee status" officially, on paper with the coat of arms and seals. He lives in Brooklyn and teaches children dancing.
At the All-American Competition on ballroom dancing in his category, he, with his partner vika, took the third place. The first was taken by Misha and Fain from Brooklyn, the second Nikolai and Katya from Los Angeles.
But the fourth, fifth, and so on, took John, Peter, Jose and others.
And danced Marat with his girlfriend really very good!
They were shown on TV ...
Transcript.
1 School stage of the All-Russian Olympiad of Schoolchildren in Mathematics Grade 8 1. The gardener wants to plant six gooseberry bushes so that at a distance of 2m from each of them grew exactly three bushes of the gooseberry. Can he do it? Answer: Yes. For example, if on two sides of the ABCD square to construct the correct AEB and DCF triangles, then for each point the condition will be performed, since de \u003d ec
2 Answer: Redhead had 7 tails, in gray 7, in white 9. Only the answer, without explanation 1 point; 4. Marat boy can rise from the first floor to the fifth floor, and Dasha's girl has time to reach only to the fourth. Dasha is twice as fast as it rises, and Marat descends at the same speed as Dasha. The children decided to compete and coming from the first floor to 25, starting at the same time. Marat, reaching 25 floors, began to descend to meet the loser Dasha. How much time will take place from the beginning of the competition until the meeting? Solution: For a minute, Marat rises to the 4th floor upwards, and Dasha on the 3 floors up. For the same moment both can go down on 6 floors down. In order to defeat the marat to overcome 24 floors. After 6 minutes, Marat reaches the finish, and Dasha rises only on 18 floors (up to 19). Now the distance between them is 6 floors, and the rate of approximation 3 + 6 \u003d 9 floors per minute. To meet them will need 40 seconds. Answer: 6 minutes and 40 seconds only the answer, without explanation 1 point; 5. In the ABC triangle, all parties are equal to 2017 cm. Points M, N, P, K are located as shown in the figure. It is known that CK + PC \u003d MA + AN \u003d 2017 See Find the angle of Kon. Solution: Note that CK + PC \u003d AP + PC and MA + AN \u003d MA + MC. Then CK \u003d AP and AN \u003d MC. Consequently, the triangles APN and MKC are equal. ANP \u003d CMK and APN + ANP \u003d 120 o. Then MPO + PMO \u003d 120 o. Kon \u003d pom \u003d 60 o. Answer: Kon \u003d 60 Only an answer, without explanations 0 points;
3 School Stage of the All-Russian Olympiad of Schoolchildren in Mathematics Grade 9 1. A natural number is called Palindrome if it does not change when recording its numbers in the reverse order (for example, 626 Palindrome, and 2017 is not). Imagine the number of 2017 as the sum of two palindromes. Solution: for example, \u003d 2017. The presence of any faithful example 7 points. 2. Airat and Dina together weigh 84 kg, Dina and Tanya 76 kg, Tanya and Sasha 77 kg, Sasha and Masha 67 kg, Masha and Airat 64 kg. Who is heavier all and how much does he weigh? Solution: A + D \u003d 84, D + T \u003d 76, T + C \u003d 77, C + M \u003d 67, M + A \u003d 64. Mix all equations and obtain 2 (a + d + t + s + m) \u003d 368. Then a + d + t + c + m \u003d 184. Using the second and fourth equality from the condition, we obtain A + 76 + 67 \u003d 184. Consequently a \u003d 41, d \u003d 43, T \u003d 33, C \u003d 44, m \u003d 23. Answer: The Heavy Sasha. Sasha weighs 44 kg. Only the answer, without explanation, without specifying weight 0 points; Only the answer, without explanation, indicating the weight of 3 points; 3. Damir drew a square 5 5 on the airtal sheet and the same cell paints each minute. Lesha considers the number of previously bordering (on the side) of previously painted cells and records this number on the board. Prove that when all the cells are painted, the amount of numbers on the board will be equal to 40. Proof: Note that Lesha considers the number of borders of this cell, for which both neighboring cells are painted. Performing your operations, Lached every border considers one and only once. Then, the sum of all numbers is equal to the number of boundary segments, namely 2 * 4 * 5 \u003d Locate the area of \u200b\u200bthe painted part of the parallelogram, if the area of \u200b\u200bthe large parallelogram is equal to 40 (the vertices of all parallelograms except the largest in the middle of the respective parties? Solution: In the ABCD parallelogram, we will spend the segments of EG and FH. They are parallel to the sides. Then 4 smaller parallelogram are formed. In each of them, the diagonal divides parallelograms into two equal parts. Consequently, total
4 Area of \u200b\u200b"angular" triangles AEH, EBF, FCG, GDH is equal to the EFGH parallelogram area. The task is given that all the quadrangles of the parallelogram. It is not necessary to prove it! Then the area of \u200b\u200bthe "angular" triangles of the largest parallelogram is 20. In the second 10, in the third 5. I will subtract the entire parallelogram of the area of \u200b\u200b"angular" triangles of the first and third parallelograms \u003d 25. Answer: 25. Only the answer, without explanation 1 point; 5. Instead of skips, insert such numbers so that the expression x + x + 6 (x + 4) \u003d (x +) (x + x + 8) becomes the identity. Solution: Let the numbers a, b, c be missed. (x + a x + 6) (x + 4) \u003d (x + b) (x + c x + 8). Substitute x \u003d 0 to the equation. We obtain 24 \u003d 8b, b \u003d 3. We will substitute x \u003d 4. We obtain 0 \u003d (4 + 3) (16 + C (4) + 8). Then c \u003d 6. We will substitute x \u003d 3. We obtain (9 3 a + 6) (3 + 4) \u003d 0. Then a \u003d 5. Answer: (x + 5 x + 6) (x + 4) \u003d (x + 3) (x + 6 x + 8). Only the answer, without explanation 4 points;
5 School stage of the All-Russian Olympiad of Schoolchildren in Mathematics Grade 10 1. Is it divided into 19? Solution: \u003d 7 () \u003d answer: Yes. Only the answer, without explanation 0 points; 2. In the ABCD rectangle on the CD side, the middle M was noted on the middle of the AD side of the middle of N. CN and AM segments intersect at point K. How many times the square of the AKCB quadrangle is larger than the MDNK quadrangle area? Solution: ED Median Triangle ACD. It is known that the medians of the triangle divide it to six areometric. Then area of \u200b\u200bthe triangles AEK, CEK, CMK, DMK, DKN, ANK are equal. And the area of \u200b\u200bthe ACD triangle is equal to ABC Square. Then the attitude \u003d answer: 4 times. Only the answer, without explanation 1 point; 3. Build the graph of the function y \u003d (x + 1) + x. y \u003d x x solution: Let's see the mind. We convert x at x 0 y \u003d 2x 1 with x< 0. Тогда график примет вид x 1
6 Only the right schedule, without explanation 4 points; 4. In the village of Hobbits, everyone either always tells the truth, or always lies. The wizard invited several of the hobbits to himself and asked each of them about each of the others, the "belly" of one or "liar". A total of 54 responses of "Pravdolub" and 56 liar answers were received. How many times could the wizard hear the truth? Solution: If n hobbits are invited, then n (n 1) \u003d \u003d 110 of responses, from where n \u003d 11. Let from these 11 hobbits T truthlyubs and (11 t) liars. The answer "Liar" can only give a liar about the belt and the belt about the liar, such phrases were 2t (11 t) \u003d 56, from where T \u003d 4 or T \u003d 7. If the beluters are four, then they gave 4 10 \u003d 40 truthful answers. If the beluters are seven, they gave 7 10 \u003d 70 truthful answers. Comment. Pay attention to the fact that it follows from the condition that half of the answers "Liar" are truthful. But it is not immediately clear what the proportion of the truthful responses is "Pravdolub". Complete solution of 7 points. Both cases (how many truthful and liars) are properly found, but the number of truthful responses of 4 points is incorrectly calculated. There are 2 situations described in the task. If only one is correctly disassembled, then put 3 points. Both answers are given without explanation 1 point. Only one of the answers 0 points is given. Answer: 40 or 70
7 5. The merchant jewelry has 61 weight gains 1g, 2g, 61g. He put them in a row so that the weight of each, starting with the second, is a divider of the sum of the scales of all previous giri. The first hother weighs 61g, the second 1g. Find the weight of the Third Giri. Answer. 2. Decision. The sum of all numbers, in addition to the latter, is divided into the last number, it means that the sum of all numbers is also divided into the last number. The sum of all numbers from 1 to 61 is equal to, the last number is 1, 31 or 61. Since 1 and 61 stand in the first and second places, the last number 31. Third-number divider number \u003d 62, that is, it is equal to 1, 2 or 31. We know that the numbers 1 and 31 are not located in third place, so there is a number 2. Comment in third place. To give an example, as the numbers are located on the other cards (or prove its existence) is not required. Complete correct solution of 7 points. It is argued that on the third card number 2 or number 19, but there are no other progress 1 points.
8 School Stage of the All-Russian Olympiad of Schoolchildren in Mathematics Grade 11 1. Find some pair of natural numbers a and b, more than 1, satisfying equation A 13 B 31 \u003d solution. It is enough to bring one example. Since 2017 \u003d, suitable a \u003d 6, b \u003d 6. Comment: Many different answers are possible with all sorts of combinations of degrees of bodies and triples. At least one pair of values \u200b\u200bA, B is shown and it is shown that it satisfies this condition 7 points. There is a couple of numbers, nothing is justified (and the jury knows how to show that the couple is suitable) 5 points. The main idea of \u200b\u200bsolving is true, but an arithmetic error is allowed (for example, it is written that 2017 \u003d) 2 points. 2. Does the COS2015x + TG2016X SIN2017x \u003d 0 equation? At least one root? Justify the answer. Answer: for example,. Solution: COS + TG SIN \u003d + 0 \u003d 0 The correct answer is shown, and it is shown that with the value of the equality correctly, 7 points. Only the right answer is 3 points. 3. Dan cube. A, B and C of the middle of his Röbeber (see Figure). What equal to cosine Angle ABC? Solution: Do not detracting the generality, we take the side of the cube for 2. Then AC \u003d 2, AB \u003d CB \u003d calculated by the three sides of the ABC angle cosine. cosα \u003d \u003d. The right answer is obtained with all rationale for 7 points. The decision is correct, but the answer is incorrect because of the arithmetic error 5 points.
9 Received 4 points. Only the answer (including correct) 0 points. Answer: 4. On the coordinate plane (x, y), depict the set of all points for which Y 2 + Y \u003d x 2 + x. Answer: Solution: y + y \u003d x + x x y + x y \u003d 0 (x y) (x + y + 1) \u003d 0. Then. A faithful schedule is built with all rationale for 7 points. A faithful schedule is built without justifying 3 points. 5. In the penalty, Ravil has 9 pencils. He noticed that among any four pencils at least two color. And among any five pencils, no more than three have one color. There are many different color pencils at Ravil, and how many pencils of each color? Answer. Three colors of three pencils. Decision. Not one color is not more than three, since otherwise the condition "among any five pencils no more than three have one color" would not be fulfilled. Total pencils 9, therefore, there are no less than three colors. On the other hand, among any four pencils, at least two color, therefore, there are less than four colors. Thus, the colors of the pencils are three, and each of no more than three pieces, and the entire pencils 9. So, each color is 3. A full response with a faithful explanation of 7 points.
10 It is reasonable that the children are three 5 points. Faithful considerations, but the decision was not brought to the end of the 1-2 points. The answer is without justifying 0 points.
Mathematics. Class. Option --5-7 Criteria for estimating tasks with expanded response C (SINX) (COS X +) Decide equation \u003d. TGX The left part of the equation makes sense at TGX\u003e. We equate the numerator to zero: (SINX
All-Russian Schoolchildren Olympiad in mathematics. 016 017 Uch. G. School Stage. 10 class of tasks, answers and evaluation criteria 1. (7 points) point o ABCD square center. Find some seven pairwise
8th grade. March 017. 8-1. Let a be a set of integers having at a division of 3 residues; B Many of the integers having in the division of 8 residue 6. Find all the numbers that are included simultaneously in A and B. Answer.
All-Russian Olympiad Schoolchildren in Mathematics 2015 2016 Uch. G. School Stage 9 Decision class and evaluation criteria 1. A natural number is called Palindrome if it does not change when recording
Reduce the fraction: A A A A. Grade 9 answer: a a. We will find the definition area of \u200b\u200bthis expression: A A A 0 0 A 0. Using the identity XY X Y, we obtain: a (a) 0 (a) (a) 0 A A A A A \u003d A (A) (A) (A)
Grade 8 First Day 8.1. In the state, each resident either knight or liar. Knights always speak the truth, and liars are always lying. All residents are familiar with each other. President once expressed two statements:
2016 2017 academic year Grade 51 Establish in Record 2 2 2 2 2 brackets and signs of action so that it turns out 24 52 Anya lies on Tuesdays, Wednesdays and Thursdays and tells the truth in all other days of the week
Mathematical Olympiad "Future Researchers Future of Science" Final Tour 9.03.015 Tasks with decisions Grade 7.1. Before the competition for running Petya planned to escape the entire distance at a constant speed
Department of Education of the Yaroslavl region All-Russian Olympiad of schoolchildren 07/08 of the school year Mathematics, class, municipal stage General principles Checks and estimates olympiad work Mathematical tasks
Tournament them. A.P.Savina, 2016 Mathematical battles, 3 Tour Grade 6, Higher League 1. Starting from some natural number, Petya increases each step or natural Petya may receive 218?
1 Tour Task 1. Is it possible to place 12 12 in half cells in half the cells so that in one square 2 2, composed of the cell cells, was the odd number of chips, and in the rest of the one? Task 2.
Grade 6.1. Kolya had two wooden cubes. On the first cube, on the same face, he wrote the letter A, on the other on three sides, he wrote the letters E, Yu, I. Show how to add letters on the verge of cubes
Zonal Olympiad Schoolchildren in Mathematics Krasnodar Territory, December 10, 2013 Grade 5, Fedorenko I.V. Text, Telephone for references +7 918 225-22-13 1 There are 2013 apples and scales on which
MSTU them. N.E. Bauman Olympiad Schoolchildren "Step into the Future" Tour, Grade 9, February 15, 015, option 1 1. A decimal notation of a natural number n contains 1580 digits. Among these figures there are three tops, tops
Grade 11 Task 11.1. Petya and Vasya participated in the position of the President of the Chess Club. By noon, Petit had 5% of the vote, and Vasi has 45%. Only friends petition came to the vote
Grade 9 first round (0 minutes; each task of points) ... is it true that if b\u003e a + c\u003e 0, then quadratic equation A + B + C \u003d 0 has two roots? Answer: Yes, right. The first way. From this inequality it follows
Grade 6.1. Replace in the example decimal fractions Each sprocket is 2 or digit 3 so that it turned out true equality: 0, + 0, + 0, + 0, \u003d 1. 6.2. Graduate students went to
In mathematics (2016-2017 uch. Year), grade 5 5.1. In the example, the addition of the figures was replaced by letters: the same are the same, different different. It turned out ABBB + A \u003d VGGG. Restore an example. 5.2. Two notebooks cost
Penza state University Physical and Mathematics Faculty "Employment of Physics and Mathematics School" Mathematics identical transformations. Solving equations. Triangles Task 1 for
Combined Interuniversity Mathematical Olympiad 0004 I Option (Answers and Brief Solutions) X + X + X + Task from x \u003d It follows that x + x \u003d x x for all this means that the sequence arithmetic
Interregional Schoolchildren Olympiad "Higher Sample", 2017 Mathematics, 2 Stage p. 1/10 Solutions and criteria for estimating tasks of the Olympics 10-1 in a company of 6 people some companies in three walked
All-Russian Schoolchildren Olympiad in Mathematics, Municipal Stage, 2016, Grade 11 1. The angle X satisfies the equality to calculate. Answer: 6. Decision. 1st way .. 2nd way.,. Then reasonably received
Grade 5.1. In the record 2 0 1 0 2 0 1 1 1, arrange the signs + between some numbers, so that the result is the number of 2013. The solution. For example, so 2010 + 2 + 0 + 1 1 + 1 or 2010 + 2 + 0 + 1 + 1 1. 5.2. Can
Grade 11 First round (10 minutes; each task is 6 points). 1.1. Solve inequality: X + Y 2 + 1. Answer: (1; 0). The first way. I rewrite this inequality: x + 1 y 2. Since x y 2 1 0, then x y 2 +
All-Russian Schoolchildren Olympiad in mathematics. 2016 2017 uch. G. School Stage. 8 class of tasks, answers and evaluation criteria 1. (7 points) in frame 8 8 in 2 cells (see Figure) of only 48 cells.
C mathematics class version of ma- (without logarithm) criteria for estimating tasks with an expanded answer a) decide the equation sin + cos + \u003d b) Find all the roots of this equation belonging to the segment π; π.
Mathematics. Grade 11. Embodiment MA10511 1 Criteria for estimating tasks with an expanded answer 13 SIN X COS X a) Decide equation + \u003d 3. B) Find all the roots of this equations belonging to the gap 5π; π.
Tasks of the municipal stage of the All-Russian Olympiad of schoolchildren in mathematics in the 0-0 school year solving tasks 0 class General notes for verification: almost all tasks are written in the criteria on the basis of "given"
Grade 7.1. Square in the square 7 7 four figures shown in the figure so that in any square 2 2 it turned out to be painted at least one cell. 7.2. In the family of funny dwarves dad, mom and baby. Names
Algebra 1. Build sketches of graphs of the following functions: y \u003d 2 (x + 2) / (3 2x); y \u003d y \u003d () (4 x) / (x + 1) 1; y \u003d 5 (x) / (x 1); y \u003d 3 x2 5 x +2; y \u003d 3 () 2x 1 1; 2 1 x 2 x 2; y \u003d 1 x 2 3 x + 2; y \u003d.
Tasks of the absentee tour of mathematics for grade 9, 2014/2015 uch. year, first level of complexity Task 1 solve equation: (x + 3) 63 + (x + 3) 62 (x - 1) + (x + 3) 61 (x - 1) 2 + + (x-1) 63 \u003d 0 answer: -1 Task 2 Amount
The All-Russian Olympiad of schoolchildren "Mission is fulfilled. Your vocation-financier! " Selection (correspondence) stage Mathematics 8 and 9 classes Option 1 Task 1. (10 points) ABCD rectangle diagonal
Grade 8 1. Prove that for any natural number, it is possible to choose such a natural number A so that the number A (n + 1) (+ n + 1) is focused on. 2. Two participated in the city Olympiad in mathematics
Grade 7 1. Decipher the numeric rus ( different Figures Match different numbers are the same identical). +. Cut the rectangle 4 x 8 to 8 equal parts so that in each part there was an asterisk. *
Object class Time (min) All-Russian Olympiad Schoolchildren Total Points September 22, 2017 Mathematics Number of points for the task Task Task Task Task Task Task Problem 1 2 3 4 5 6 7 Mathematics
Solving the tasks of the municipal stage of the All-Russian Olympiad of Schoolchildren in Mathematics 017 Year 8 Class Task 1. Red hat decided to go to her grandmother, whose house was 1 km walk from her house.
Grade 6.1. Crossed from among 987654321 as many digits as possible so that the remaining number is divided by 15. 6.2. Fold out five-taped figures, among which there are no two identical, some checkered
XL tournament named after M. V. Lomonosov October 1, 017, a contest in mathematics. Answers and solutions (preliminary version of 01.10.017) in brackets indicated how the class is recommended (solving problems more
Grade 5.1. If mom enters the room, then the total age of the in the room will increase by 4 times, and if the dad will enter the total age in 5 times. How many times the total
Mathematics. Grade 9. Embodiment MA90901 1 Criteria for estimating tasks with an expanded response The "Algebra" module 1 Decide equation x 6 6 x 8 3. x 6 6 x 8 3; x from where x or x 4. Answer :; 4. 6x 8; x x 4 0, conversion
Stage II Grade 7 3.12.2017 Work is designed for 180 minutes 1. Inscribe the four beams of OA, OB, OC and OD with a general start so that the corners of 100, 110, 120, 130 and 140 are found on this drawing. Write what exactly
0 class ... (points) Find all solutions of inequality: X + Y 5,5 + x Y 005 00. Answer: (;). Let (x; y) be the solution of inequality. Then from the terms of the task it follows that x y
Decisions of the tasks of the second absentee tour of the Olympiad "Applicient - 06" Olympics calculate 5, the solution to split both parts of equality 5 to this is possible, if 0, then 0, those 0, which contradicts the condition we have: 5 0 means
United Inter-University Mathematical Olympiad 000 General Evaluation Criteria According to the results of verification of each task, one of the following estimates are exhibited (listed in descending order): The task is solved
Grade 9 First round (10 minutes; each task is 6 points). 1.1. Straight y \u003d k + b, y \u003d k + b and y \u003d b + k are different and intersect at one point. What could be its coordinates? Answer: (1; 0). From the equation first
Open Olympiad Schoolchildren in Mathematics Solutions Grade 8 1 Option 1. (2 points) Whose methods can be broken down the portigid figure to rectangles 1 h 3? Answer: 6 on each side of the figure is
Mathematical Olympiad "Future Researchers Future of Science" First Tour. Option.0.0. In each parallel, 5 tasks were offered, the maximum rating of each task 0 points 9 class. To the number 0 send
Conditions of tasks 1 Municipal stage 8 Class 1. Two numbers are written on the board. One of them increased 6 times, and the other was reduced by 2015, while the amount of numbers did not change. Find at least one pair of such
The solutions and exemplary test criteria of the OMM-010 The following criteria, of course, cannot cover all cases and if some decision is not falling under the criteria, it is worth assessing it in common
Solutions and you will show all the data on the number on the numeric axis, which is the left of the left of all and is the smallest number 4 answer: 5 And we analyze the inequality on the numerical axis of the set of numbers satisfying
The second (final) stage of the Olympiad of schoolchildren "Step into the Future" for 8-0 classes on the general educational subject "Mathematics", Grade 8, Spring 08. Option 3 Task. (5 points) find all natural
Class Number 890 has such a property: By changing any of its digit to (increasing or decreasing), you can get a number, to find the smallest three-digit number, which has the same answer: 0
All-Russian Olympiad of Schoolchildren 013-014 in the city of Moscow typical tasks I (school) stage of the Olympiad in Mathematics Grade 5. Brief decisions. 1. Vasya can get the number 100 using ten bobs,
I. V. Yakovlev Materials on mathematics Mathus.ru Theorem of Pythagora We are ready to derive the most important geometry theorem theorem of Pytagora. With the help of the Pythagora theorem, many geometric calculations are performed.
The All-Russian Olympiad of schoolchildren "Mission is fulfilled. Your vocation-financier! " Tasks, solutions, criteria Final (full-time) Stage Mathematics 8-9 Class, 2016/2017 School year Task 1. (10 points)
Topic I. parity Task 1. Square table 25 25 painted in 25 colors so that in each row and in each column all colors are presented. Prove that if the location of the colors is symmetrically relative
7 Class 71 Increase in the circles in the number of numbers from 2 to 9 (without repetitions) so that no number shall be aimed at any of its neighbors 72 rectangles cut into several rectangles,
Terms and solutions to the tasks of the Interregional Olympiad of schoolchildren on the basis of departmental educational institutions In mathematics 0-0 of the school year, the given tasks were offered in three age categories
1.1. (6 points) How many roots have an equation x 6 x cos (x) 0? Answer: Five. x 6 xxx 6 0, x 6 x 0, x or x 15, cos (x) 0 cos (x) 0, xk, k z x 0, 5 k, k z. x 6 x 0 xxx 6 0 1, 5 Thus, roots
Options for the final (full-time) stage of the MSTU OLYMPIAD AM NE Bauman "Step into the Future" in mathematics for 8-0 classes 03-04 academic year Option (0 class) Petya, Vasya and Tolya compete in running on
The Olympiad "Path to Olympus", 8 K L and C 1. To a black number N, its largest divider was added, different from N. Can the amount obtained equal to 018?. The assembled honey fills several 50 liter bids.
School stage of the All-Russian Olympiad of schoolchildren in mathematics, 2018/2019 academic year. Answers 8 Class 1. The ball weighs more cat Matroskin on half the weight of Uncle Fedor, Uncle Fedor as much as a ball
Tasks and solutions for district (urban) olympiads in mathematics 7-8 school year 9 class Locate the lowest integer X, satisfying the inequality of the answer -7 x 7 8 x square three decreases P (x) AX BX C (A,
Mathematics. Class. Option M06 Task assessment criteria with an expanded response A) Decide equation 0. COS X π SIN X B) Find all the roots of this equation belonging to the segment A) converting the equation:
XXVI Interregional Olympiad Schoolchildren in Mathematics "Sammat-8" Final Tour Class It is known that the function f () is continuous at point 0 and for any valid satisfies the equation F (8) F () +
Science 1 3 7 Class Table of Contents Grade 7 ... 2 8 Class ... 3 9 Class ... 4 10 Class ... 5 11 Class ... 6 7 Class Science 1 3 7 Class 1. In a circle in random order put numbers from 1 to 31 and calculated all
Municipal education "Guryevsky City District" All-Russian Olympiad Schoolchildren in Mathematics ( school stage) 216-217 academic year 11 class maximum number of points 2 execution time 4
Mathematical Olympiad "Future Researchers Future of Science" Final Tour. 03/6/06 7 Class 7 .. There are kg cereals. Is it possible using three weighing on the cup scales to measure kg, if there is one three kilogram
Grade 7 First round (10 minutes; each task is 6 points). 1.1. Are there such integers a and b, which the number B is a natural degree of number A and the number B 16 times more than any? Answer: Yes,
Mathematics for all Yu.L. Kalinovsky table of contents 1 median, bisector, height ................................. 5 1.1 Medians triangle 5 1.2 triangle bisector 7 1.3 triangle heights 10 medians
7th grade February 8 06g Time for writing 4 astronomical hours Each task is estimated at 7 points 7 .. Prove that if an integer, then an integer. 7 .. Is it possible to paint the plane in 06 colors
