Objective 1:
Find all triples of nonzero numbers a, b and c that form an arithmetic progression and such that from numbers, you can also make an arithmetic progression.
Solution: By the property of the arithmetic progression, we have a + c = 2b and one of the following equationsThe first case leads to the equation b² = 2ac, which has no solutions for a + c = 2b; the other two lead to the same answer: all triples of the form - 2t, - 0.5t, t, where t ≠ 0.
Answer: - 2t, - 0.5t and t at t ≠ 0.
Objective 2:Find triples of numbers a, b and c, which are powers of five with non-negative integers, such that by attributing the decimal representation of one of them to the decimal representation of the other, we get the third number.
Solution: Let a = 5 n, b = 5 m, c = 5 k and in the number b, exactly t decimal places. We have the equation: 5 n • 10 t + 5 m = 5 k. Obviously, m< k. Сократив уравнение на 5 в наибольшей степени, получим либо 2 t + 5 m - n - t = 5 k - t , либо 5 n - m + t 2 t + 1 = 5 k - m . Первое уравнение имеет единственное решение в целых числах t = 2, m - n - t = 0, k - t = 1, откуда b = 25, m = 2, n = 0, k = 3 и искомые числа - 1, 25, 125. Второе уравнение выполняется только при n - m + t = 0, что приводит к предыдущему случаю.Answer: 1, 25 and 125.
Objective 3:Zeros are written at the vertices and intersection points of the diagonals of a regular pentagon. In one move, it is allowed to add + 1 or - 1 simultaneously to all numbers located on any of the diagonals of the pentagon. Which of the pentagons indicated in the figures can be obtained after several moves?
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((Proposed by S.E. Nokhrin.))
Solution: Let's number the diagonals of the pentagon with numbers from 1 to 5, and let x i be the number of units added to the i-th diagonal. The number at any vertex (the point of intersection of the diagonals) is equal to the sum of the numbers x i over all i such that the i-th diagonal passes through this vertex (the point of intersection of the diagonals). We have a system of ten equations with five unknowns, which turns out to be inconsistent in all cases shown in the figures.Answer: no pentagon can be obtained.
Task 4:In an acute-angled triangle ABC the heights are drawn: AH, BK and CL. Find the perimeter of the triangle HKL if the height AH = h and the angle ∠ BAC = α are known.
((Proposed by V.N. Ushakov.))
Solution: Lines KL, KH and HL (see fig.) Cut off triangles similar to ∆ ABC from ∆ ABC. Indeed, ∆ CHA ∽ ∆ CKB according to the I criterion of similarity of triangles (2 equal angles). From here. But then ∆ KHC ∽ ∆ BAC according to the II sign of similarity of triangles (proportionality of the sides and equality of the angles between these sides). It can be proved similarly that ∆ AKL ∽ ∆ ABC and ∆ BHL ∽ ∆ ABC. So, we have ∠ HLB = ∠ ALK = ∠ C, ∠ AKL = ∠ CKH = ∠ B. Then the points H ′ and H ″, symmetric to the point H with respect to lines AB and AC, respectively, lie on the line KL. Indeed, ∠ HLB = ∠ H′LB (since ∆ HLO ′ = ∆ H′LO ′), but ∠ HLB = ∠ ALK, hence ∠ ALK = ∠ H′LB, and hence the points K, L, H ′ lie on one straight line. It can be proved similarly that H ″, K, L are collinear. The segment H ″ H ′ is equal to the perimeter ∆ KLH (KH = KH ″, and LH = LH ′). Consider now ∆ H ″ AH ′. It is isosceles because AH ′ = AH = AH ", and ∠ H ″ AH ′ = 2 • (∠ CAH + ∠ BAH) = \ = 2 α. Hence H ″ H ′ = 2AH ′ sin \, α. Thus, the perimeter of ∆ KLH is 2h sin \, α.1. Solve the number puzzle.
2. Ignat is now four times older than his sister was when she was half his age. How old is Ignat now, if in 15 years he and his sister will be together for 100 years?
3. Children in pairs leave the forest where they collected nuts. In each pair there are a boy and a girl, and the boy either has twice as many or half as many nuts as the girl. Could it be that they all have 2011 nuts?
4. Cut the rectangle with sides 4 and 9 into the smallest number of pieces so as to make a square.
5. On the island of O there are knights who always tell the truth, and liars who always lie. The traveler met two natives - A and B. Native A said the phrase:
At least one of us (A or B) is a liar.
Can you tell who A is and who B is (a knight or a liar)?
Olympiad tasks municipal stage mathematics
1. Find all such three-digit numbers that the sum of the digits of the number is 11 times less than the number itself https://pandia.ru/text/78/035/images/image003_105.gif "width =" 27 "height =" 17 "> square points are taken so that the straight line crosses the side at the point, the straight line crosses the side at the point and https://pandia.ru/text/78/035/images/image013_32.gif "width =" 104 "height =" 21 ">.
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5. At the inspection 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 soldiers in the line said: "My neighbors in the line are liars." (The warriors at the ends of the line said: “My neighbor in the line is a liar.”) What is the largest number of knights in a line if 2011 soldiers were on display?
Olympiad tasks municipal stage All-Russian Olympiad for schoolchildren mathematics
1. Vasya wrote several integers on the board. Petya signed his square under each of Vasya's numbers. Then Masha added up all the numbers on the board and got 2011. Prove that one of the guys was wrong.
2. The cooperative receives apple and grape juice in the same cans and produces an apple-grape drink in the same cans. One can of apple juice is enough for exactly 6 cans of drink, and one can of grape juice is exactly 10. When the recipe of the drink was changed, one can of apple juice was enough for exactly 5 cans of drink. How many cans of drink is enough for one can of grape juice now? (The drink cannot be diluted with water.)
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4. 
Prove that for all positive https://pandia.ru/text/78/035/images/image023_20.gif "width =" 13 "height =" 15 "> the difference between the roots of the equation is 3?
3. Given NS points, no four of which belong to the same plane. How many planes can be drawn through the various triplets of these points?
4..gif "width =" 12 "height =" 15 src = ">, forming an arithmetic progression and such that the numbers and can also be used to make an arithmetic progression.
5. The diagonals of the parallelogram meet at a point. Let and be the intersection points of the circles, one of which passes through the points https://pandia.ru/text/78/035/images/image031_14.gif "width =" 16 "height =" 17 src = ">, and the other through and https://pandia.ru/text/78/035/images/image002_138.gif "width =" 19 "height =" 19 "> if the point lies on the line segment and does not coincide with its ends.
Olympiad tasksmunicipal stage mathematics
7th grade
At least one of us (A or B) is a liar.
Can you tell who A is and who B is (a knight or a liar)?
Olympiad tasksmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
8th grade
Olympiad tasksmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
Grade 9
- Which five-digit numbers are more: those with numbers in strictly ascending order, or those with numbers in strictly descending order? (For example, the first group includes 12,459, but excludes 12,495 and 12,259).
Olympiad tasksmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
Grade 10
- Numbers from 21 to 30 are written in a row. Is it possible to place signs "+" and "-" between them so that the value of the resulting expression was equal to zero?
- At what valuesdifference of the roots of the equation is equal to 3?
- Given n points, no four of which belong to the same plane. How many planes can be drawn through the various triplets of these points?
- Find all triples of non-zero numbers and forming an arithmetic progression and such that the numbers and you can also make an arithmetic progression.
- Parallelogram diagonalsintersect at the point... Let u - points of intersection of circles, one of which passes through the points and, and the other through and ... Find the locus of points if point lies on the segmentand does not coincide with its ends.
Olympiad tasksmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
Grade 11
- What is the smallest natural is that divisible by 770?
- Prove that if, then the equation
- Find if; ; ,,.
- At the base of a regular pyramid is a polygon with an odd number of sides. Is it possible to arrange arrows on the edges of this pyramid (one on each edge) so that the sum of the vectors obtained is equal to
?
- There are 20 students in the class. Each is friends with at least 10 others. Prove that in this class it is possible to choose two threes of students so that any student from one triple is friends with any student from the other triple.
Preview:
Grade 7 (solutions and answers)
Answers and solutions to problemsmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
- Answer: 2222 – 999 + 11 – 0 = 1234.
- Answer: 40 years old.
Solution: Let's use the table to solve the problem.
The equation: ... Now Ignat is 40 years old.
- Answer: it couldn't.
Solution: Note that the number of nuts for each pair of children is divisible by 3. This means that the total number of nuts must be divisible by 3. However, 2011 is not divisible by 3.
- Solution:
- Answer: A is a knight, B is a liar.
Solution: If A is a liar, then his statement is false, i.e. both must be knights. Contradiction. So A is a knight. Then his statement is true and B is a liar.
Grade 8 (solutions and answers)
Answers and solutions to problemsmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
- Answer: 198.
Solution: Three-digit numbercan be written as
... From the condition
follows that
... On the right is a two-digit (single-digit, if c = 0) number, which is divisible by 89, which means
... But then
- Answer: part of a circle with a diameter OP
Solution: Let O - the center of this circle, M - the midpoint of the chord cut off from the circle of the straight line passing through the point P. Then PMO = 90 o ... Therefore, the desired set is a part of a circle with a diameter OP lying inside the given circle.
Solution: The condition implies the equality of the triangles), where
... Besides,
... Therefore the triangles
are equal, and therefore
.
- Answer: 31 11 14
Solution:
- Answer: 1006 knights
Solution: Note that the two warriors standing next to each other could not be knights. Indeed, if they were both knights, they would both be lying. Select the warrior on the left and divide the row of the remaining 2010 warriors into 1005 groups of two warriors standing side by side. Each such group contains no more than one knight, i.e. among the considered 2010 warriors no more than 1005 knights, i.e. in total in a line no more than 1005 + 1 = 1006 knights.
Consider the line of RRLRL ... RRLRL. There are exactly 1006 knights in such a line.
9 class (solutions and answers)
Answers and solutions to problemsmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
- Answer: more than those with numbers in descending order.
Solution: 1) Let's write the number of the first group in reverse order. We will get the number of the second group, and different numbers of the second group are obtained from different numbers of the first group. At the same time, the numbers of the second group ending in 0, for example 98 760, could not be obtained by a “coup” from the numbers of the first group (the number 06789 = 6789 is not five-digit). This means that there are more numbers in the second group.
2)
The numbers of the first group are obtained from the number 123 456 789 by crossing out four digits, i.e. their, and the numbers of the second group - from the number 9 876 543 210 by deleting five digits, i.e. their
.
Grade 10 (solutions and answers)
Answers and solutions to problemsmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
Expressing and from equations (1) and (3) and substituting into equation (2), we obtain, after simplification, the equation... Solving it, we will find.
- Answer:,, where. Solution: By condition and one of the equalities holds:, or ... In the first case, having solved the system,, we get ... In the second case, we get or , ... The third case is similar to the second.
- Answer: segment without its ends, where is the point lies on the beam and.
Solution: Let - a circle passing through the points and and intersecting at the point ... Then, according to the property of inscribed angles, therefore the points,,, lie on the same circle; iflies on the segment, then if lies outside this segment (pointon the image). Thus, since both , i.e. circle passing through the points and ... So, we have shown that the pointshould lie on a segment... Let us now show that any point of this segment, except and , is included in the required locus of points. Indeed, let... Then, choosing the point so that, we get that and.
Grade 11 (solutions and answers)
Answers and solutions to problemsmunicipal stageAll-Russian Olympiad for schoolchildren mathematics
Let's consider the first case. Because, then the branches of the parabola given by the formulapointing up. And since, then there are points of the parabola lying below the axis... Hence, the parabola crosses the axisat 2 points. Therefore the equationhas two valid roots.
In the second case, the branches of the parabola are directed downward, and, so the parabola crosses the axisat 2 points. Then the equationagain has two real roots.
Method 2. Consider the inequality... Expanding the brackets on the left, multiplying the inequality by -4, then add to both sides of the inequality, we get: ... We transform this inequality to the form:... Since then ... Therefore the equationhas 2 real roots.
Solution: Obvious solutions,
,
... It is clear that other triplets of numbers with zero components are not solutions of this system. It remains to consider the case when... Then obviously- corners of a right-angled triangle with legs (- natural). Therefore, the triple- one more solution.
4. Answer: It is impossible.
Solution: Let the arrows be placed somehow. Project all the resulting vectors onto the line containing the height SO pyramids. The projections of vectors lying in the plane of the base are equal, and the projections of the vectors lying on the lateral edges are equal to or - ... Since the number of vectors lying on the lateral edges is odd, it follows that the sum of their projections cannot equal, therefore cannot equaland the sum of all the vectors obtained.
5 ... Let us number all the students in the class using natural numbers from 1 to 20 and denote bynumber of mutual friends and th students, and the sum of all such numbers across ... Then, to prove the statement of the problem, it suffices to show that for some and the inequality holds.
Total numbers will be ... Since each student has at least 10 friends in the class, when counting the numbereach student we take into account at least times therefore.
Thus, the sum of 1140 integers is at least 2400, so one of the numbersat least 3, as required.
