LXIII Moscow Mathematical Olympiad

Conditions and objectives

Grade 8 | Grade 9 | Grade 10 | Grade 11

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List of Problems

Grade 8

1. two different numbers x and y (not necessarily integer) such that x 2 -2000x = y 2 -2000y. Find the sum of x and y. ( Zlobin ) Solution

2. In the elections to the 100-seat parliament attended by 12 parties. In the parliament held the party in favor of which is strictly greater than 5% of the voters. Between the parties in parliament seats are assigned in proportion to the number of points they vote (that is, if one of the parties gathered in x times more votes than the other, then the seats it will get in the x-fold increase). After the election, it was found that each elector voted for exactly one of the parties (the invalid ballots, votes "against all," and so on were not), and each party received a whole number of seats. The Party lovers of mathematics gained 25% of votes. What is the largest number of seats it could get? (Response to explain.) ( I. Yashchenko ) Solution

3. trapeze equal lengths m and n cm cm (m and n - natural numbers, m is not equal to n). Prove that a trapezoid can be cut into equal triangles. ( Shapovalov ) Solution

4. In the triangle ABC is equal to the length of the median length of the BM side AC. On the extensions of sides BA and AC chosen points D and E, respectively, so that the equalities AD = AB and CE = CM (see Fig.).Prove that DM and BE perpendicular. ( R. Zhenodarov ) solution 

to the problem of Figure 4 (8 cl.)

5. of the cards in the deck is "face up". From time to time Peter takes out a deck stack of one or more consecutive cards, in which the top and bottom cards are "face up", turns the whole pack as a whole and puts it in the same place the deck. Prove that in the end all the cards will fall "face down", no matter how acted Peter. ( Shapovalov ) Solution

6. What is the largest number of horses can be arranged on the board 5 * 5 cells, so that each of them exactly beat the other two? (Give an example and explain why you can not assign more horses.) ( M. Gorelov )Solution

Grade 9

1. Solve the equation

(X +1) 63 + (x +1) 62 (x-1) + (x +1) 61 (x-1) 2 + ... + (x-1) 63 = 0.

( R. Smith ) Solution

2. In line written out 23 natural numbers (not necessarily distinct). Prove that between them so you can add parentheses, signs of addition and multiplication, the value of the expression will be evenly divisible by 2000. (Shestakov ) Solution

3. Given a circle and a point A inside it. Find the set of vertices C of various rectangles ABCD, where the points B and D lie on a circle. ( M. Panov ) Solution

4. Gregory wrote in to chess numbers 1, 2, 3, ..., 63, 64 in some order. He said Loesche only sum in each rectangle of the two cells, and added that 1 and 64 are on the same diagonal. Prove that this information Lesch can accurately determine which cell number which is recorded. ( Shapovalov ) Solution

5. ABCD - a convex quadrilateral. Circles constructed on the segments AB and CD as diameters relate externally at the point M, is different from the point of intersection of the

diagonals of the quadrilateral. Circle passing through the points A, M, and C, again crosses the line joining the point M and the middle of AB, the point K, and the circle passing through the points B, M and D, again crosses the same line at the point L. Prove that | MK-ML | = | AB-CD |. ( Sharygin ) Solution

6. fortification system consists of the dugouts. Some of the bunkers are connected trenches, and from any shelter can run across in any other. In one of the hidden bunkers infantryman. The gun can be covered with one shot every dugout. Each interval between shots infantryman always crosses one of the trenches to a nearby shelter (even if on a nearby dugout just shoot guns, infantry can run across the back). A system reliable if the gun is not a guaranteed strategy defeat infantry (ie, a sequence of shots, thanks to which the gun hit infantry regardless of its initial location and subsequent movement.) Figure 6 to the problem (9 cl.) a) Show that fortification system, as shown in the figure, is reliable. b) Find all the reliable system of fortifications, which are no longer reliable after the destruction of any of the trenches. ( Shapovalov, A. Spivak ) Solution

Grade 10

1. Points A and B are taken on the graph of y = 1 / x, x> 0. Are omitted perpendiculars to the x-axis, the surface normal - H A and H B ; O - origin. Prove that the area of the figure bounded by the lines OA, OB and the arc AB, equals the area of the figure bounded by straight AH A , BH B , the x-axis and the arc AB. ( R. Anno, Kirichenko ) Solution

2. Suppose that f (x) = x 2 +12 x +30. Solve the equation

f (f (f (f (f (x))))) = 0.

( M. Evdokimov ) Solution

3. On paper "in the box" is drawn in a convex polygon, so that all its vertices are the vertices of the cells and none of the parties is not vertically or horizontally. Prove that the sum of the lengths of the vertical lines of the grid lines, enclosed within a polygon is the sum of the lengths of the horizontal segments of the grid lines inside the polygon. ( Galperin ) Solution

4. ABCD - a convex quadrilateral. Circles constructed on the segments AB and CD as diameters relate externally at the point M, is different from the point of intersection of the

diagonals of the quadrilateral. Circle passing through the points A, M, and C, again crosses the line joining the point M and the middle of AB, the point K, and the circle passing through the points B, M and D, again crosses the same line at the point L. Prove that | MK-ML | = | AB-CD |. ( Sharygin ) Solution

5. A sequence x 1 , x 2 , ..., x n , ... denoted {x n }. From the available sequence {B n } and {C n } (possibly, B { n } coincides with C { n }) are allowed to receive a sequence {B n + C n }, {B n -C n }, {B n * C n } and {B n / C n } (if all C n different from 0). In addition, any of the available sequences can get a new, deleting some early members. First, there is only a sequence { n }. Can I get out of it the above operation sequence {n}, ie, 1, 2, 3, 4, ... if a) a n = n 2 , b) a n = 2 n 1/2 ; a) a n = (n 2000 1) / n? ( Shapovalov, V. Dotsenko ) Solution

6. taken out from the pack of 7 cards, showed everyone shuffled and distributed Lesha Grisha and 3 cards and the remaining cards a) hid; b) gave Kohl. Gregory and Alex can take turns to report aloud any information about their cards. Can they tell each other their cards so that while Kohl was unable to calculate the location of any of the cards, which he has not seen?(Gregory and Alex did not agree on any specific method of communication, and all negotiations are taking place in plain text .) ( Shapovalov ) Solution

Grade 11

1. the greatest common divisor (GCD) of positive integers m and n is 1. What is the largest possible value of the GCD of numbers m +2000 n and n +2000 m? ( Zlobin ) Solution

2. Calculate

(| Sin (1999x) | - | sin (2000x) |) dx.

0( Folklore ) Solution

3. chords AC and BD of the circle with center O intersect at point K. Let M and N - center of the circle circumscribed about the triangles AKB and CKD, respectively. Prove that OM = KN. ( A. Zaslavsky ) Solution

4. Fedya has three sticks. If one can not make a triangle, Fyodor shortens the longest of the sticks on the sum of the lengths of the other two. If the length wand to vanish again and the triangle can not be folded, Fedor repeats the operation, etc. Can this process continue indefinitely? ( Shapovalov ) Solution

5. A circular chess tournament, each participant played with each other once. We call a lot wrong when it won the player eventually scored fewer points than the loser. (Victory is worth 1 point, draw - 1/2, defeat - 0.) Could be wrong party 

a) more than 75% of the total number of games in the tournament, b) more than 70%? ( Tokarev ) Solution

6. Can I place an infinite number of equal convex polyhedra in a layer bounded by two parallel planes, so that no polyhedron can not be removed from the bed without moving the rest of us? ( A. Kanel-Belov ) Solution

Solving problems

Grade 8

1. This in equality rewrite as follows:

2000 (xy) = (xy) (x + y).

Since x is not equal to y, can be reduced by xy. Hence x + y = 2000.

2. A: 50 seats. If the 10-party rack up exactly 5% of the vote, and two, including the opposition fans Mathematics (PLA), - 25%, the representatives of the PLA will receive exactly 50 seats in parliament. We prove that a greater number of places to get the PLA can not. Indeed, the number of seats won by the PLA in the parliament, as well

100 * ( number of votes cast for PLA ) / ( number of votes received by all the parties in parliament ) = 100 * ( the number of votes received by the PLA ) / (( total number of voters ) - ( number of votes cast for all parties, not in parliament ))

This shows that the largest number of seats in the PLA will receive if the total number of votes cast for the party Not Last, maximum. If the Parliament does not last 11 games, they also would not have scored more than 55% of the vote, but 55% 25% <100%. So not passed in parliament more than 10 games, and they scored a total of no more than 50% of the vote. Therefore, the PLA will receive in parliament to 50 seats.

3. Suppose m> n. Trapezoid sides continue to intersect and divide each side of a triangle for m equal parts. Through the points of division draw lines parallel to the sides of the triangle (see Fig. 8.1), and by the theorem of Thales obtain a partition of the triangle into equal small triangles. Upper base of the trapezoid is one of the lines held, since its length is an integer centimeters. Thus, we obtain a partition of the trapezoid into equal triangles.Fig. 8.1

4. Suppose K - point symmetric with respect to point A. M Isosceles triangle KMB (KM = MB), EMB also isosceles triangle. Denoted by  value of the angle at the base of the triangle KMB, a  - the angle at the base of the triangle EMB. Since the sum of the angles of any triangle is 180 o , then / KBE =  +  = 90 o (see Fig. 8.2), ie, KB   |    BE. Note that the quadrilateral DKBM - parallelogram, ie, DM | | KB. Therefore, DM   |    BE.Fig. 8.2

5. Located cards in the deck is comparable number in which the numbers as much as a pack of cards, and on the k-th place on the right is "1" if the k-th card from the bottom is face down, and "2" otherwise. Then after each conversion is the number decreases. (Indeed, we compare that number with the previous. Among all the numbers that have changed, we choose the leftmost, ie find the most senior level changed. Obviously, in this bit "2" changed to "1".) As the number of n- digit numbers of ones and twos of course (no more than 2 n ), in the end we get a number consisting of one unit that corresponds to the location of all the cards face down.

6. Fig. 8.3 shows the location of 16 horses, satisfying task. We show that a greater number of horses to place impossible. Color the squares of the board, as shown in Fig. 8.3. Note that the number of horses, located on the dark squares, equal to the number of horses, located on the white squares. So, if the number of empty white cells is n, the number of empty black cell has n +1. Fig. 8.3 In any optimal location horses central square is empty.Otherwise, out of eight cells, which has a horse standing on a central field, exactly six empty white. Hence n > 6, and the number of horses not exceed 25-n-(n 1) < 12. white cells divide into four groups, as shown in Fig.8.4 (one group of cells marked with the same numbers). We show that the optimal location of at least one cell of each group is empty (it will follow that n > 4). Assume the contrary, for example, that all cells of group 3 are horses. Denote their letters a, b and c (Fig. 8.5). Horse standing on a cell a, beats cell f, d and center. But, as shown above, the central cell is empty, it means that the cells of the f and d are the horses. Similarly, we can show that on e and g are also horses. But then the horse is standing on a cell c, has four horses, located at d, e, f and g, which is impossible. 

Fig. 8.4 Fig. 8.5

So, n > 4. Hence, the number of horses up to 25-n-(n 1) < 16.

Grade 9

1. A: x = 0. Multiply both sides by (x +1) - (x-1) = 2. Then it takes the form

(X +1) 64 - (x-1) 64 = 0

and is easily solved:

(X +1) 64 = (x-1) 64 , | one x | = | x-1 |,

since x +1 is not equal to x-1, we get x +1 = - (x-1), where x = 0.

2. Divide the number of data 23 into eight groups of standing in a row of numbers, three groups of five numbers and four groups of two numbers (the order in which these groups are - no matter). Each group is in parentheses, and between groups arrange the multiplication

sign. If you place the signs in each group so that the results of operations in a group of two numbers divided by 2, and in the group of five numbers - 5, then the entire expression will be divided into 2 4 5 * 3 = 2000. claim that such an alignment marks Group exists. If the numbers in a group of two numbers with different parity, then you need to put in between the multiplication sign, if the same parity - addition. The result is likely to be divided by 2. Consider a group of numbers, a 1 , a 2 , a 3 , a 4 , a 5 , going in that order. We write out the remnants of the division by five five amounts:

a 1 , a 1 + a 2 , a 1 + a 2 + a 3 , a 1 + a 2 + a 3 + a 4 , a 1 + a 2 + a 3 + a 4 + a 5 .

If one of the residues is 0, the corresponding sum is divided by 5. In this case, you must add addition signs between the numbers included in this amount, the principal of (if required) to bracket, and any remaining gaps between the numbers of characters to fill in the multiplication. If none of the residue is non-zero, then by the pigeonhole principle, among them there are two identical residue. Suppose, for example, the amount of a 1 + ... + aI and a 1 + ... + a j (I <j) give the same remainder when divided by 5. Their difference, which is the sum of numbers in a row facing a one I + ... + a j , divided by five, and again we mark out the addition, we conclude that amount in brackets, and the remaining positions are filled with signs of multiplication. Thus, in any case, we will be able to place the signs in a group of five numbers so that the result was divided by 5.

3. prove that point, the equality C OC 2 = 2R 2 -OA 2 , where the O - center of the circle, R - the radius. We introduce the notation, as shown in Figure 9.1. OC segment is the hypotenuse of a right triangle with legs x and y (see Fig. 9.1), by the Pythagorean theorem, we obtain

OC 2 = x 2 + y 2 = (x 2 + t 2 ) + (y 2 + z 2 ) - (z 2 + t 2 ).

Observing that x 2 + t 2 = y 2 + z 2 = R 2 , z 2 + t 2 = OA 2 . It follows that the required locus is a circle with center O and radius (2R 2 -OA 2 ) 1/2 . Fig. 9.1

4. Suppose that there is an adequate supply of Leschi dominoes $ 1 \ times 2 $. If he puts on the board two dominoes so that they overlap exactly one cell, then subtracting from the sum

of the numbers under one domino sum under the other, he knows the difference between the numbers, covered by exactly one domino (these numbers are written in the cells of the same color). The third domino to put so that one half of it was setting one of the two numbers, and the second - is not covered by the cell. Adding previously obtained difference with the amount of numbers under the new dominoes, Alex will receive the sum of those two numbers, which are covered by exactly one of the three dominoes (these numbers are written in the cells of a different color). So, adding more and more dominoes can construct a chain connecting any two squares of the board, and if the cells of the same color, to know the difference, and if different - their sum. It is known that the numbers 1 and 64 are located on the same diagonal, ie, the cells of the same color. Their difference is equal to 63, and the difference between any two other integers between 1 and 64 is less than 63. So Alex could identify which cells recorded 1 and 64. He just does not know which of these cells which number is recorded. But he knows the sum of the numbers in each of these two cells with any number recorded in a cell of a different color. For the cell with the number 1 all of these amounts do not exceed 64, and for the cell with the number of 64 - at least 66. So Alex can identify exactly where the recorded 64. Now, knowing the sum (or difference) of the number 64 to the number in any other cell, Alex easily determine where what number is recorded.

5. Let P and Q - mid AB and CD, O 1 and O 2 - centers of circles passing through the points A, M, C and B, M, D, respectively, H 1 and H 2 - projection O 1 and O 2 on line PQ (Figure 9.2). 1 o . Point of M, P and Q are collinear. In fact, the direct PM and QM contain the radii of circles tangent at the point M, and, therefore, perpendicular to the total internal tangent to these circles. 2 o . P and Q lie on a circle with a diameter of O 1 O 2 . Indeed, PO 1   |    PO 2 , as these lines - the perpendicular segments MB and MA, the angle between them direct (M lies on the circle with diameter AB). Similarly, QO 1   |    QO 2 . 3 o . Clearly, KH 1 = H 1 M, LH 2 = H 2 M (diameter perpendicular to a chord, bisects it.) Fig. 9.2 4 o . PH 1 = QH 2 , since the projection of the midpoint of O 1 O 2 divides the segment H 1 H 2 in half, but this projection and bisects the segment PQ. Finally, | MK-ML | = 2 | MH 1 -MH 2 | = 2 | MP-MQ | = 2 ((1/2) AB-(1/2) CD | = | AB-CD |. 

6. a) We prove that the condition specified in the fortification system (call it the trefoil) reliable. Denote shelters, as shown in Figure 9.3. Limit the initial positions of infantry bunkers O, A 2 , B 2 and C 2 (these shelters will be called even, and the others - odd). We believe that an infantryman so lucky that saved if at all possible. Fig. 9.3 Note that the shot gun on one of the odd bunkers certainly useless. If the gun is shot by dugout O (lucky infantryman certainly not there), then the remaining three even dugouts infantryman can run across in any of the six odd. Now, if the gun is shot by one of these odd bunkers, the grunt of the other five will be able to run across any of the four even. Thus, repeating the initial placement. 

Struck dugout

Infantryman

O any shelter

B 1 , B 3 , C 1 , C 3

O, B 2 , C 2

B 2 A 1 , B 1 , C 1 , C 3

(for C 2 - the same) A 1

(for B 1 - the same)O, B 2 , C 2

B 2

(for C 2 - the same) C 1 or C 3

A 1 , B 1 , C 1 , C 3

O, A 2 , B 2 , C 2

If his first shot gun will cover one of the bunkers on the branches even shamrock, say A 2 (first bunkers A 2 , B 2 and C 2 nothing of each other do not differ), the infantry of the remaining bunkers O, B 2 and C 2 can to defect in any of the bunkers A 1 , B 1 , B 3 , C 1 , C 3 . If after this gun fires not dugout A 1 , the grunt available any of the four even vertices. Again, the initial situation. If the second shot gun produced by A 1 , then the remaining four dugouts infantryman able to cross into the dugouts O, B 2 and C 2 . Subsequent bust in the table. In the left column it - shelters, which fires a gun in the right - shelters, where then can run across infantryman.The horizontal lines separate the various options. Some cases are not analyzed (about them says "the same"). This means that the exact same situation up to a permutation of the branches. Each case is handled to the repetition of the situation that is already dated. So, after each shot infantryman can be more than one dugout. So every time he has a choice, by which he will be saved. Therefore, the system is reliable, shamrock. b) Answer: these (minimal) reliable systems strengthening is a shamrock and all systems consisting of only one cycle dugouts A 1 -A 2 - ...-A n -A 1 . If the system has a cycle several bunkers A 1 -A 2 - ...-A n A- 1 , then the system of fortifications secure. After all, an infantryman, running only on this cycle, every time can choose one of two available to him dugouts, which will not be covered in the following shot. Note that if any part of the fortification itself is reliable, the infantryman can be saved only in this part Thus the whole system is reliable. Therefore, if in addition to the loop A 1 -A 2 - ...-A n A- 1 has other trench, then the system is not minimal. We show that any cycle is minimal. Destroying, say, trench A n A- 1 , we obtain a linear maze of A 1 -A 2 - ...-A n . Here's how to operate a gun. It consistently shoots dugout A 2 , A 3 , ..., A n . If the first was an infantryman in the dugout with an even number, then one of these shots it will cover. If it does not, the gun produces another series of shots from the dugout A 1 or A 2 and consistently moving in increasing numbers dugout. Then, from what shelter she begins a second series, depending on the parity of numbers dugout, which at this moment is Footman (this is easy to calculate, as the first room was odd and after each perebeganiya parity number varies). remains to consider the system of fortifications without cycles. We show that only minimal reliable among them - it is the shamrock. Take any system that does not contain any cycle, or trefoil, and we show how to operate a gun. (We describe a strategy for the communication system. If the system is connected, ie, consists of several parts, unconnected trenches, the gun must consistently implement this strategy for each site.) Call dugout intersection, if there go three or a trench. Trench leading from the dugout, called pass-through, if, after running through it, an infantryman can run across two more times, not having visited twice in any dugout. For example, in trench A shamrock 1 -A 2 , leading from the dugout A 1 , not sequentially, and trench A 2 -A 1 from A 2 -through. Finally, the shelter called deadlock, if there is only one trench. Since our system does not contain a shamrock from any shelter goes no more than two cross trenches. Determine where the gun will attack. Take any intersection. If out of it are two cross trenches, each leading to a different intersection, then we choose one of them and shall proceed through it to the nearest intersection. If this new crossing out another pass-through trench leading to the intersection,

turn on the trench up to the next nearest intersection. By doing so, until we come to a crossroads with a single coming out of it through the trench, or two, in one of which can be reached deadlock dugout without going through any intersection. In the first case will go to any non-through to a nearby trench dug in the second - go through this self ' through my trench dug up adjacent to a dead end. Thus, we determine when the shelter opened fire. We split the bunkers on the even and odd, so that every time the infantry rush from one dugout to the dugout another parity parity (this is possible because the system has no cycles). We will show how to shoot, that is guaranteed to hit infantry, provided that it is initially in the dugout with the same parity as the shelter from which to start shooting. (If making all these shots, the gun will be covered and not grunt, then he was in the other dugout parity now exactly known parity dugout with the infantryman.) Let the gun bunkers consistently hit from the selected and ending intersection. Then on the linear portion of shelled infantryman not. We have chosen the initial shelter to tranches of leading to other parts of the system, it was not more than one through. (If only two of them, the second is to just shelled area.) Any trench is not through either deadlock dugout or in the dugout, from which you can get in a few dead-end or back to the intersection. In both cases, only one of the trench dug parity opposite considered intersection. Once the gun has covered the intersection, ran as an infantryman in the dugout again, parity, which is opposite to the parity of the intersection, so that if it is out of the trench, then hitting the dugout, which does this trench gun hit infantryman. If not, the gun has a crossroad again, not giving infantryman run on land already proven system of fortifications. Checking all non-through trench gun starts only through, striking the dugout, which does this trench, and then successively all the bunkers to the nearest intersection. There is repeated non-through inspection of passes, etc. Thus you can check the entire system. Thus, any system that does not contain any cycles, or trefoil, is unreliable. Destroying any trench shamrock, we get an insecure system, so that the shamrock is the minimum system reliability. Finally, any system consisting of a trefoil and anything else deemed reliable, but not minimal.

Grade 10

1. can assume that the abscissa of A is less than the abscissa point B (Fig. 10.1). Consider the point of intersection of K segments AH A and OB. Then the difference of the area under consideration is equal to the difference between the area of a triangle and quadrilateral OAK H A KBH B , which, in turn, is equal to the difference between the areas of the triangles OAH A and OBH B . Since OH A * AH A = OH B * BH B = 1 (recall that A and B lie on the graph), these areas are equal. Fig. 1.10

2. Note that

f (x) = (x +6) 2 -6.

This shows that

f (f (f (f (f (x))))) = (x +6) 32 -6.

It remains to write the answer: x = -6 + 6 1/32 .

3. prove that each of the quantities considered equal to the area of a polygon. Check it for the sum of the lengths of the horizontal segments. We draw these segments. Then the polygon is partitioned into two triangles and trapezoids few, and the height of these figures will be equal to 1. It remains to express the area of these figures in the base and the height of a known formula and add. We see that each horizontal segment will amount to two times, that is, by a factor of one, as required.

4. Refer to the solution of five options in grade 9 .

5. A: a), c) can be, and b) can not. explain how you can act in a) and c). Obviously allowed operations can be derived from a sequence { n } {a sequence of n 1 -a n $. Such a transformation is denoted by T, and m-fold application of the transformation T is denoted by T m . Note that if P (n) - a polynomial of degree n m-1, the use of T to the sequence {P (n)} gives a sequence {Q (n)}, where Q (n) - a polynomial of degree m-2. Hence, in particular, that the use of T m to {P (n)} has zero sequence. More useful to us the sequence I, all of whose members - one (it can be obtained by dividing the sequence itself). Now it is easy to show the sequence of operations for each of the items:

{N 2 }T

{2n +1}-I

{2n}

/ (I + I)

{N}a) ;

c) {

n 2000 1

}

T 2000

{

2000!

}

I /

{

n (n +1) * ... * (n +2000)

}

* (2000! * I)

T 2000

{N (n +1) * ... * (n +2000)}

{An + b}

,

nn (n +1) * ... * (n +2000)

2000!

where a and b - integers, a is 0. Further actions are clear. b) We prove that the sequence {n} can not be obtained. For this we note that all the sequences that can be obtained from {n +2 1/2 }, have the form {P (n +2 1/2 ) / Q (n +2 1/2 )}, where P and Q - polynomials with integer coefficients. (In fact, the original sequence of this type has. Termwise In addition, subtraction, multiplication or division of such sequences, obviously, again in an order of this kind, and the ejection of several members is equivalent to replacing P (x) / Q (x) to P ( x + r) / Q (x + r) for some integer r, which can be represented in the desired form, revealing all the brackets in the numerator and the denominator.) If this form is a sequence {n}, then this form is submitted and the sequence all the members of which are 2 1/2 . But from the equation P (n

+2 1/2 ) / Q (n +2 1/2 ) = 2 1/2 that the ratio of the leading coefficients of the polynomials P and Q is 2 1/2 , which is impossible. Contradiction.

6. a) Let Gregory says, "I called his or {map} or {names three cards, which he does not}." After that, Alex has to say: "I have called his or {map} or {names three cards Grisha if the second set, named Grisha, does not match the set, and any other three cards, which he does not otherwise} ". After that, each of them, apparently, knows all the cards. Kohl's nothing to be seen. Indeed, named three sets of cards: A, B and C. Sets B and C intersect in two cards, Gregory said: "I have either A, or B", Alex said: "I have either A, or C". This means that either Grisha set A, and Lesha - C, or Grisha - B, and Lesha - A. Of course, these are different decomposed, and even down card can not be determined. b) Note that the previous method does not work: knowing the hole cards, Kohl's can all identify. We index cards numbered from 0 to 6. Let Gregory and Alex in turn call the remains of dividing the numbers of their cards at 7. Then they find out the balance of each one of them should only add to his total amount of the other and find the remainder, the opposite of the total modulo 7 (that is, one which when added to the total of number divisible by 7). This will be the number of closed card. After this alignment of recovery is easy. Check that Nick did not recognize him. Consider the map with the number s. We show that it could be brought to Grisha if he called the amount a. For this it is necessary to supplement the map with the amount of the other two rooms as. It is easy to see that there are three different pairs of numbers whose sum is as. Two of them may have corrupted that there is a card number or s hole card, but at least one pair remains. And add to it we set Grisha. The same argument shows that any map could be and Leschi.

Grade 11

1. A: 2000 2 1. Let a = 2000m + n, b = 2000n + m, d - the greatest common divisor of a and b. Then d also divides the number

2000a-b = (2000 2 -1) m and 2000b-a = (2000 2 -1) n.

Since m and n are relatively prime, then d divides 2000 2 1. On the other hand, when m = 2 000 2 -2000-1, n = 1, we obtain a = (2000 2 -1) (2000-1), b = 2 000 2 -1 = d.

2. A: 0. The graph of | sin (kx) | on the interval [0;  ] consists of k identical "caps", which are obtained from the graph of sin x on the same segment by compression to the ordinate in k times. The area under the graph is also decreasing in k times. As a result, the area under the k "cap" is the same for all k.

3. Suppose X - Mid KB. Then / KMX = (1/2) / KMB = / = KAB / KDC. Since MX   |    BD, then KM   |    CD. Since in this case ON   |    CD, ON | | KM. Similarly, OM | | KN. Fig. 11.1 If the points O, K, M, N lie on a straight line, then OMKN - parallelogram and OM = KN. Otherwise we consider the orthogonal projection of the segment OM and KN on AC. Since the points O, M, N, respectively, in the middle of the projected segments AC, AK, KC, then the projections of the two parallel lines are equal to (1/2) KC. Therefore equal length and the segments themselves. 

4. A: can. The polynomial P (x) = x 3 -x 2 -x-1 has a root t, greater than 1, since P (1) <0, and $ P (2)> 0. Then t 3 = t 2 + t +1> t 2 + t. Take a long stick to t 3 , t 2 , t. After the first sawing obtain sticks with lengths t 2 , t, 1. Since the ratio of the lengths is not changed, the process will continue indefinitely.

5. A: a) can not, and b) they can. a) Let N - the number of players, M = [N / 2]. Players taking the first M locations, called the strong, and the rest - weak (between members of the same amount of points the distribution of seats is arbitrary). Let X - number of regular games between strong and weak. The amount of points scored strong in meetings among themselves, is M (M-1) / 2, and in meetings with weak - less than X. Therefore, the average result of the strong no more ((M-1) / 2) + (X / M). Similarly, the average of at least weak ((NM-1) / 2) + ((M (NM)-X) / (NM)). If there is a wrong party, that not all players scored equally, and the average result of the strong than weak. Hence X> M (NM) / 2 > N (N-1) / 8. Since the total number of lots is equal N (N-1) / 2, the percentage of correct parties is greater than 1/4. b) Take the first 2k +1 tournament player, in which each participant with the number I < k lost participants numbered i +1, ..., i + k, and beat the others, and each participant with a number i> k beat participants with numbers ik, ..., i-1 and lost the rest. Obviously, all players scored k points, and in the tournament table above the main diagonal ones are only k (k +1) / 2 cells from 2k (2k +1) / 2. Now "be fruitful" every player, replacing his unit of the new n, and let participants from different units play with each other as well as the corresponding former members, and the members of a single block play with each other in a draw. Obtain a new table that is still all players equally points. Correct this table so that the amount of points the players are no longer equal. To do this, we will change the results of the participants of the block k +1: in their meetings against members of the bloc k +1- i replace i * n wins draws, so that the total score of each member bloc k +1 decrease, and each party bloc k +1 -i will increase by i / 2. On the contrary, the party with the players block k +1 + i replace draws

i * n losses. Number of improper parties will be n 2 (2k (2k +1) / 2)-n 2 (k (k +1) / 2)-2n (k (k +1) / 2). The total number of lots is equal n (2k +1) (n (2k +1) -1) / 2. When n = k = 20 Party is wrong 235.600 / 335.790> 0.7, the total number of parties.

6. between the planes will place regular tetrahedra, the distance between the opposite edges of which is equal to the distance between the planes. Let one of the edges of each tetrahedron lies in one of the boundary planes, as opposed to it - in another. Two tetrahedra can be arranged so that the end of the "top" edge first coincides with the middle of the "top" edge of the second, and the middle of the "lower" edge of the first - with the end of the "lower" edges of the second, and in this case as the "upper" and "lower "edges are perpendicular to the two tetrahedra. By extending this process to the entire layer, we find that each tetrahedron is surrounded by four others (Fig. 11.2), two of which do not allow him to push "up", the other two - "down." Fig. 2.11

Last Updated: January 31, 2001

   

LXII Moscow Mathematical Olympiad

Conditions and objectives

Grade 8 | Grade 9 | Grade 10 | Grade 11

Archive (zip) of this page with pictures (23Kb) is here .

List of Problems

Grade 8

1. comparing fractions 111110/111111, 222221/222223, 333331/333334, arrange them in ascending order. Solution

2. Show how any quadrilateral cut into three trapezoid (parallelogram can also be considered a trapezoid.) Solution

3. Locate any four distinct positive integers a, b, c, d, for which a number of 2 +2 cd + b 2 and c 2 +2 ab + d 2 are perfect squares. Solution

4. Petin bank account contains $ 500. Bank authorizes the operation of only two types: $ 300 to remove or add $ 198. What is the most Peter can withdraw if the other he has no money? Solution

5. In a right triangle ABC point O - the middle of the hypotenuse AC. On the segment AB to take a point M, and on the segment BC - Point N so that the angle MON - straight. Prove that AM 2 + CN 2 = MN 2 . Solution

6. chess tournament, each participant played every two parties: one with white pieces, the other - black. At the end of the tournament was that all the participants scored the same number of points (for a win is worth 1 point for a draw - 1/2 points for a loss - 0 points.) Prove that there are two players who have won the same number of games white. Solution

Grade 9

1. on the board in the laboratory written two numbers. Every day, a senior fellow Peter erases the board, both in the number and writes instead of the arithmetic mean and the harmonic mean. On the morning of the first day on the board were written the numbers 1 and 2. Find the product of the numbers written on the board in the evening of the day of 1999. (Arithmetic mean of two numbers a and b is the number of (a + b) / 2, and the harmonic mean - number 2 / ((1 / a) + (a / b))).Solution

2. two play the following game: the first writes to a number at will the letter A or B (left to right, one at a time, one letter per turn), and the second after each turn first to swap any two of the letters written out or nothing changes (this is also considered a move). After both players to 1999 turns, the game ends. Could the second game, so that in any action the first player in the result is a palindrome (ie, a word that reads the same from left to right and right to left)? Solution

3. diagonals of a parallelogram ABCD intersect at O. The circle passing through the points A, O, B, tangent to BC. Prove that the circle passing through the points B, O, C, is tangent to CD. Solution

4. Determine all positive integers such k, the number 1 ... 12 ... 2.2 ... 2 is a perfect square. (in the first term (reduced) only 2000 figures, of which the last place are the figures " 2 "in the amount of k units, and the remaining -" 1 ", the second term (the subtrahend) of 1001 standing order numbers "2") Solution

5. incircle of the triangle ABC (AB> BC) touches the sides AB and AC at the points P and Q respectively, RS - average line parallel to AB, T - intersection point of lines PQ and RS. Prove that T lies on the bisector of the angle of the triangle B . Solution

6. competition in n-boryu involved two n people. For every athlete known for his strength in each of the types of programs. Competitions are as follows: First, all athletes participating in the first appearance of the program and the better half of them go to the next round. This half is involved in the following form and half of them go to the next round, and ~ t \, etc., while in the $ n $-th kind of program will not be a winner. `` We call athlete possible winner'', so to place sports in the program, he will be the winner. a), prove that it can happen that at least half of the athletes is "possible winners", b) prove that it is always the number of "potential winners" at most 2 n -n; c) prove that it can happen that the "potential winners" exactly 2 n -n. Solution

Grade 10

1. know that (a + b + c) c <0. Prove that b 2 > 4ac. Solution

2. Two circles intersect at points P and Q. The third circle centered at P meets the first at points A, B, and the second - in the points C and D. Prove that the angles and AQD BQC equal. Solution 

3. Find all pairs of positive integers x, y, that the number of x 3 + y and y 3 + x divided by x 2 + y 2 . Solution

4. 2n radius circle divided into 2n equal sectors: n n blue and red. In the blue sector from some consecutive record counterclockwise numbers from 1 to n. In the red zone, from some recorded the same number and the same way, but in the clockwise direction. Prove that there is a semi-circle, which contains all the numbers from 1 to n. Solution

5. Grasshopper jumps on the interval [0,1]. In one leap he can get from point x to point or x / 3 1/2 , or at the point x / 3 1/2 + (1 - (1/3 1/2 )). On [0,1] is a point a. Prove that, starting from any point, grasshopper may be a few jumps at a distance of less than 1/100 of a point a. Solution

6. For the numbers 1, ..., 1999, arranged in a circle, calculated sum of the products of all sets of 10 numbers, consecutive. Find the balance of numbers in which the sum is the greatest. Solution

Grade 11

1. a, B, C - the triangle. Prove the inequality ((a 2 +2 bc) / (b 2 + c 2 )) + ((b 2 +2 ac) / (c 2 + a 2 )) + ((c 2 +2 ab) / (a 2 + b 2 ))> 3. Solution

2. A plane convex figure bounded by segments AB and AC and BC arc of a circle. Draw any line that divides in two: a) the perimeter of the figure, b) its area. Solution 

3. Facets of a regular octahedron are painted in white and black. However, any two faces with a common edge, painted in different colors. Prove that for any point within the octahedral sum of the distances to the planes of white faces is the sum of the distances to the planes of black faces. Solution

4. the meadow, having the shape of a square, a round hole. The meadow jumping grasshopper. Before each jump, he chooses the top and jump towards it. Length of the jump is equal to half the distance to the vertex. Whether grasshopper bunker? Solution

5. graph - a set of vertices, some of which are connected by edges (each edge connects exactly two vertices of the graph). Coloring of a graph is correct, if the vertices of the same color are connected by an edge. A graph correctly colored in k colors, and it can not be properly colored in fewer colors.

Prove that in this graph, there is a path along which there are k vertices of all colors exactly once. Solution

6. Solve in positive integers the equation (1 + n k ) L = 1 n + m , where l> 1. Solution

7. Prove that the first digits of the numbers of the form 2 2 n form a non-periodic sequence. Solution

From the last three tasks in grade 11 are taken into account two.

Solving problems

Grade 8

1. Consider the numbers 1-x = 1/111111, 1-y = 2/222223, 1-z = 3/333334, as well as their inverses 1 / (1-x) = 111111, 1 / (1-y) = 111111 1/2, 1 / (1-z) = 111 111 +1 / 3. We see that 1 / (1-x) <1 / (1-z) <1 / (1-y). As all the above are positive, 1-x> 1-z> 1-y. Therefore, x <z <y.

2. Suppose B - the largest interior angle of the quadrilateral ABCD. BM make a cut from the top of B, parallel sides AD (point M gets inside the quadrangle). From point M hold sections MN and MK, parallel to the sides BC and CD, respectively (Figure). 

3. Suppose that ab = cd. Then a 2 +2 cd + b 2 = a 2 +2 ab + b 2 = (a + b) 2 , c 2 +2 ab + d 2 = c 2 +2 cd + d 2 = (c + d) 2 . Thus, it is enough to find four distinct positive integers a, b, c and d, for which ab = cd. To do this, we find a number n, into a product of two factors in different ways. For example, this number is n = 6, in which case we can take a = 1, b = 6, c = 2, d = 3.

4. Since the 300 and 198 are divisible by 6, Peter can withdraw only the amount divisible by $ 6. Maximum multiple of 6 and not exceeding 500, - is 498.

We prove that the rent 498 dollars possible. Perform the following operations: 500-300 = 200, 200, 198 = 398, 398-300 = 98, 98, 198 = 296, 296, 198 = 494. Money you have in the bank decreased by $ 6.

Doing the same procedure 16 times, Peter would remove $ 96. Then he can take off 300, 198, and put again to remove 300. As a result it will be 498 dollars.

5. Consider a 180 o clockwise around the center O (see fig.) This will turn the triangle into a triangle ONC ON'A. Consider the quadrilateral MAN'O; therein corners MAN 'and MON' straight. In fact, / MAN '= / + BAC/ BCA, and / MON '= / MOA + / NOC = 180 o - / MON. Therefore, MN ' 2 = AM 2 + AN ' 2 = OM 2 + ON ' 2 . Next, ON '= ON, OM 2 + ON 2 = MN 2 . On the other hand, AN '= CN, - and the desired equality is proved. 

 

6. total in the tournament were played in n (n-1) parties, and each was played 1 point. Therefore, when the results of the equality of all participants scored n-1 point. Each player played a white n-1 party, and the number of won games they white n equal to one of the numbers 0, ..., n-1. Assume that the problem is not true: all the different number of games won white. Then realized all the options from 0 to n-1. Consider two players in the tournament: A, n-1 winning party white, and B, have not won a single party. Take a look at what could be the result of the game, which is played against A B black. On the one hand, A scored n-1 points, playing white, so that all of their games with Black, including this one, he had to lose. B but did not win a single game with white, so he could not win this one. Contradiction.

Grade 9

1. the product of numbers on the board does not change. Indeed, ((a + b) / 2) * (2 / ((1 / a) + (1 / b))) = ab. Therefore, the desired product is 2.

2. A: Yes. We present the strategy of the second player. The first 1000 moves he misses. Progress with the number n >> 1000 he does so:

1) if n-m (2000-n)-th place are identical figures - does nothing;

2) if in these places - different numbers, one of them is not the same as the one that is on the 1000-m site. The second player to change its 1,000 th digit.

3. 's theorem about the angle between the tangent and a chord / CBO = / BAC. On the other hand, / = BAC / ACD; hence, / CBO = / OCD. But / CBO - angle inscribed in the circumcircle of the triangle OBC and relies on the arc of the circle OC. In this case / OCD - the angle between the chord and the OC ray CD. Consequently, direct CD coincides with the line tangent to the circle at point C.

Remark. considered in problem parallelogram has the property that if we combine the middle of its adjacent sides, you get a parallelogram similar source. Is sufficient to note that the triangles BCD and COD are similar.

4. A: k = 2.

Denote n = 1000. We have two cases:

1) k> 1000. Then

k k-(n +1) ------ / \ N +1 / \1 ... 12 ... 2 - 2 ... 2 * 1 = 10 ... 12 ... 2\ / \ / \ / -------------- 2n n +1 2n-k

Obviously, this number is not a square integer: n is even, so the expansion of the number is an odd number of fives.

2) k < 1000. Then

k --- / \ K1 ... 12 ... 2.2 ... 2 = 1 ... 10 ... 0 - 2 ... 20 ... 0 = 10 (1 ... 1-2 ... 2)\ / \ / \ / \ / \ / \ / \ / \ / ----------------------------- n2 n +1 2n-kk n +1- kk 2n-k n +1- k

Obtained: k = 2l, and enough to find all l <n, the number ofA = 1 ... 1 - 2 ... 2 - \ / \ / ------ 2n-2l n +1-2 l

a perfect square. Note that the number x is a perfect square if and when, and 9x. We have:9A = 9 ... 9 - 19 = 9 ... 98 ... 980 ... 01 \ / \ / \ / \ / ------------ 2n-2l n-2l n-2 n-2l

"Close" to the number 9A full square - number B = (10 NL ) 2 . Obviously, B> 9A. It is also obvious that when Y> Z to Y 2 -Z 2 > Y 2 - (Y-1) 2 = 2Y-1. And now we find the difference between B-9A: 2n-2l n-2l +1B - 9A = 10 - 9 9 ... 980 ... 01 ... 9 = 19 = 2 * 10 - 1 \ / \ / \ / --------- n-2 n-2l n-2l +1

Clearly, 2 * 10 n-2l one -1 < 2 * 10 NL -1, with equality precisely when l = 1, we immediately get the answer and tasks.

5. Assume that R is on AC, S - for BC. Then

RQ = RC-QC = (b / 2) - ((a + bc) / 2) = ((ca) / 2). As AQP and RQT triangles are similar, and isosceles triangle AQP, the RQ = RT. Consequently, the

ST = RS-RT = RS-RQ = (c / 2) - ((ca) / 2) = (a / 2) = BS. TSB This triangle is isosceles and / SBT = / = STB / TBA, and BT - bisector angle B of triangle ABC. 

6.

a) Decision by induction. The base is clear: one winner only competition of the two - it is half. Suppose that there is competition for example n 2 n 2 athletes with n-1 possible winners. We prove then that there is a (n +1) competition for 2 n +1 athletes with 2 n possible winners. Divide athletes into equal two groups A and A '. We assume that some form of a competition of any of A 'is stronger than any of the A, and the remaining options - on the contrary. If the first event to conduct a, there remains a group of A ', if any other - will remain the group A. Applying the induction hypothesis to the groups A and A 'separately, we obtain an example with 2 n-1 2 n-1= 2 n possible winners.

b) Specify the type of competition for each athlete that at any order of the competition is eliminated in this form or sooner. Building induction.

For 1 type of competition - it is the weakest in the 1st form.

Suppose we have built a lot of A k = {a 1 , ..., a k } athletes such that a I disposed of the i-th kind of competition or earlier.

Denoted by a k 1 athlete who (k +1)-th as the weakest non many A_k. We prove that a k one is disposed of (k +1)-th form of competition or in any order, before the competition. Suppose that after the (k +1)-th type of competition is the r-th order, and from a variety of A k in the first (r-1) competitions w retired people. In the r-th order as competition eliminated two nr people. Therefore, a k one takes place in the next competition only if the condition 2 nr < kW. But after the (k +1)-th type of competition must pass at least kw competition with the numbers 1, ..., k. Therefore kW < nr-1 <2 n-R . Thus a k one is disposed of (k +1)-form of competition or earlier.

c) We note that the construction described in a), is in the choice of the competition, were selected group A. The maximum number of possible winners of 2 n athletes competing in some of the kinds of competitions n (n +1)-th potential is 2 n -1.

It is easier to prove by induction a stronger statement: for all n > 1, there is an example (n +1) with 2 events n participants that choosing n competitions and sequencing, it can be two winners n -1 participants, and the only exclusive party (let's call him an outsider) can provide access to the final.

Base when n = 1 is obvious (two competitions, each with the same strength athletes).

The induction step. Building the sample (n +2) competition with 2 n +1 members, assuming known cases (n +1) and 2 events n participants. Again divide athletes into equal two groups B and B '. We assume that in some kind of competition b any of B 'is stronger than any of the B, and the remaining species - on the contrary. Each of the groups separately in the kinds of competitions, other than b is an example of execution assertion. Further suppose that outsider in B - the most powerful in the form of B b.

Spend the first b, we obtain 2 n -1 possible winners of B ', and for some order of outsider B' will be released in the final by the induction hypothesis.

If you do not hold b, then in the first competition eliminated all of B ', and then carried out n-1 competition. By the induction hypothesis, the choice of the first event and the order of the rest of the winners can be two n -1 athletes from B.

It remains to explain how to make the winner of the outsider in B. To do this, first perform the kind of competition, which is in the final order, providing the final output of the outsider. After that will be only athletes from B.Then hold the competition in a manner that provides an output of an outsider in the B final, and complete - the competition b. It outsider wins to build the example.

Complete the solution of c), using the induction step more accurate estimate proved above. The induction base is still evident. For the inductive step we repeat the argument of a) and note that in the group A 'winners can be two n -n people, and of the A - 2 n -1 people. Total obtain 2 n -n 2 n -1 = 2 n 1 - (n one) potential winners.

Grade 10

1. Consider the quadratic polynomial f (x) = x 2 + bx + ac. From the condition that f (c) = c 2 + bc + ac = (a + b + c) c> 0. At point c the function f (x) takes a negative value, therefore, the parabola y = f (x) crosses the axis Ox at two points, ie, has two distinct roots. Hence, the discriminant of this quadratic polynomial is positive: b 2 -4ac> 0.

2. triangles APB and DPC isosceles. Denote the angles at their bases / ABP = / BAP =  , / DCP = / CDP =  . Quads and ABQP DCQP inscribed, hence / = AQP / ABP =  and / DQP = / DCP =  . We get: / AQD = / + AQP / DQP =  +  . Next, / BQP =  - / BAP =  -  , as / CQP =  -  and / BQC = 2  - / -BQP / CQP =  +  .

3. A: (1,1).

We first prove that x and y are relatively prime. Suppose the contrary. Then x and y are divisible by a prime p; let p in the factorization of x and y, respectively, in the degrees of a > 1 and B > 1, for definiteness we put a >B. Then the maximum degree p, which divides x 3 + y, is equal to b (since x 3 divided by P 3A } and especially on P B 1 , and y is divisible by P B and P is not divisible by one B ). But x 2 + y 2 is divisible by P 2b , therefore, x 3 + y is not divisible by x 2 + y 2 . This contradiction shows that x and y are relatively prime.

Furthermore, the condition that the number of x (x 2 + y 2 ) - (x 3 + y) = y (xy-1) is divided by x 2 + y 2 . Note that the y and x 2 + y 2 have no common factor greater than 1 (since x and y are relatively prime), then, xy-1 is divisible by x 2 + y 2 . But if xy> 1, then this is impossible, since x 2 + y 2 > 2xy> XY-1.

4. shall call the red numbers on the red zone, and blue - standing in the blue sector. The distance between the two numbers a and b is called the number of numbers arranged in a smaller arc between the numbers a and b.Number, arranged in a circle, divided into pairs equal. We choose the pair of equal numbers, the smallest distance between them (if there are several pairs, choose any of them.) For definiteness, let the chosen pair - red one and a blue one, and a smaller arc  between them goes from red to blue units counterclockwise. The arc  or not the numbers (ie, two units are side by side), or all of one color, or red and blue number n would be at a distance less than the distance between units. Let all the numbers on this arc (if any) --- blue (the case when they are red, similar). Draw the diameter of separating the blue one from the number following it clockwise, show that the desired diameter. Indeed, consider the half-circle containing a blue one. Read blue numbers recorded in a semi-circle from the unit counter-clockwise - it is the number 1, 2, ..., l (l - a number). Now read the red numbers. Since the arc  no red numbers, then it will be the numbers n, n-1, ..., nm (m -

number). Since the total number in the half-n, then it contains all the numbers from 1 to n once.

5. Let f: [0, 1] -> [0, 1], f (x) = x / 3 1/2 and g: [0, 1] -> [0, 1], g (x ) = 1 - ((1-x)) / 3 1/2 - functions corresponding jump grasshopper. The range of the f - [0, 1/3 1/2 ], the range of g - the interval [1 - (1/3 1/2 ), 1]. Each of these segments has a length of 1/3 1/2 , and together they cover the interval [0, 1].

Let n - a positive integer. Consider all the functions h 1 (h 2 (... (H n (x)) ...)): [0, 1] -> [0, 1], where each function H I - or f, or g . It is easy to see that the range of each of these functions is a segment of length (1/3 1/2 ) n .Induction on n, that these segments cover the interval [0, 1]. For n = 1, this has already been verified. Suppose that the range of possible functions h 1 (h 2 (... (H k-1 (x)) ...)) covering the interval [0, 1]. Fix any of the functions h 1 (h 2 (... (H k-

1 (x)) ...)). The range of values of this function is covered by the ranges of the functions h 1 (h 2 (... (H k-1 (f (x))) ...)) and h 1 (h 2 (... (H k-1 ( g (x))) ...)). Thus, we are done.

Now suppose that on the interval [0, 1] is a point a. Consider the interval (a-0, 01; a +0,01) and show that the grasshopper can get into it. Choose n so large that the inequality (1/3 1/2 ) n <0.01. By the above, it is possible to choose the function h 1 (h 2 (... (H n (x)) ...)), such that a point belongs to the range of its values. Then the whole range of values of the function (the segment length (1/3 1/2 ) n ) lies in the interval (a-0, 01; a +0,01). This means that from any point in [0, 1] grasshopper gets inside the interval (a-0, 01; a +0,01), performing consistently jumps corresponding to the functions H n , H n-1 , ..., h 1 .

6. A: desired alignment in Figure 4 (or get out of it by turning or axial symmetry).

Lemma. Suppose circumferentially arranged 1999 distinct positive integers a 1 , a 2 , ..., a 1999 and let a 1 > a 1998 . Consider the following operation: a number of I and a 1999-I , where i = 1, 2, ..., 999, swap, if a I <a I-1999, and does not change otherwise. If at least one pair of numbers changed places, the sum of the products dozen numbers of consecutive increase.

Proof. Consider symmetrical groups of 10 numbers a I , ..., a nine I and a 1999-I , ..., a 1990-I . Let z - the product of the numbers contained in both the first and the second group (the product of the zero number of factors considered to be equal to unity); x and x '- products of the numbers, contained only in the first and only in the second group left in place after surgery; y and y '--- product of numbers, contained only in the first and only in the second group, are exchanged in a transaction. Then the sum of products of the numbers in these two groups before surgery is s 1 = zxy + zx'y ', and after the operation s 2 = zxy '+ zx'y. We have: s 1 -s 2 = z (x-x ') (y-y'). It is easy to see that this difference is not positive. In addition, if the operation is not all the numbers have remained in place, at least one pair of symmetric groups of 10 numbers, the difference is strictly negative, which proves the lemma.

Solution. consider the numbers 1, 2, ..., 1999 apart so that the edges between adjacent numbers are equal. Let the numbers are placed optimally, ie, so that the sum of the products of adjacent numbers ten highest. Draw a diameter of one of the numbers. Of the lemma that for all pairs that are symmetric with respect to this diameter, smaller numbers are located on one semicircle, and more - on the other. Up to rotation and axial symmetry there is a unique arrangement of the numbers satisfying this property. Indeed, the number 2 should be close to the number 1. Otherwise there diameter separating 2 of 1, and the numbers 1 and 2 are not

symmetrical with respect to that diameter. Denote the number of symmetrical numbers 1 and 2 on this diameter, respectively, by A and B. Then A> 1 and 2 <B, which contradicts Lemma.

Then, a desired balance of induction. Suppose we have shown that the numbers 1, 2, ..., 2k, with 1 < k < 998 are arranged as in the answer, ie, in the order (to be specific clockwise) 2k, 2k-2, ..., 2 , 1, 3, ..., 2k-1 row.Denoted by A and B, respectively, the number after 2k anticlockwise following the 2k-1 clockwise. Suppose that the number of 2k +1 is different from A and B. Then let C - following the 2k +1 Clockwise rotation. C is different from 1, 2, ..., 2k. Number of C and 2k-1 and 2k +1 and B are symmetric with respect to a diameter but C> 2k-1, 2k +1 <B - this is a contradiction. Suppose now that the number of 2k +2 is different from A and B. Then let C - following the 2k +2 Clockwise rotation, D - followed by 2k +2 CCW number. Suppose A is not equal to 2k +1. Then 2k +2 <A, but D> 2k - this contradicts Lemma. If A = 2k +1, then B is not equal to 2k +1, and we obtain a similar contradiction (2k +2 <B, but C> 2k-1). Thus, we find that either A = 2k +1 and B = 2k +2, or A = 2k +2 and B = 2k +1. It is easy to see that the lemma is not only contrary to the second case. This completes the proof of the induction step. 

Grade 11

1. Due to the triangle inequality, a 2 > (bc) 2 . Hence a 2 +2 bc> b 2 + c 2 . The right-hand side is positive, and it can be divided. We find that the first term on the left side of inequality is greater than 1. The same is true for the other two. So their sum is greater than 3.

2.

a) It suffices to draw a line through the middle of the arc and the center sloping BAC.

b) Let A - angular point, B and C - the ends of the arc, D - its middle. Segments, based on the chord BD and DC, are equal. Therefore it is sufficient to hold the line through the point D, which bisects the area of the quadrilateral ABDC. 

Through the middle of the diagonal BD line l, parallel to AC. Suppose, for definiteness, l intersects the segment AB (l case of intersection with the segment AD is similar). Let E -

intersection point l and AB; straight CE - unknown. This is evident from a consideration of areas of triangles ACD, ACE and ACB (with a common base AC).

3. planes, which belong to the brink of each color, form equal regular tetrahedra. Statement of the problem follows from the fact that the sum of the distances from an internal point of a regular tetrahedron to its faces is constant and equal to three times the volume of a tetrahedron, divided by the area of the face. (To prove this, connect the point to the vertices of a regular tetrahedron and consider the amount of formed parts, which are the bases of the original faces of the tetrahedron.)

4. necessary to prove the following statement. Let each side of the square has a length of 1 and divided into 2 n equal parts (n > 0), and through the points of division draw a line parallel to the sides. Then the grasshopper can get into any of the 4 n of these cells.

When n = 0 is trivial fact. We carry out the induction step from n to n +1. Consider some of the cells of a 4 -n-1 . Choose closest to the top of her original square and run the homothety with center at the top and by a factor of 2. Then the selected cell goes into one of the cells of a 4 -n . By the induction hypothesis, the grasshopper can hit it. If he jumps now half way to the top of this, he gets into the right cell.

5. Colors, which is colored graph, we number from 1 to k. Those top two colors that do not coexist with any of the vertices of color 1, repainted in a color 1. The new coloring will be correct, so it k colors. Hence, any two vertices of color is not repainted and therefore coexist with the vertices of color 1. Similarly, the top three colors that can not coexist with the vertices of color 2, repainted in color 2, and so on until the last color.

After this, we consider some vertex color k. She was not repainted, and so is adjacent to the vertex color k-1. This peak is also not repainted, because otherwise it would be the original color of k, and it could not coexist with the top of the same color. Once top is not repainted, it is adjacent to the vertex color k-2, etc. Continuing this process, we construct a path from vertex k colors that have not been repainted.

6. only solution: n = 2, k = 1, l = 2, m = 3. Prove it.

Let p - prime factor l. Since n m = (1 + n k ) L -1, then n m divisible by (1 + n k ) P -1. But this expression is equal to n k * P + n 2k * P (P-1) / 2 + n 3K * R, where r - a non-negative integer. Dividing by n k , we obtain P n + k * P (P-1) / 2 + n 2k * R. If n is not divisible by p, then this expression is prime to n, n and m can not share it. Hence, p - divisor n. Then 1 + n k * (P-1) / 2 + (n 2k / P) * R - a natural number greater than one. If k> 1 and p is odd, the second term is divisible by n (third split always), the amount is relatively prime to n, n and m can not share it. Therefore, k = 1 and l has the form 2 s .

Now recall that n m = (1 + n k ) L = -1 (1 + n) L -1 = ln + ... The right side of all the members, beginning with the second, divided by n 2 . Of this, as m> 1, it follows that l is divisible by n. Hence, n, as l, is a power of two.But (1 + n) L -1 = [(1 + n) L / 2 +1] * [(1 + n) L / 2 -1] = = [(1 + n) L / 2 +1]. .. (N +2) n,} where n +2 is also a power of two. Therefore, n = 2. Factor decomposition preceding n +2 = 4, would be equal to 3 2 +1 = 10 and would not be a power of two. So, l = 2, so m = 3.

7. Consider the circle of length 1 as the interval [0, 1] with endpoints identified. Then, the fractional part of f k * lg2 can be considered as a point of the circle. The first digit of the number 2 k controls the position of f with respect to the division points 0, lg 2, ..., lg 9. (For example, if 2 k begins with 7, 7 * 10 s <2 k <8 * 10 s , for a natural s. fractional part of k * lg 2 is equal to k * lg 2 - s, and it is between 7 and lg lg 8.)

Suppose that the first digits of the numbers 2 2 n repeat with period k. Then for any n fractions of numbers 2 n * LG 2 and 2 n + k * 2 LG fall into the same interval of the circle, the length of any of these intervals does not exceed lg 2 <1/3.

Let the circle plotted the fractional parts of two positive numbers A and B; these fractional parts are distinct and are not diametrically distant points of the circle, the length of the smaller of the two arcs into which these points divide the circle is equal to x. Then it is easy to show directly the length of one of the arcs connecting the fractional parts of the numbers 2A and 2B, is 2x. Now suppose that the fractional parts of the numbers A and B are in the same range, and consider a pair of 2A and 2B, 4A and 4B, etc. From the above it follows that at some stage one of the arcs connecting the fractional parts of a pair will be more third but less than 2/3. Hence, these fractional parts belong to different intervals of the circle. Applying these considerations to the numbers A = 2 n 

0 * lg 2 and B = 2 n 0 + k * LG 2, where n 0 - a fixed

positive number, we get a contradiction to the assumption of periodicity.

   

LXIV Moscow Mathematical Competition

Mathematics holiday 18/02/2001 - Grade 6 , Grade 7 Olympics 04/03/2001 - Grade 8 , Grade 9 , Grade 10 , Grade 11 Solutions: Grade 6 , Grade 7 , Grade 8 , Grade 9 , Grade 10 , Grade 11 Conditions and solving the district round 28/01/2001 (grades 5-11)

Closing ceremonies for 6-7 classes (math holiday) and award winners ( see list here ) took place on the day of its February 18, 2000

Closing ceremonies for 8-11 classes will be held March 18, 2001 in the main building of Moscow State University on Sparrow Hills.

12.00 - problem analysis, members of the jury will tell tasks, the main criteria for testing and will answer your questions.

13.00 - featuring works, students will be able to see their work and find out what it is placed such estimates.

14.00 - lecture for the students read Academician DV Anosov.

16.00 - Closing Ceremony and award winners.

Issuance nerozdannyh at the closing awards and prizes - Wednesdays (starting March 21), from 16.00 to 18.00 in the cab. 202 MCCME (Vlas'evski per., 11). Tel. for information 241-12-37.

Contact phone 241-12-37.

Terms

Grade 6

1. Solve the riddle: AX * YX = 2001. (A. Blinkov)

2. Ofen (seller in dressing, peddler) purchased on the wholesale market game handles and offers buyers or one knob for 5 rubles or three pens for 10 rubles. From each customer receives the same profit Ofen. What is the wholesale price of a pen? (A. Sablin)

3. Natasha and Inna bought on the same box of tea bags. It is known that one packet is enough for two or three cups of tea. Natasha box enough for a cup of tea, only 41, and Inna - only for 58 cups. How many bags were in the box? (A. Spivak, I. Yashchenko)

4. Arrange in a circle 6 different numbers so that each of them is equal to the product of the two neighbors. (A. Mityagin)

5. Vifsla, Tofsla Hemul and playing snowballs. The first snowball thrown Tofsla. Then, in response to each hit him Vifsla snowball throwing six snowballs Hemul - 5, and Tofsla - 4 snow. After some time, the game is over. Find one in whom much snowballs hit if off target flew 13 snowballs. (In itself not throw snowballs.) (T. Golenishchev-Kutuzov, VA Kleptsyn)

6. Fields checkered board size 8 * 8 will take turns in red paint, so that after each new tile painting figure consisting of the filled cells, was the axis of symmetry. Show how you can paint a) 26, b) 28 cells, observing this condition. (In response to the stand with those cells that are to be painted over, the numbers from 1 to 26 or up to 28 in the order in which to conduct brushing.) (I. Akulich)

Grade 7

1. the Guinness World Records says that the largest known prime number is 23 021 377 1. Not a typo is it? (S. Markelov)

2. Coming to the shooting range, a player bets $ 100 in cash. After each successful shot of his money increased by 10%, and after each slip is reduced by 10%. Could it after a few shots had to be 80 rubles 19 kopecks? (I. Yashchenko)

3. In order to build a model house is not enough space. Changed the project architect: removed two entrance and added three floors. The number of apartments increased. He was delighted and decided to take another 2 entrance and add another 3 floors. Could it be apartments with even less than in a typical project? (In every section the same number of floors, and at all levels in all doorways same number of apartments.) (T. Golenishchev-Kutuzov, VA Gurovits, P. Kozhevnikov, I. Yashchenko) 

4. in the wall there is a small hole (point). The owner is a checkbox in the form below (Figure 1). Show in the figure all the points at which you can drive a nail, so to check the hole closed. (A. Shen)

5. Mark on board 8 * 8 more cells so that any (including any marked) cell bordered by side with exactly one marked cell. (A. Spivak)

Grade 8

1. on checkered paper to draw a rectangle of width 200 and height of 100 cells. Paint over it to cells, starting from the top left and going in a spiral (before reaching the edge or already shaded area, turn right, Fig. 2).Which cell will be drawn last? (Enter the number of its row and column. Example, the lower right-hand cell is in the 100-th row and column of 200 m.) (A. Khachaturian) 

2. Can I put 100 points on the plane (first the first, then the second and so on to the hundredth), so that no three points lie on a straight line and at any time to figure consisting of points that have been set, had the axis of symmetry? ( I. Akulich)

3. Given six words:

SplinterCoatCASINO

MulletSandbankSHELEST

One step can be replaced by any character in any of these words in any other (for example, a step can be derived from the word splinter word ZKNOZA. How many steps it takes to do all the words of the same (pointless allowed)? Give an example and show that the lower number of steps do not. (V. Dotsenko, A. Shen)

4. In triangle ABC the bisector held AK , median BL and the height of CM . Triangle KLM is equilateral. Prove that the triangle ABC is equilateral. (R. Zhenodarov)

5. Lesch planned two-digit number (10 to 99). Gregory tries to guess, calling double figures. It is believed that he guessed, if one number he called correctly, and the other was wrong for not more than one (for example, if you think about the number 65, 65, 64 and 75 are suitable, and 63, 76 and 56 - no). Invent a way to ensure the success of Grisha in 22 attempts (or whatever number conceived Lesch). (Folklore)

6. (Continued) Show that there is no way to guarantee the success of Grisha in 18 attempts. (Folklore)

Grade 9

1. Can I place it on a football field four players so that the pairwise distances between them is 1, 2, 3, 4, 5 and 6 meters? (A. Mityagin)

2. In some countries, the total salary of the highest paid 10% of employees is 90% of the salary of all employees. Could it be that, in each of the regions into which the country is divided, pay 10% of all employees is not more than 11% of the wages paid in the region? (M. Slack)

3. Inside the angle with vertex M marked point A . From this point, let a ball which was reflected from one side of the angle at point B , then from the other side of the point C and back to A ("AOI" is "angle of reflection", Fig. 3). Prove that the center O of the circle circumscribed about the triangle BCM , lies on AM . (A. Zaslavsky) 

4. stones lie in heaps three: one - 51 stone, in other - 49 stone, and in the third - 5 stones. Allowed to combine all into one pile, and to share a handful of an even number of stones in two equal. Can I get 105 piles, one for each stone? (V. Kleptsyn)

5. A positive integer N to 999 ... 99 ( k nines) times the amount svoix digits. Specify all the possible values of k , and for each of them give an example of such a number. (Galperin)

6. Participants chess tournament played each other in the same party. For each participant A was estimated number they dialed points (for a win is worth 1 point for a draw - 1/2 points for a loss - 0 points) and power coefficient as follows: total score of those participants who have A win, minus the sum of points those to whom he has lost. a) Can factors force all participants to be greater than 0? b) Can the power coefficient of all participants to be less than 0? (A. Tolpygo)

Grade 10

1. Are there three square trinomial, such that each of them has a root, and the sum of any two trinomials has no roots? (A. Kanel)

2. possible to arrange a guard around a point object so that neither the object nor the hour it was impossible to sneak up? (Every time stands still and see the 100 m straight ahead.) (V. Kleptsyn)

3. Give an example of a polynomial P ( x ) degree in 2001, for which the identity

P ( x ) + P (1 - x ) = 1. (V. Senderov)

4. In acute triangle ABC held the height AH A , BH B and CH C . Prove that the triangle with vertices at the points where the heights of triangles AH B H C , BH A H C , CH A H B is a triangle H A H B H C . (Hakobyan)

5. On the two cells are checkerboard black and white chips. In one move, you can move any of them to the adjacent horizontal or vertical cells (two chips can not be on the same cell). Whether as a result of moves to meet all the possible options for the location of the two chips, and exactly once? (Shapovalov)

6. In the game "Descent" two armies occupy the country. They take turns, each turn taking one of the free cities. First Army captures the city from the air, and every next move, it can capture any city roads into some already occupied this army town. If there are no cities, the army stops fighting (in this case, perhaps another army continues its actions). Will any such scheme cities and roads that the army, walking the second, be able to capture more than half of all cities, no matter how acted first

army? (The number of cities of course, every road connects exactly two cities.) (P. Grozman, Shapovalov, D. Shapovalov)

Grade 11

1. Are there three square trinomial, such that each of them has two distinct real roots, and the sum of any two trinomials has no real roots? (A. Kanel)

2. Given a geometric progression. It is known that her first, tenth and thirty members are natural numbers. Is it true that it is also a member of the twentieth is a natural number? (Folklore)

3. In triangle ABC point I - center of the inscribed circle, I '- center of the circle tangent to side AB and the extensions of sides CB and CA ; L and L '- the point at which the side AB for these circles. Prove that IL ', I' L and height CH triangle ABC intersect at one point. (A. Zaslavsky)

4. Prove that there is a polynomial of degree at least two and a non-negative integer coefficients, the value of which, for any prime p is a prime number. (A. Kanel)

5. Prove that there exists a location in space in 2001 of a convex polyhedron such that no three of the polyhedra have no common points, and any two relate to each other (ie, have at least one boundary point, but no common interior points). (A. Kanel)

6. Circling placed several boxes. Each of them can be based on one or more balls (or it may be empty). In one move is allowed to take all the balls out of any boxes and put them, moving clockwise, starting with the next box, put in each box one ball. a) Prove that if every time you take a course of balls out of the box, which hit the last ball in the previous course, that at some point in the initial placement of the balls again. b) Prove that a few moves from any initial placement on cases a ball can get any more. (V. Gurovits)

Solutions

Grade 6

1. A: AH = 29, YX = 69, or, conversely, AH = 69, YX = 29. Since 2001 = 3 * 23 * 29, number 2001 can be represented as the product of a two-digit numbers in the following ways: 69 * 29 or 23 * 87.

2. A: The wholesale price of the handle - 2 rubles 50 kopecks. If the wholesale price of the handle - x rubles, 5 - x = 10/3 x , where x = 2.5.

3. A: 20 bags.

First solution. As Inna drank 17 cups of tea over Natasha, then at least of the 17 bags she made three cups of tea. The remaining 7 = 58-17 * 3 cups were available in only one way: 2 bags of 2 cups each and 1 sachet per 3 cups. Hence, the box was 17 +3 = 20 bags. While Natasha of 19 bags of 2 cups cooked, and from twenty - three cups of tea.

The second solution. Note that bags could not be more than 20: if in a pack was at least 21 bag, Natasha would not have to drink at least 2 * 21 = 42 cups of tea. But the bags could be no less than 20, otherwise Inna drank not more than 3 * 19 = 57 cups. So, in each packet could be only 20 bags. Inna used 3 times 18 sachets, and Natasha - only 1.

4. If there are numbers a and B , then the following is the number of B / a , followed by 1 / a , then 1 / B , finally, a / B . These six numbers satisfy the conditions of the problem. Of course, the poor choice of the numbers a and b any of these numbers match, but we will not stop: to solve the problem it is sufficient to show an example. For example, take a = 2, B = 3.

5. A: in Hemul, Vifslu Tofslu and hit once. If the Vifslu, Tofslu Hemul and got x , y and z , respectively, snowballs, then all was thrown 13 + x + y + z snowballs (since 13 snowballs did not reach the goal). On the other hand, gave up six Vifsla x , Hemul - 5 y , and Tofsla - 4 z 1 snowballs (with the first snow.) We obtain the equation

6 x 5 y four z one = 13 + x + y + z , where 5 x four y 3 z = 12. Since x , y , z - non-negative integers, x can be 0, 1 or 2, y - 0, 1, 2 or 3, z - 0, 1, 2, 3 or 4. Bust find solutions (1,1,1), (0,3,0) and (0,0,4). But, as in itself not to throw snowballs, then one of the numbers x , y , z can not have two zeros. Therefore can only be the first case.

6. A is shown in Fig.

1 20 13

2 21 12

3 22 11

14 15 16 4 17 18 19 27

10 23 5

9 24 6

8 25 7

26 28

Grade 7

1. A: Of course, it's a typo. Any degree of numbers ending in 1, also ending in 1. Therefore, the difference 23 021 three hundred and seventy-seven -1 ending in 0 and, therefore, is not a prime number.

In fact, the largest known prime number is now number 2 3021377 -1. Primes of the form 2 n -1 are called Mersenne numbers (named after the 17th century mathematician M. Mersenne, who studied them). It can be proved that the composite n number of 2 n -1 is composite. Therefore, the numbers correspond to simple Mersenne n . For example, 2 2 1 = 3, 2, 5 1 = 31, 2 7 -1 = 127 - prime. But we can not say that every primep corresponds to a prime 2 P -1. For example, 2 11 -1 - a composite number. Search for Mersenne numbers occupied many mathematicians, for example, Euler proved that the number 2 31 -1 simple. Finite or infinite set of numbers Mersenne - a question that has no answer.

2. A: Yes, it could, if it once and hit three times missed. The solution is easier to find, if we divide 8019 into factors: 8019 = 9 3 * 11. After a successful shot of money multiplied by 1.1, and after an unsuccessful - 0.9 and 100 * 1.1 * 0.9 3 = 80.19.

3. A: Yes, it could. For example, if the original project was five entrances, 4 floors and each floor one apartment: 5 * 4 = 20, 3 * 7 = 21, 1 * 10 = 10.

4. A is shown in Fig. 

5. example is shown here. 

We argue, using the chessboard. Note that the white border on the side of the cage only to black and vice versa. Therefore, we first note a few white cells, so that each black cell has been marked exactly one neighbor (figure). Next, we note some of the black cells, so that each and white cells appeared marked neighbor (figure), with a new black cells marked neighbor appears.

Grade 8

1. A: The cells in the line 51, column 50. First will be painted over the outer layer of cells, and then stay rectangle 98 * 198 cells. This rectangle will paint over the spiral, after painting its outer layer will be a rectangle 96 * 196 cells, and so on. After painting 49 layers remain unfilled rectangle 2 * 102, located in the rows and columns 50-51 50-151. The latter will be painted over the lower left cell of the rectangle.

2. A: Yes. You can, for example, to put points on the circle at regular intervals small enough (as in the figure, only smaller). 

3. A : 25. Write the words in a column:

SplinterCoatCASINOMulletSandbankSHELEST

After all the changes the letters in each column should be the same. Number of changes will be the smallest, if in each column to keep most of the letters (any of them, if some of these letters). For example, in the first column, you can leave the letter E or K, they both require four rotations. The minimum number of replacements is 4 +4 +5 +4 +4 +4 = 25.

Among the words that can happen as a result, there is a meaningful, like green, drops or tile.

4. Recall that in a right triangle median drawn to the hypotenuse is equal to the half of it, and conversely, if the median is half the side to which it is carried out, a rectangular triangle. 

Median ML right triangle AMC is equal to half the hypotenuse AL and LC , as well as segments KL and KM , as the two sides of the triangle KLM are (see Fig.). In

triangle AKS median KL is half the side AC , so the angle AKC line and in a triangle ABC bisector AK is high. Therefore, the triangle ABC is isosceles ( AB = BC ) and AK is also the median: BK = KC . So, MK - median right triangle BMC , so BC = 2 MK 2 =KL = AC . So, AB = BC = AC , as required.

5 and 6. arrange the two-digit numbers in the cells rectangle width and height of 9 10 (horizontal axis units, vertical - tens).

Every attempt to match the cross Grisha of five cells (see Fig.) In the center of which he called the number, and on the sides of four numbers, which differ in one digit by one (if called number contains the number 0 or 9, some cells cross beyond the edge of the rectangle , such cells do not match any numbers). Grisha task - to cover a rectangle 9 * 10 such crosses. We show that the 22 crosses him enough, and 18 crosses - no.

Coverage of the 22 crosses are easy to find, if we notice that the crosses can be laid flat without overlapping (but will have to add a few crosses on the edges of the rectangle). For example, Gregory can call the number 11, 13, 17, 25, 29, 30, 32, 37, 44, 49, 51, 56, 63, 68, 70, 75, 82, 87, 89, 90, 94, 97. 

In the problem of the total area of 6 X's is 18 * 5 = 90, ie, equal to the area of the rectangle. However, covering the corner squares, we inevitably come out of the rectangle, and this loss hurt cover the entire rectangle.

Grade 9

1. A: Yes, you can. Put them on the line so that the distance between the first and second players were 2 m, between the second and third - 3 m, between the third and fourth - 1 m (players called in the order they appear on the line).

2. A: Yes, it can. Assume that each region of all get the same pay and there is a region in which they live those 10% of employees who receive 90% of the salary.

3. perpendiculars to the sides of the angle, vosstavlennye at points B and C , meet at a point M ', diametrically opposite M . From the equality of the angles of incidence and reflection that M '- the point of intersection of bisectors of the triangle ABC . On the other hand, the point M is the intersection of the bisectors of the

angles B ' BC and BCC '(see figure). therefore equidistant from the lines AB and AC . Therefore M also lies on the bisector of the angle BAC . Hence, the entire diameter of MM ', including the point O , lies on the bisector AM angle BAC . 

4. A: No, you can not. Note that if at some point the number of stones in each pile is divided into an odd number a , then all allowed actions received heaps of stones will be divided into a . After the first course you can get three options for placing stones pile of 100 stones and 5 stones (common factor 5), of 56 stones and 49 stones (common factor 7), of 51 stone and 54 stone (common factor 3).

5. A: There is a number for any k : N k = 9 k * (10 k -1). 9 Let k = s   1  ...   s   t  0 ... 0  ( s t is not 0, the zeros on the end might not be). Check that the sum of the digits of N k is 9 k . We write the difference between the number 9 k * 10 k and 9 k , given that 9 k <10 k , for any k :  

-

s 1 s 2 ... s t -1 s t 0 ... 0 0...

0 0 0 ... 0

s 1...

s t -1 s t 0 ... 0

s 1 s 2 ... s t -1 s t -1 9 ... 9 (9 - s 1 )...

(9 - s t -1 ) (10 - s t ) 0 ... 0

 The bottom line is written the number N k . It is easy to see that the sum of the digits is

s 1 + ... + s t -1 ... nine nine + + (9 1) - s 1 - ... - s t = 9 k . entries in the left side of k nines)

6. A: a), b) no, they can not. The sum of the points scored by party A , a S A , and its coefficient of power - through F A . Consider the amount of S A S A F A . If we substitute a definition of F A and open the brackets, we get the sum of numbers of the

form +   S A S B , where players A and B played a draw. Each of these terms is a time with the "+" and once with "-". Therefore, the entire amount is 0. Hence, the signs of the coefficients of the forces of all participants may not be the same.

Grade 10

1. A : Yes, there are. For example, these are the polynomials x 2 , ( x -1) 2 and ( x -2) 2 . The sum of any two of them is greater than zero for any x .

2. A: Yes, you can. The figure shows an example of the location of eight time satisfying. Circles denote hour, each of the arrows indicate the direction in which the watch looks. 

3. For example, P ( x ) = 2 2000 ( x - (1/2)) 2001 + (1/2). The graph of this function (Fig.) can be obtained from the graph of an odd function y = x 2001 shift to the right by 1/2, tensile and shear up to 1/2, and therefore has a center of symmetry of the M (1/2, 1/2). Now it is easy to see that P (1 - x ) = 1 - P x ). 

4. Let H 1 , H 2 , H 3 - orthocenters (intersection point of heights) triangles AH B H C , BH A H C , CH A H B , respectively, H - orthocenter ABC , M 1 , M 2 , M 3 - mid H B H C , H C H A and H A H B . We show that the point H I symmetric point H with respect to M I ( I = 1, 2, 3). Consider, for example, the point of H 2 . Since H C H 2   |     BC and AH   |     BC ,

segments H C H 2 and HH A parallel (see Fig.). 

Similarly, H A H 2 | | HH C . Hence, H C H 2 H A H - point parallelogram and H 2 symmetric H relative to the middle part of H A H C . The same argument holds for the points H 1 and H 3 .

Obtain the configuration shown in the figure. 

Since M 1 M 2 - total average line triangles H A H B H C . and H 1 H 2 H segments H A H B and H 1 H 2 equal. The equalities H B H C = H 2 H 3 and H A H C = H 1 H 3 . Therefore, triangles H 1 H 2 H 3 and H A H BH C are equal.

5. A: can not. We say that the location of one color chips when the chips are on the cells of the same color, colorful - if the cells of a different color. Note that when the chips monochrome and colored alternate location, then there should be equal. But the total number of colorful locations is 2 * 32 2 , and the plain - 2 * 32 * 31, because the two chips can not be on the same cell. Hence, all possible arrangements can not meet.

6. A: Yes, there is. Consider a country whose maps are shown in the figure (points - the city, the segments - roads). 

We show that the Second Army can always take at least two points A I . Indeed, if the first army first move is a point on the "thread" of k points, the second army should take appropriate this "branch" point A I ; if the first is A I , the second - B I ; selects if the first point B I , then second - one of the points of A j , connected line with B I . Further actions are obvious.

Since the cessation of hostilities second army is at least two points A I , first in less than k three points. Therefore, the proportion of cities captured by the second army, at least (2 k three) / (3 k 6). Even at k = 1, this number is greater than 1/2. Note that in the problem instead of 1/2, you can take any number  <2/3: for sufficiently large k the number of (2 k 3) / (3 k six) will be more  .

Grade 11

1. A: Yes, there are. For example, these are the polynomials ( x -10) 2 -1, x 2 -1 and ( x 10) 2 -1.

2. A: Yes, right. Let a 1 , a 2 , ..., a _ n , ... - This geometric progression, q - the denominator. By hypothesis, a 1 , a 10 = a 1 q 9 and a 30 = a 1 q 29 - positive integers. Therefore q 9 and q 29 - positive rational numbers. It follows that q 2 = q 29 / ( q 9 ) 3 - a positive rational number q = q 9 / ( q 2 ) 4 as a positive rational number.

Let q = m / n , where m and n - natural coprime. The number of a 30 = a 1 m 29 / n 29 natural, m 29 and n 29 are relatively prime, hence, a 1 divided by n 29 . Hence we find that a 20 = a 1 q 19 = a 1 m 19 / n 19 - the number of natural.

3. denote the M point of the inscribed circle diametrically opposite point of L , in M '- point excircle diametrically opposite point of L '(Figure). Thus, ML and M ' L '- diameters, respectively, of the inscribed and escribed circles, they are parallel to the height CH triangle ABC . Incircle goes into escribed under the homothety with center C . With this dilation diameter ML passes in diameter parallel to it, ie, M ' L '. Therefore, the points C , M , L 'lie on one line, and the point C , M ', L are

collinear. It follows that the triangles L ' LM and L ' HC combined homothety centered at L '. Since L ' I - median triangle L ' LM , line L ' I intersectsCH in the middle. 

Furthermore, triangles LL ' M 'and LHC combined homothety centered at L . Since LI 'is the median in a triangle LL ' M ', straight LI 'also intersects CH in the middle, which proves the problem.

4. Suppose that a polynomial Q ( x ) exists. Let Q ( x ) = a n x n + a n -1 x n -1 + ... + a 1 x + a 0 , where a 0 , a 1 , ..., a n - non-negative integers, a n is 0, n > 2.

If a 0 = 0, then Q ( x ) = x ( a n x n -1 + a n -1 x n -2 + ... + a 1 ), therefore, the simple p the number of Q ( P ) divided by the p and more P (here we use the fact that the degree of Q is at least 2), so that Q ( P ) - a composite number. Let's say, a 0 > 2. Denoted by p be a prime divisor of a 0 . Then Q ( P ) = ( a n P n -1 + a n -1 P n -2 + ... + a 1 ) P + a 0 is divisible by p , and more P , hence, Q ( P ) - the number of composite .Thus, there is a unique opportunity: a 0 = 1.

If, for any prime p the number of Q ( P ) is prime, then the number of Q ( Q ( P )) is simple for any prime P ). Hence, the constant term of the polynomial Q ( Q ( x )) must be 1. However, since Q (0) = a 0 = 1, Q (Q (0)) = a n + a n -1 + ... + a 1 +1> 1. We have a contradiction, which concludes the proof.

5. put N = 2001, and describe the construction of location N of convex polyhedra, no three of which are disjoint and any two relate.

Consider an infinite circular cone, the vertex O which is at the origin, and the axis is directed along the axis Oz . On the circle, is a section of the cone plane z = 1, we take the point A 1 , A 2 , ..., A N - vertices of a regular N -gon. Denoted by B 1 , B 2 , ..., B N mid smaller arcs A 1 A 2 , A 2 A 3 , ..., A N A 1 , respectively. Through C ( t ) denote the disc, which is a section of the cone plane z = t , in O t - center of the circle through A t I , B t I - point of intersection of

the OA I , OB I to the circle C ( t ), where t > 0, 1 <   I <   N . 

Prove an auxiliary assertion.

Lemma. Suppose that a convex polygon M lies inside the circle C ( t 0 ), t 0 > 0. Consider the infinite upward "prism" P , whose base is a polygon M , and the lateral "edge" direct parallel OB I for some I . Polygon (equal to M ), which is obtained by the intersection of the prism P and the circle C ( t ), denoted by M ( t ) (thus, M ( t 0 ) is equal to M ). Then there is a positive T > t 0 , for all t > T polygon M ( t ) will be contained within the segment S I ( t ) = A t I B t I A t I 1 circle C ( t ).

Proof. polygon M ( t ) is the image of M under parallel translation by a vector parallel to the image OB I . Therefore polygon M ( t ) is contained within a circle of equal circle C ( t 0 ), relating to the circle C ( t ) at the point B I ^ t. The distance from point B I t ^ to the line A I tA_ ^ {1} ^ I t is directly proportional to the length of the forming OB t I ; therefore, there is a T , that for all t > T is a distance greater than the diameter of the circle C ( t 0 ) (Figure xx). This means that the polygon M ( t ) falls within the segment S I ( t ) range of C ( t ). Lemma.

We turn to the inductive construction of our location.

We choose t 1 > 0. Take a convex polygon M 1 in the circle C ( t 1 ). Consider the infinite upward "prism" P 1 whose base is a polygon M 1 , and lateral "edge" direct parallel OB 1 .

Suppose that for some n (1 < n < N ) is defined positive numbers t 1 , t 2 , ..., t n -

1 , t 1 < t 2 <... < t n -1 , convex polygons M 1 , ..., M n -1 , lying inside the circle C ( t 1 ), C ( t 2 ), ..., C ( t n -1 ), endless "prism" P1 , P 2 , ..., P n -1 with the bases of M 1 , M 2 , ..., M n -1 and lateral "edges", parallel to OB 1 , OB 2 , ..., OB n -1 , respectively.

According to the lemma, there exists t n > t n -1 , such that the polygon M 1 ( t n ) is contained within the segment S 1 ( t n ) circle C ( t n ), polygon M 2 ( t n ) is contained within the segment S 2 ( t n ) the circle C ( tn ), ..., polygon M n -1 ( t n ) is contained within the segment S n -1 ( t n ) the circle C ( t n ).

Shift each of the lines A t n I A t n I 1 (1 <   I <   n -1) in the direction of O t n B t n I , until it coincides with the line L I ( t n ), related to polygon M I ( t n ). Now we construct a convex polygon M n , which is contained in the circle C ( t n ), for each polygon M I ( t n ), and lies on the opposite side of him straight L I ( t n ): take, for each i one of the points of tangency polygon M I ( t n ) and direct L I ( t n ); when n > 3, these points are the vertices of a convex ( n -1)-gon, which can be selected as M n ; when n = 2, 3 must be added a few points).

Consider the infinite upward "prism" P n , the base of which is a polygon M n , and the lateral "edge" direct parallel OB n . Build is done, when we get the prisms P 1 , P 2 , ..., P N . Finally, "srezh" prism plane z = T , where T > t N .

We prove that the resulting (ordinary) prism satisfy task. It suffices to show that the prism P n intersects the prism P I , I < n , only in the plane circle C ( t n ). Suppose, on the contrary, there is a common point of RC C ( t ) of two prisms, where t > t n . Through the point R draw lines parallel to OB n and OB I ; denote their point of intersection with the plane of the circle C ( t n ) in R n and R I , respectively. Then the point R nmust belong to a polygon M n ( t n ), and the point R I should belong to the polygon M I ( t n ). Clearly, the (non-zero) vectors R I R n and B t n I B t n n opposite directions.

But this is impossible, since the point of R I and B t n I are on the same side of the line L I ( t n ), and the points of R n B and t n n - on the other. Completes the proof.

6. a) The current state of the system described in the problem defined by the number of balls in each box and specifying the box, which should begin to lay out the balls in the next time. Therefore, the possible states of the system is finite. Each state can, laying balls, move to another state of the system, which is uniquely defined. Conversely, knowing the state of the system at the moment, we can uniquely determine the state of the system before the final unfolding of balls. Indeed, the last unfolding would end on a dedicated box, so to restore the previous state, you need to take one ball from the selection box and then, going against theclockwise direction to take the ball from each box as possible. When we encounter an empty box, we put it all collected balls and declare it checked. (In fact, we have given a description of the operation, return to the course described in the problem.) If we denote the state of the system points, and the ability to move from one state to another - an arrow connecting the corresponding points (ie, construct a graph of the system states), then each point will be published exactly one arrow at each point will be to exactly one arrow. Start moving the arrows from a given state A 1 . Obtain a sequence of states of A 2 , A 3 ... As the number of states is finite, at some point in the sequence { A I } there recurrence. Suppose, for example, A k = A n , where n < n . Since the point A k contains exactly one arrow from the equation A k = A n be A k -1 = A n -1 , ..., A 1 = A k - n 1 Thus, a n - k moves we returned to the state of A 1 .

b) In contrast to the problem as well) is now state of the system is determined only by how spread out balls on boxes. At any moment there are several possible approaches, namely, as much at the moment is not empty boxes. Obviously, there are as many ways to do the reverse operation, as described in the decision of a). In terms of the state graph of the system, this means that each point of the leaves as many arrows as it includes.

Prove two assertions.

1) For any initial state can ensure that all the balls were in the same pre-assigned box M .

In fact, if each operation to take a non-empty boxes of balls closest to M moving counter-clockwise, then either the number of balls in M increases, or closest to the M non-empty box will be even closer, he could not continue indefinitely. Assertion.

Denoted by K state where all the balls are collected in a box M , and through B some arbitrary state.

2) From B could go to K , moving the arrows against their direction (back arrow).

Indeed, according to the previous statement, there is a sequence of (different) states B = A 1 , A 2 , ..., A n = K , the points A 1 , A 2 , ..., A n series connected by arrows. We move to the back arrow from A n to A n-1 , of A n -1 in A n -2 , ..., of A 2 to A 1 and then from A 1 , while it is possible, by choosing at each step backward arrow, on which we have not yet passed. We can only stay in K . Indeed, suppose that we erase the motion direction. Cnachala at every entry point shooter as much as out of it. If you erase the arrows at a time in each location, except for the initial point of K , is no less shot than leaves. Consequently, if we got to a certain point (except K ) on the reverse direction, and then we can get out of it. It follows that of B could go to K on the back arrow, ie, from K could go to B by the usual arrows.

It is now easy to complete the proof. Let there be two arbitrary states A and B . In order to pass, moving the arrows from A to B , you can first go from A to K , and then from K to B . As we have shown, both is possible.

       

65 Moscow Mathematical Competition

Competition was held on 3 March 2002.

See the list of winners here , checking work (list of scores on the registration numbers of participants) - here .

Information about the formation of the All teams from Moscow Mathematical Olympiad (Maikop, 21-29.04.2002) is here .

Conditions and objectives

Grade 8 | Grade 9 | Grade 10 | Grade 11 (to perform tasks in all classes have 5 astronomical clock).

Grade 8

(Solving 8th grade is here )

1. the island two thirds of all the men are married and three fifths of all women are married. What proportion of the population of the island is in a marriage?

2. squared sum of the digits of A is equal to the sum of the digits of A 2 . Find all such double figures A .

3. Dana kruzhnost diameter AB . Another circle centered at A intersects the segment AB at point C , and AC <(1/2) AB . Common tangent of two circles for the first circle at D . Prove that CD is perpendicular to AB .

4. Two players in turn put on the board 65 * 65 for one piece. In this case, no single line (horizontal or vertical) should not be more than two pieces. Who can not make a move - lost. Who wins in a particular game?

5. In triangle ABC medians AD and BE intersect at M . Prove that if the angle AMB a) direct, b) sharp, then AC + BC > 3 AB .

6. in check box m * n , each cell can be either alive or dead. Every minute together all the living cells die, and those dead who have had an odd number of live neighbors (on the side), come to life. Specify all pairs ( m , n ), for which there is an initial placement of live and dead cells, that life in the rectangle would last forever (ie at any given time at least one cell is alive)?

Grade 9

(Problem solving in grade 9 is here )

1. ranks recruits stood face to sergeant. At the command of "left", some have turned to the left, some - to the right, and the rest - around. Always a sergeant can get in order, so that on either side of him was equally recruits, standing to face him?

2. Suppose a , B , C - the triangle, prove the inequality a 3 + B 3 three abc > C 3

3. bisectors of angles A and C of a triangle ABC intersect described near circle at points E and D , respectively. Segment DE intersects the sides AB and BC at points F and G . Suppose that I - intersection point of bisectors of the triangle ABC . Prove that the quadrilateral bfig - diamond.

4. Find all the integers x and y , satisfying the equation x 4 -2 y 2 = 1. 

5. placed in a number of n lights and lit some of them. Every minute after that all the bulbs at the same time change the state of the next rule. Those lights that burned in the last minute, will go out. Those who at the last minute did not burn and neighbors with exactly one lighted lamp, light up. For which n can be so light a few bulbs in the beginning, so that later was found at any time, at least one lighted lamp?

6. acute triangle cut rectilinear cut into two (not necessarily triangular) part, then one of these parts - again in two, and so on, at each step, choose any of the existing parts and cut it (straight) into two. After a few steps it was found that the original triangle split into several triangles. Can they all be obtuse?

Grade 10

(Problem solving in Grade 10 is here )

1. tangents of the angles of the triangle - the natural numbers. So that they can be equal.

2. about positive numbers a , B , C is known that (1 / a ) + (1 / B ) + (1 / C ) >   a + B + C . Prove that a + B + C > 3 abc .

3. In convex quadrilateral ABCD points E and F are the midpoints of the sides BC and CD respectively. Segments AE , AF , BF divide rectangle into 4 triangles, squares which are sequences of natural numbers. What is the most important area of the triangle ABD ?

4. Each viewer who bought a ticket in the front row theater, took one of the seats in the front row. It turned out that all the seats in the first row are busy, but every viewer is sitting in the wrong place. Taker can swap neighbors if both do not sit in their seats. Always whether it can all sit in its place?

5. yields of milk in the city mayoral election are as follows. If in the next round of voting no candidate won more than half of the votes shall then be the next round with all the candidates, except for the last number of votes.(Never two candidates do not gain a tie, if the candidate won more than half the votes, he becomes mayor and elections are over.) Each voter in each round of voting for a candidate. If the candidate is left to the next round, the voter shall vote for him again. If the candidate was eliminated, then all voters vote for the same candidate from among those remaining. At the next election in 2002 running for the candidate. Bender became mayor, who finished the first round of k -th largest number of votes. Determine the largest possible value of k , if Bender was elected a) in 1002-th round, b) in the 1001-th round.

6. possible to color all points of the square and the circle in black and white so that the set of white points of these figures were similar to each other and a set of black dots were similar to each other (possibly with different coefficients of similarity).

Grade 11

(Tasks in grade 11 is here )

1. tangents of the angles of the triangle - the natural numbers. So that they can be equal.

2. Prove that on the graph of y = x 3 can be noted a point A , and the graph of the function y = x 3 + | x | 1 - a point B , the distance AB is less than 1/100.

3. Each viewer who bought a ticket in the front row theater, took one of the seats in the front row. It turned out that all the seats in the first row are busy, but every viewer is sitting in the wrong place. Taker can swap neighbors if both do not sit in their seats. Always whether it can all sit in its place?

4. in increasing sequence of natural numbers, each number, starting from 2002, is a factor of the sum of all previous numbers. Prove that there is a sequence number, starting with which each number is the sum of all the previous ones.

5. Suppose AA 1 , BB 1 , CC 1 - high-angled triangle ABC ; O A , O B , O C incenter of the triangle A B 1 C 1 , B C 1 A 1 , C A 1 B 1 , respectively; T A , T B , T C - the point of contact of the inscribed circle of triangle ABCwith sides BC , CA , AB , respectively. Prove that all the sides of the hexagon T A O C T B O A T C O B are equal.

6. yields of milk in the city mayoral election are as follows. If in the next round of voting no candidate won more than half of the votes shall then be the next round with all the candidates, except for the last number of votes.(Never two candidates do not gain a tie, if the candidate won more than half the votes, he becomes mayor and elections are over.) Each voter in each round of voting for a candidate. If the candidate is left to the next round, the voter shall vote for him again. If the candidate was eliminated, then all voters vote for the same candidate from among those remaining. At the next election in 2002 running for the candidate. Bender became mayor, who finished the first round of k -th largest number of votes. Determine the largest possible value of k , if Bender was elected a) in 1002-th round, b) in the 1001-th round.

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65 Moscow Mathematical Competition

Conditions and objectives grade 8

(Olympics held 03.03.2002, work allotted five astronomical clock.)

1. the island two thirds of all the men are married and three fifths of all women zamuzhem.K Which percentage of the population of the island is married? (V. Gurovits)

Answer: 12/19 of the population of the island.

Solution. Suppose that the number of couples on the island of the same N , that is married to N female, married N men. Married women account for three fifths of all women of the island, so the island (5/3) N women.Married men make up two thirds of all the men of the island, so the island (3/2) N men. Only on the island (5/3) N + (3/2) N = (19/6) N residents, and the marriage is two N residents. Seeking share equal to (2 N ) / ((19/6)N ) = 12/19.

2. squared sum of the digits of A is equal to the sum of the digits of A 2 . Find all such double figures A . (A. Blinkov)

Answer: 10, 11, 12, 13, 20, 21, 22, 30, 31.

Solution 1. Note that A 2 < 99 2 = 9801 <9999. Therefore the sum of the digits A 2 less than 9 * 4 = 36. Since it is the square of the sum of digits of A , the sum of the digits of A less than 36 1/2 = 6, that is less than or equal to 5. There are 15 options: 10, 11, 12, 13, 14, 20, 21, 22, 23, 30, 31, 32, 40, 41, 50, out of which condition is satisfied only 9 options in response.

Decision 2. denote S ( n ) the sum of the digits of n . Note that adding two numbers "column" can only transfer units in the high end, and each such transfer reduces the amount of numbers by 9. Therefore the sum of digits of the sum of any number of components does not exceed the sum of numbers of terms. As a consequence, S ( n ) < n , since n = 1 + one ... one ( n units), with equality only for single digits, after the addition of another unit to the sum of the nine units already there transfer. In addition, by appending numbers to the right of zero sum does not change, that is, S (10 n ) = S ( n ).

Suppose that the required two-digit number A = AB = 10 a + B . with equality if and only if a 2 , 2 ab and b 2 - digit numbers (if this is the case, then the addition of 100 a 2 , 10 * 2 ab and b 2 hyphenation does not happen, because they contain all of its non-zero numbers in the various categories.) So, it is necessary and sufficient that a 2 , B 2 and 2 ab were less than 10, that is, 1 < a < 3, 0 < B < 3 and AB < 4. 

Hence, we find all of the options in the response.

3. Given the circle with diameter AB . Another okruzhnosts center at A intersects AB at point C , and AC <(1/2) AB . Common tangent of two circles for the first circle at D . Prove that CD is perpendicular to AB . (A. Zaslavsky)

Decision. denote E the point of contact of the second circle with common tangent through O - center of the first circle (see Fig.). Triangle AOD - isosceles, so / ODA = / OAD . Since OD | | AE , then / DAE = / ODA , where DEA = DCA (on two sides and the angle between

them). Hence, / DCA = / DEA = 90 o and straight DC and AB perpendicular. 

Remark. easy to verify that the problem statement is true for any radius of the second circle, smaller AB .

4. Two players in turn put on the board 65 * 65 for one piece. In this case, no single line (horizontal or vertical) should not be more than two pieces. Who can not make a move - lost. Who wins in a particular game? (A. Buccini, A. Ivanischuk)

A: wins second.

Solution 1. Suppose the first set on the board first piece. Note that the permutation contours board does not change anything. The same applies to the permutation of verticals. Therefore, we assume that the second player in addition allowed to swap any of the horizontal and the vertical.

After the first course of the first player, the second player rearrange horizontally and vertically so that the first block was in the middle of the vertical, but not in the center.

He went on to do the moves of the first player moves symmetrically about the center of the board. Then the second block would also be in the middle of the vertical, and the first will not be able to occupy the central square.It is easy to check that the second will always be able to make a balanced course (separately consider the case should move to the middle horizontal).

Decision 2. describe this game is different. There are two rows of 65 points each (one set point indicate the horizontal boards, another point - the vertical).

Setting checkers on the horizontal and vertical intersection corresponds to straight lines connecting the points that represent these horizontal and vertical.

Thus, the rules prohibit the conduct of a single point for more than two segments.

The second player has to play (except for the last move), so that after each stroke to straight lines formed unclosed polygon (possibly self-intersecting). Then, after the k -th move of the second player polygon will consist of 2 k links and go through 2 k 1 point ( k one row of pixels and k 1 - the other). Therefore, the first 64 course the second player can always continue broken or combine it with a separate segment, first held in the previous course. The last, 65th move, the second player must connect the beginning and end of a segment of a polyline, making it a closed curve passing through all 130 points. Then, the first player can not make a move.

5. In triangle ABC medians AD and BE n eresekayutsya at M . Prove that if the angle AMB a) direct, b) sharp, then AC + BC > 3 AB . (Bogdanov)

Solution 1. draw a triangle AMB median A and F BG and let N point of intersection (see Fig.). Note that MF = (1/2) MB = ME , as M - the point of intersection of the medians - divides the BE in the ratio of 2:1. Since /AMF < 90, The base of the perpendicular from the vertex A to the line BE , lies on the line of MB . Therefore AF < AE (equality is achieved in the case of / AMB = 90 o ), so that AN = (2/3) AF < (2/3) AE = (1/3) AC .Similarly, BN < (1/3) BC . Using the triangle inequality, we obtain the desired inequality: AC + BC > 3 ( AN + NB )> 3 AB . 

Decision 2. draw median CK and extend it to the segment of the same length, we get the point F , such that KF = CK (see Fig.). Quadrangle ACBF - parallelogram. Of ACF have: AC + AF > FC = 2 CK , mean, AC+ CB > 2 CK .

Built on the side AB as diameter circle. Since the angle AMB is not stupid, the point M does not lie inside the circle (this can be understood if we consider the point M 1 intersection CK with the circle, the angle AM 1 B - direct and / AMB < / AM 1 B ). So, MK > M 1 K = AK = (1/2) AB . But CK = 3 MK , so AC + CB >

2 CK = 6 MK > 3 AB . 

6. in check box m * n , each cell can be either alive or dead. Every minute together all the living cells die, and those dead who have had an odd number of live neighbors (on the side), come to life. Specify all pairs ( m , n ), for which there is an initial placement of live and dead cells, that life in the rectangle would last forever (ie at any given time at least one cell is alive)? (A. Gorbachev)

A: for all pairs of integers ( m , n ), except for the pairs (1, 1), (1, 3) and (3, 1).

Decision. enough to consider the case m < n . Suppose that for some n in the box 1 * n is eternally living arrangement. Then such arrangement exists in all rectangles m * n . It suffices to consider the following option: take forever living arrangement 1 * n and copy it to all rows rectangle m * n (see Fig., F - a living cell, M - dead). 

ZHMMZHZHMMZHMMZHMMZHZHMMZHMMZHMMZHZHMMZHMMZHMMZHZHMMZHMMZHMMZHZHMMZHMMZHMMZHZHMMZHMMZHMMZHZHMMZHMM

^ ^ ^ ^ ^ ^ ^ ^ ^ ^ZHMMZHZHMMZHMM

Then it will evolve columns as well as the cells in the corresponding column rectangle 1 * n . Indeed, all the cells of a living column die as the corresponding cell rectangle 1 * n . Each column of the dead cells at the top and bottom neighbors are dead, so the number of its neighbors is the number of live living adjacent columns, and, hence, the number of live neighbors for the corresponding cell in the box 1 * n . Because it is always at least one living cell, then built the alignment m * n is always at least one living column.

Note that for n = 2 and n > 4 ever-fresh arrangement of 1 * n there: each of the following setups has period 2, that is, every other minute is reset.

n = 2: LM <-> MF

even n > 4: ZHMMZHZHMMZH ... <-> MZHZHMMZHZHM ...

odd n > 4: ZHMMMZHZHMMZH ... <-> MZHMZHMMZHZHM ...

Also, there is the balance (as of period 2) in a rectangle 3 * 3: 

FFM MMZHMMM <-> ZHMZHMZHZH zhmm

Thus, in all the boxes except 1 * 1 * 1 and 3, the living arrangements of examples ever been built. We show that in boxes 1 * 1 * 1 and 3, any arrangement will sooner or later die. Case 1 * 1 is obvious. Case 1 * 3 can be considered (perhaps, counting from the second step), that the first cell is dead. Then there's 3 options that have at least one living cell. Evolution version MZHZH includes MZHM: MZHZH -> zhmm -> MZHM -> ZHMZH -> MMM, and the option MMZH zhmm symmetrical version, which is also included in the above example.

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65 Moscow Mathematical Competition

Conditions and objectives grade 9

(Olympics held 03.03.2002, work allotted five astronomical clock.)

1. ranks recruits stood face to sergeant. At the command of "left", some have turned to the left, some - to the right, and the rest - around. Always a sergeant can get in order, so that on either side of him was equally recruits, standing to face him? (folklore, language - Shapovalov)

A: always.

Decision. agree when sergeant stands in line, denoted by m the number of people standing in the line to the left of the sergeant to face him, and the letter n - the number of people standing on the right of the sergeant to face him.

Let the first sergeant will fall to the left edge of the line. Then his left no one will be ( m = 0). If and to the right of the sergeant, no one will be around to face him ( n = 0), then the problem is solved. Otherwise, ( n > 0), let the sergeant goes from left to right from person to person. If it passes rookie standing back to him, the number m is increased by 1, and the number n is not changed. If the sergeant is a rookie who stood facing him, the number n is reduced by 1, and the number m is unchanged. Otherwise both of m and n are not affected. Thus, the number m - n first negative, and while driving down the line sergeant may be increased by not more than one during the passage of each recruit. But when the sergeant comes to the right side, is the number n is zero, which means that the number m - n is non-negative. So, starting with a negative number, m - n , and adding to it a few times on one, we

got a negative number. So, at some point we had to get a zero, that is, at the moment m = n , and on both sides of the sergeant to face him was equally recruits.

2. Suppose a , B , C - the triangle, prove the inequality a 3 + B 3 three abc > C 3 (V. Senderov)

Solution. Using the triangle inequality a + B > C and the fact that a 2 - AB + B 2 > 0, we obtain the chain of inequalities:

a 3 + B 3 3 abc = ( a + B ) ( a 2 - AB + B 2 ) 3 abc> > C ( a 2 - AB + B 2 ) 3 abc = C ( a 2 2 AB + b 2 ) = C ( a + B ) 2 > C * C 2 = C 3 .

3. bisectors of angles A and C of a triangle ABC intersect described near circle at points E and D , respectively. Segment DE intersects the sides AB and BC at points F and G . Suppose that I - intersection point of bisectors of the triangle ABC . Prove that the quadrilateral bfig - diamond. (V. Zhgun)

Decision. Reflect triangle ABC in the line DE (see Fig.). By property inscribed angles / AED = / ACD = / BCD = / BED , so direct AE at this reflection becomes a straight line BE , similarly, direct CD on reflection becomes a straight line BD . Then the point of intersection of AE and CD , then there is a point I , the reflection goes to the point of intersection of BE and BD , that is, the point B . Thus, the triangleIFG will at our reflection in the triangle BFG , where IF = BF , IG = BG and direct BI is perpendicular to DE . But a ray of BI - bisector FBG , and therefore the triangle FBG - isosceles, that is, FB = BG .

So all sides of the quadrangle BFIG equal, then it is a rhombus. 

4. Find all the integers x and y , satisfying the equation x 4 -2 y 2 = 1. (V. Senderov)

Answer: x = 1, y = 0 or x = -1, y = 0.

Decision. signs x and y can be chosen arbitrarily, so we will look for only non-negative solutions. It is clear that x - an odd number, x = 2 t 1. erepishem P equation in the form

x 4 = -1 ( x -1) ( x 1) ( x 2 +1) = 2 t * (2 t 2) * (4 t 2 4 t two) = 2 y 2 .

Now we see that y - an even number, y = 2 u . An equation nan eotritsatelnye t , u :

t ( t one) (2 t ( t 1) one) = u 2 .

Number t , t one and 2 t ( t one) 1 are relatively prime, and the Institute of Chemistry value you - a perfect square. Hence, each of them is also a perfect square. This is possible only when t = 0 (the only pair of consecutive perfect squares - is 0 and 1). Then u = 0.Z nachit, x = + 1, y = 0.

5. placed in a number of n lights and lit some of them. Every minute after that all the bulbs at the same time change the state of the next rule. Those lights that burned in the last minute, will go out. Those who at the last minute did not burn and neighbors with exactly one lighted lamp, light up. For which n can be so light a few bulbs in the beginning, so that later was found at any time, at least one lighted lamp? (A. Gorbachev)

A: for all n , but 1 and 3.

Decision. See solution 8.6 , where this problem corresponds to the case rectangles 1 * n (burning bulbs - they are living cells).

6. acute triangle cut rectilinear cut into two (not necessarily triangular) part, then one of these parts - again in two, and so on, at each step, choose any of the existing parts and cut it (straight) into two. After a few steps it was found that the original triangle split into several triangles. Can they all be obtuse? (Galperin)

Answer: they can not.

Decision. induction on the number of available units following statement: after each cut there of (polygon) with at least three non-obtuse angles.

The induction base. First, there is one part, and a triangle with three acute angles.

Induction. Let the next step of a polygon M with at least three non-obtuse angles and we make another cut.

If we do not cut the polygon M , then the induction step is proved. Suppose we cut M in a straight line into two parts. Denote the point of intersection of this line with the polygon M letters A and B .

If point A is inside one of the sides of the polygon M , then cutting line forms with the party the two adjacent corner, at least one of them - not dumb. If the point A - the top of the polygon, then the corresponding angle of the polygon is divided into two angles cut, the amount of which is less than 180 o . So, the more of them are not stupid. If the angle of A itself was not stupid, it is obvious that both of the resulting angle is also not stupid. Thus, after the cut to the point A is no longer adjacent obtuse angles (at least one). Similar arguments about the other end of the section indicates that no additional obtuse angle appears there.

By the induction polygon M has at least three non-obtuse angles. So after cutting the two corners of the resulting polygons in an amount not less than five. But according to the pigeonhole principle, at least one of them is obtuse angles less than three. The induction step is proved.

Thus, we are done. So, after the original triangle split into triangles, at least one of them will be at least three non-obtuse angles, ie, all the angles of this triangle are not stupid.

65 Moscow Mathematical Competition

Conditions and objectives class 10

(Olympics held 03.03.2002, work allotted five astronomical clock.)

1. tangents of the angles of the triangle - the natural numbers. So that they can be equal. (A. Zaslavsky)

Answer: The tangents of the angles of a triangle are equal to 1, 2 and 3.

Solution 1. denote the angles of a triangle (in decreasing order) by  ,  ,  . Then 0 < <  / 3, so 0 <TG < 3 1/2 . Since TG  - positive integer, then it is 1,  =  / 4. Then  +  = 3  / 4, whence

-1 = TG (  +  ) = (TG  TG +  ) / (1 - TG  TG  ), TG  TG +  = TG  TG  - 1, (TG  - 1) (TG  - 1) = 2.

Number 2 - simple, so one bracket is equal to 1 and the other 2. Hence we obtain: TG  = 2, TG  = 3.

Decision 2. Draw on a checkered paper triangle ABC, as shown in the figure. Counting of cells is easy to see that TG / BAC = 3, TG / ACB = 2. We now point H and D . Again, by counting cells that ADBH - rhombus with equal diagonals, it is a square. Therefore / ABC =  / 4, and TG / ABC = 1. So, draw a triangle satisfies. 

We show that other values may not be tangents. As proved in the decision 1, the smallest of the angles of a triangle always equals  / 4. While the other corners strictly more if the second angle is also equal  / 4, the third angle would be equal to  - (2  / 4) =  / 2 and his tan was not defined. Thus, one of the tangent is equal to 1 and the other two - more than 1.

Since the sum of the angles of a triangle is equal to  , you can not change the value of the tangent of one angle, without changing the values of the other two. So, apart from option 1, 2, 3, it remains to consider the case when one of the tangent is equal to 1 and the other two - more than 3. In this case it was necessary to increase the two angles of the triangle ABC , leaving the third unchanged, it is impossible for the same theorem, the sum of the angles of a triangle. So, the option in the figure - the only one.

2. about positive numbers a , B , C is known that (1 / a ) + (1 / B ) + (1 / C ) > a + B + C . Prove that a + B + C > 3 abc . (Zlobin)

Decision. Multiplying inequality (1 / a ) + (1 / B ) + (1 / C ) > a + B + C on the common denominator, we obtain the equivalent inequality bc + AC + AB > ( a + B + C ) abc .

Now we prove an auxiliary inequality ( a + B + C ) 2 > 3 ( AB + bc + ca ):

( a - B ) 2 + ( B - C ) 2 + ( C - a ) 2 > 0 <=> <=> 2 ( a 2 + B 2 + C 2 ) > 2 ( AB + bc + ca ) <= > <=> a 2 + B 2 + C 2 > AB + bc + ca <=> <=> a 2 + B 2 + C 2 2 ( AB + bc + ca ) > 3 ( AB + bc + ca ) <=> <=> ( a + B + C ) 2 > 3 ( AB + bc + ca ).

Is easy to obtain the desired inequality: ( a + B + C ) 2 > 3 ( AB + bc + ca ) > 3 ( a + B + C ) abc , a + B + C > 3 abc . 

3. In convex quadrilateral ABCD points E and F are the midpoints of the sides BC and CD respectively. Segments AE , AF , BF divide rectangle into 4 triangles, squares which are sequences of natural numbers. What is the most important area of the triangle ABD ? (Shestakov)

A: 6.

Decision. vacancy triangles are n , n 1, n 2, n three. Then the area of the quadrilateral ABCD is 4 n six. Area of the triangle BCD four times the area of a triangle ECF . Hence, S ABD = S ABCD - S BCD < (4 n 6) -4 n = 6.Equality is achieved, if the triangle ECF has the smallest area of the four triangles. 

It remains to prove that the value of 6 is possible. An example is an isosceles trapezoid with bases AD = 6, BC = 4 and height 2 (the numbers in the figure indicate the areas of the triangles).

4. Each viewer who bought a ticket in the front row theater, took one of the seats in the front row. It turned out that all the seats in the first row are busy, but every viewer is sitting in the wrong place. Taker can swap neighbors if both do not sit in their seats. Always whether it can all sit down soon? (Shapovalov)

A: always.

Decision. enumerate all viewers numbers of tickets 1, ..., n . Suppose, to be specific, it's the right place has a number n , and the far left - the number 1. We solve the problem of induction on the number of viewers. We reduce the problem to the same problem with a smaller number of spectators, viewers grafting n in place.

Let chopper operates as follows. He transplants the viewer n the right if its right neighbor, it does not sit in his place. If the viewer n can not transplant the right place to k , then the ( k 1)-th place sits a spectator k . Choose the largest m such that the audience numbered k , k one, ..., k + m -1 are sitting on the ground k one, k two, ..., k + m , respectively. There are two cases: k + m < n and k + m = n .

In the first case ( k + m 1)-th place sits a spectator j , j is not equal to k + m . So safely transplanted j left, then left again, and so on, until he swapped with the audience n . As a result, can usher transplant viewer n one place to the right, and he can repeat this procedure as long until one of two things: the audience n be in place, or will meet the second case.

In the second case could usher sit in their places viewers k , k 1, ..., n , transplanting viewer n the right to its place.

Thus, some spectators at the right end of the row to sit in their seats, and the rest - no. The problem is reduced to the original with a smaller number of viewers.

5. yields of milk in the city mayoral election are as follows. If in the next round of voting no candidate won more than half of the votes shall then be the next round with all the candidates, except for the last number of votes.(Never two candidates do not gain a tie, if the candidate won more than half the votes, he becomes mayor and elections are over.) Each voter in each round of voting for a candidate. If the candidate is left to the next round, the voter shall vote for him again. If the candidate was eliminated, then all voters vote for the same candidate from among those remaining. 

At the next election in 2002 running for the candidate. Bender became mayor, who finished the first round of k -th largest number of votes. Determine the largest possible value of k , if Bender was elected a) in 1002-th round, b) in the 1001-th round. (B. Frenkin)

Answer: a) k = 2001, b) k = 1.

Decision. a) Ostap could not take the last place in 2002 in the first round, because otherwise he would retire immediately from the candidates. Therefore k < 2001.

Let all the candidates in the first round with almost equally, Ostap took the last place in the next round, each received all of the votes of the candidate retired. Then Ostap win at a time when the number of candidates to reach half of the retired. This happened just in 1002-th round.

Perform accurate count when the candidates in the first round to score 10 6 10 6 1, ..., 10 6 2001 vote. Then in 1001-th round at Ostap still less than half the votes, namely, the votes of all the candidates who took the last place in the 1001 first round. However, in the 1002-th round him more than half of the votes. Indeed, the and all check that it is less than twice the number of votes Ostap. 

b) Assume that k > 1. Candidate, who took first place in the first round, called a favorite. Those who dropped out in the first thousand rounds, called outsiders, and all the other candidates except Ostap - leaders. As the number of outsiders, 1000, and in 1001 the leaders, one of the leaders did not get a vote outsiders. In the first round (and later), he had more votes than any outsider (because ultimately dropped an outsider, but this is not the leader). Therefore, the favorite, who collected the first round of the greatest number of votes, not one of the outsiders.

The maximum number of votes that could collect Ostap to 1001-th round - is all the voices of outsiders at the time of departure of each one and the original voices of voters Ostap. Any one of the leaders in any of the first thousand rounds (and even more in 1001-m) has more votes than the outsider of this tour. Favorite certainly has more than Ostap had in the first round. Therefore, the leaders are in total in 1001-th round more votes than Ostap, and he can not be a winner.

Additional question (in the conditions of the proposed tasks in the Olympics is not included): Will the answer to the problem, if the voices of the outgoing candidate arbitrarily divided between the other?

6. possible to color all points of the square and the circle in black and white so that the set of white points of these figures were similar to each other and a set of black dots were similar to each other (possibly with different coefficients of similarity). (Galperin)

A: You can.

Decision. 

Consider a coloring of the square: inscribe a circle in a square and paint the black point of the square, lying outside the circle, inscribe a circle in the resulting square with sides parallel to those of the original square. Paint the white point of the circle lying outside the "small" square. By the same rule, let's give a small square, etc. Note that we consider the boundary points lie "inside" shape. Thus, the boundary of each square is painted black, except for the four points of contact of the circle inscribed in a square, and the boundary of each round - white, except for the four vertices of the square inscribed in a circle.

Let the side of the original square is a , then the side of the small square is a / 2 1/2 . Consequently, the length of the sides of the squares tend to 0. Therefore, all points except the center will be painted. Paint the center black.

Obviously, the set of black points of the square like so many black dots circle inscribed in this square (the second follows from the first by a homothety with center at the center of the square and a factor of 1/2 1/2 . A lot of white points of the square just coincides with the set of white points of the inscribed in a circle.

65 Moscow Mathematical Competition

Conditions and objectives grade 11

(Olympics held 03.03.2002, work allotted five astronomical clock.)

1. tangents of the angles of the triangle - the natural numbers. So that they can be equal. (A. Zaslavsky)

Answer: Tangents angles of a triangle are equal to 1, 2 and 3.

Decision. See solution 1.10

2. Prove that on the graph of y = x 3 can be noted a point A , and the graph of the function y = x 3 + | x | 1 - a point B , the distance AB is less than 1/100. (A. Spivak , Khachaturian)

Decision. put t = 100. Point B with coordinates ( t , C ), where C = t 3 + t one, obviously, lies on the graph of y = x 3 + | x | 1.

Consider a positive number R = C 1/3 - t . If ( t + R ) 3 = C , so that the point A with coordinates ( t + R , C ) lies on the graph of y = x 3 .

The distance between points A and B is equal to R . But it follows from ( t + R ) 3 = C = t 3 + t 1 that 3 t 2 R three tr 2 + R 3 = t one, 3 t 2 R < t 1, R <( t 1) / (3 t 2 ) = 101 / (3 * 100 * 100) <1/100.

3. Each viewer who bought a ticket in the front row theater, took one of the seats in the front row. It turned out that all the seats in the first row are busy, but every viewer is sitting in the wrong place. Taker can swap neighbors if both do not sit in their seats. Always whether it can all sit down soon? (Shapovalov)

A: always.

Decision. See solution 4.10

4. in increasing sequence of natural numbers, each number, starting from 2002, is a factor of the sum of all previous numbers. Prove that there is a sequence number, starting with which each number is the sum of all the previous ones. (Shapovalov)

Decision. Consider the sequence of partial k n , n > 2002. Let the quotient obtained by dividing the sum of S n -1 previous members for the next term a n is equal to k n , that is, S n -

1 = a n k n . Since the next term of a n 1divides the sum S n -1 + a n and a n one > a n , a k n 1 = S n / a n 1 = ( S n -1 + a n ) / a n 1 <( S n -1 + a n ) / a n = k n 1. private That is not increased: k n 1 < k n . Therefore, from a certain point (when n > N ), k n = k .But then, for n > N we have a n 1 = ( S n -1 + a n ) / k = a n + ( a n / k ), that is, at this point we get a geometric progression a n one = (( k 1) / k ) a n = (( k one) n - N ) / ( k n - N )) a N . We find that a N is divided by an arbitrarily large degree k . Hence, k = 1, as required. 

5. Suppose AA 1 , BB 1 , CC 1 - high-angled triangle ABC ; O A , O B , O C incenter of the triangle A B 1 C 1 , B C 1 A 1 , C A 1 B 1 , respectively; T A , T B , T C - the point of contact of the inscribed circle of triangle ABCwith sides BC , CA , AB , respectively. Prove that all the sides of the hexagon T A O C T B O A T C O B are equal. (L. Emelyanov)

Decision. Denote I the incenter of triangle ABC , and let R - the radius. Let / BAC =  (see Fig.).

We will prove that T B O A   |    AB . Find AB 1 = AB cos  , \ AC 1 = AC cos  . The triangle BAC and B 1 AC 1 angle BAC general and AB 1 / AB = AC 1 / AC = cos  . Thus, the triangles B 1 AC 1 and BAS similar with similarity coefficient equal to cos  . Let X - the projection of O A to AB . Segments AX and AT B - it stretches from the top of A to the points of

contact of the parties concerned with the inscribed circle in a similar triangleB 1 AC 1 and BAS , that is, AX and AT B - corresponding elements in similar triangles. Therefore AX = AT B cos  . But if the Y - the projection of T B to AB , then AY = AT B cos  . This means that AX = AY , that is the point of X and Y are the same. It follows that T B O A   |    AB . 

Similarly, the argument given above prove that T_AO_B   |    AB, T C O B   |    BC , T B O C   |    BC , T A O C   |    CA , T C O A   |    CA .

Obviously, IT C   |    AB , IT A   |    BC , IT B   |    CA, as well as IT A = IT B = IT C = R . In quadrangle T B O A T C I opposite sides are parallel, hence, T B O A T C I - a parallelogram in which IT B = IT C = T B O A = T C OA = R . Similarly, we find that T C O B = T A O B = T A O C = T B O C = R , so all sides of the hexagon T A O C T B O A T C O B are equal to R . 

Remark . From the obvious solutions that the hexagon T A O C T B O A T C O B is centrally symmetric. Less obvious is that its center of symmetry lies on the line connecting the centers of the inscribed and circumscribed circles of a triangle ABC . The reader is invited to prove the allegations themselves.

6. yields of milk in the city mayoral election are as follows. If in the next round of voting no candidate won more than half of the votes shall then be the next round with all the candidates, except for the last number of votes.(Never two candidates do not gain a tie, if the candidate won more than half the votes, he becomes mayor and elections are over.) Each voter in each round of voting for a candidate. If the candidate is left to the next round, the voter shall vote for him again. If the candidate was eliminated, then all voters vote for the same candidate from among those remaining. At the next election in 2002 running for the candidate. Bender became mayor, who finished the first round of k -th largest number of votes. Determine the largest possible value of k , if Bender was elected a) in 1002-th round, b) in the 1001-th round. (B. Frenkin)

Answer: a) k = 2001, b) k = 1.

Decision. See solution 05.10 </ HTML>

66 Moscow Mathematical Competition. Grade 8.

Competition was held March 2, 2003, to complete the task was given to five astronomical clock.

Tasks

1. 4 people in the family. If Masha will double scholarship, total revenue of the entire family will increase by 5% if the mother instead double the salary - 15% if the same salary will double the Pope - by 25%. By what percentage will increase revenue of the entire family when my grandfather would double pension?

2. Invent a ten-digit number, which featured no zeros, such that when added to it the product of its digits is a number of the same product numbers.

3. Can I paint some squares of the board 8 * 8, so that in any 3 * 3 square was exactly 5 shaded cells, and in each rectangle 2 * 4 (vertical or horizontal) - exactly four shaded cells?

4. In triangle ABC on the sides AC and BC to take a point X and Y such that /   ABX = /   YAC , /   AYB = /   BXC , XC = YB . Find the angles of the triangle ABC .

5. 15 cities in the country, some of them are connected by airlines belonging to the three airlines. We know that even if any of the airlines to stop flights, you can get from any city to any other (possibly with changes), using the remaining flights of the two companies. What is the smallest number of airlines may be in the country?

6. Boris conceived integer greater than 100. Cyrus calls an integer greater than 1. If Borino number is divided by this number, Cyrus won, or Boris deduct from their numbers called, and Kira calls the next number. She was not allowed to repeat the numbers mentioned above. If the number becomes negative Borino - Kira loses. Does it have a winning strategy?

Solutions

1. A: 55%.

Decision. If Masha will double scholarship, family income will increase the size of the grant. Consequently, car stipend of 5% of total revenue. Similarly, the mother's salary is 15%, and my father's - 25%. The remaining share of 100% -5% -15% -25% = 55% are retired grandfather. So, if he is to double pensions, family income increases by 55%.

The same solution can be formulated differently. If all members of the family suddenly began to pay twice as much as the total revenue would have increased by 100%. Of these 100 accounts for 5 percent of Masha, 15 - mum, 25 - the pope and the other 55 - to his grandfather.

2. A: for example, 1111111613.

Decision. * First, we find what some number such that the problem is not necessarily ten-. For example, try to find a number such that, after adding to it the product of numbers they simply reversed. If the last digit numbers - 13, to swap them, you need to add 18. That the product numbers were 18, you need to add is the number 6. The required number is 613 613 6 * 1 * 3 = 631. Similar numbers: 326 3 * 2 * 6 = 362, 819 +8 * 1 * 9 = 891, etc.

In the least-known example - 28 +2 * 8 = 44 - the numbers vary, but their work is saved! Another example: 214 2 * 1 * 4 = 222.

Now to get a ten-digit number is assigned to the left missing number of units. Product numbers will not change. For example, 1111111613 +18 = 11111111631. Can also be attributed to the beginning and end of an equal number of twos and fives, for example, 5555282222 +10 * 10 * 10 * 10 * 16 = 5555442222.

Other interesting examples encountered in the work of students: 1111159121, 1111954511, 1111332123, 1111113148, 2356478911, 1111553431, 1952219522, 1231451671, 3355211211, 5132486791.

3. A: impossible.

Decision. Suppose that a coloring is possible. In all the figures are not hatched cells, about which found that they were not painted, marked by shading cells, about which it is not known whether or not they are painted, double ruling - obviously painted.

Fig. 11

8 * 8 board can be cut into eight rectangles 4 * 2, so it's exactly 8 * 4 = 32 shaded cells.

It can also be cut into four squares 3 * 3, three boxes 2 and 4 * corner square 2 * 2 (Fig. 11). The four squares 3 * 3 and three rectangles 4 * 2 for 4 * 5 +3 * 4 = 32 cells, so the angular box 2 * 2 should not be a single shaded cells. Similarly, we can prove that in the other corner squares 2 * 2 no shaded cells.

Fig. 12

Board without corner squares 2 * 2 can be cut into six rectangles, 4 * 2 (Fig. 12). We get that all on board 6 * 4 = 24 shaded cells. Contradiction.

Note. Proving that corner squares 2 * 2 are not shaded, it is possible to act in a different way, like this. Place the four squares on the board 3 * 3, and two rectangles 4 * 2 as shown in Fig. 13. On the rest of the space is 32 - (4 * 5 +2 * 4) = 4 shaded cells, and there you can put two rectangles 4 * 2, overlapping on a square 2 * 2. In each of these four shaded cells, and together they have four shaded cells - then painted over their intersection, and a corner square 2 * 2 should be completely filled. Contradiction.

We give another solution . We show that even a board 6 * 6 can paint specified in the problem properly. Because it can be cut into four squares 3 * 3, it should be exactly 4 * 5 = 20 shaded cells. We cut it into four rectangles and 4 * 2 2 * 2 box (Figure 14). These rectangles 4 * 4 = 16 cells, then a 2 * 2 remaining focused 20-16 = 4 cells, ie, all the cells in it painted.

Fig. 13

Shifting the other four rectangles, we can prove the same thing about the 2 * 2 squares in the other corners, as well as in the center of the box (Fig. 15). The remaining cells are to be unpainted, since each rectangle 4 * 2 already has four shaded cells. Now easy to square 3 * 3, in which only four shaded cells (Fig. 16).

Figure 14. Figure 15. Figure 16.

4. A: equilateral triangle with all angles of 60 o .

Decision. For outside corners BXC and AYB triangles ABX and CAY write equality

/   BXC = /   ABX + /   BAX , /   AYB = /   CAY + /   YCA

(Fig. 17). Since, by hypothesis / BXC = / AYB , / ABX = / CAY , we find that / BAX = / YCA , ie, triangle ABC is isosceles, AB = BC .

Fig. 17

Consider the triangles XBC and YAB . They are two sides and an angle not between them : /   BXC = /   AYB , XC = YB , BC = AB . These triangles are either equal or /   XBC + /   YAB = 180 o (we prove this below), but the second case is impossible, since /   XBC + /   YAB < /   ABC + /   CAB = 180 o - /   ACB <180 o . Hence, XBC = YAB , and hence, /   ABC = /   BCA , and the triangle ABC is equilateral.

We now prove the above stated fact. So, in triangles XBC and YAB are two sides and the angle between them are not: /   BXC = /   AYB , XC = YB , BC = AB . Note on the line XB point B ' at a distance XB ' = YA . Point B 'can be both inside and outside the segment XB (Fig. 18, a , b ). Triangles XB'C and YAB are equal on both sides and the angle between them ( XC = YB , XB ' = YA , /   CXB ' = /   bya ). Then CB ' = CB . If B and B ' are the same, the triangles XBC and YAB equal. If B ' is different from B , then an equilateral triangle B'CB corners B'BC and BB'C equal, and hence, /   XBC + /   XB'C = 180 o , and /   XBC + /   YAB = 180 o .

Fig. 18

5. A 21 airline.

1) We show that less than 21 lines can not do. First we prove that if the 15 cities connected by airlines so that you can get from any city to any other, the airlines are not less than 14.

Denote the number of airlines in k . If we remove one of them, the city can break up into two separate groups between which there is no air traffic, may not dissolve. After removal of the following airlines at most one group can break up into two, the number of disparate groups will increase by no more than one. Continuing this reasoning, we find that when the last number of airlines disparate groups can not exceed k 1. Since the 15 cities without airlines - is 15 groups, k >   14 .

Denote the number of lines from the airlines through a , B and C . By the above, any two companies with at least 14 lines, ie, a + B > 14, B + C > 14, C + a > 14. Adding these inequalities, we obtain 2 ( a + B + C ) > 42, ie, the three companies in the amount of not less than 21 airlines.

Fig. 19

2) is an example with 21 airline. You can use any of the examples shown in Fig. 19, a , b .

6. A : Yes.

We present a strategy game for Kira. First, it should be called "2". If Boris is not lost, it was an odd number. After subtraction of two odd and it will remain.

Cyrus now be called "3." To understand what numbers we "weed out" with this move, consider the remains of the original Brad divided by 6 (6 = LCM (2,3)). If the balance was 0, 2 or 4, Cyrus would have won the first move. If the balance was 5, after the first stroke he would have 3, and Kira would have won the second swing. Thus, the game continues, if the initial number of Borino gave a remainder of 1 or 3 divided by 6. After two passages from it subtracted 2 +3 = 5, and the remainder when divided by 4 will be 2 or 4.

The third number is Kira - "4." Consider the possible remains of Brad numbers before this move when divided by 12 (12 = LCM (2, 3, 4)). Possible remainders when divided by 6 - 2 or 4, and then divided by 12 - 2, 4, 8 or 10. The game continues, if the remainder were 2 or 10 (otherwise divided by 4), after subtracting 4, they become respectively 10 and 6.

The fourth number Kira - "6". If the number of Borino gave remainder 6 when divided by 12, it will be divided by 6, or after subtracting 6 will remain the only option - the remainder of the "4". Cyrus calls the "16" to 16 after subtracting Borino number was divided by 12. You can then call the "12", certainly share! Table. 1 shows the moves of Kira and the numbers that could be at Bori, if he does not play on this course; k denotes a natural number (in each cell of the table, in general, his own).

Progress Kira

Possible number of Bori to its course

Possible number of Bori after stroke

2 any 2 k 1

3 6 k 1, 6 k 3, 6 k 5 6 k four, six k 2

4 12 k two, 12 k 4, 12 k eight, 12 k 10 12 k ten, 12 k six

6 12 k 6, 12 k ten 12 k 4

16 12 k 4 12 k

12 12 k -

Note that after the fifth stroke Borja could deduct from their own number a 2 +3 +4 +6 +16 = 31, ie, Borino number could be negative. Let us now look at the process of the game, knowing that it suffices to consider the remains of the fission Borini 12 (Table 2).

Progress Kira

Possible residues modulo 12 Borini numbers before its course

Possible residues modulo 12 Borini number after stroke

2 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 11, 1, 3, 5, 7, 9

3 1, 3, 5, 7, 9, 11 10, 2, 4, 8

4 2, 4, 8, 10 10.6

6 6, 10 4

16 4 0

0 12 k -

Thus, every move Kira "weeds" some leftovers. Already before the fifth course, it is clear that the number has the form Borino 12 k four, but you have to be called "16" in order to bring it to the form 12 k , and then to be called "12".

Note. Olympiad participants were invented and other sequences of moves Kira. Among them, special attention should be consistency with the minimum amount of numbers - 6, 4, 3, 2, 5, 12 (try to prove its minimality own).

66 Moscow Mathematical Competition. Grade 9.

Competition was held March 2, 2003, to complete the task was given to five astronomical clock.

Tasks

1. Bullies Jay and Bob in class drawing drawn tadpoles (four circles in the figure - the same radius, triangle - equilateral, horizontal side of the triangle - the diameter of the circle). Which of the tadpoles have bo'l `shimi area?

2. product of five numbers is not equal to zero. Each of these numbers is reduced by one, and their work has not changed. Give an example of such numbers.

3. In the shop three floors, which can be moved between only by elevator. Study attendance floors store revealed that since the beginning of the day and before the store closed: 1) from buyers outside the elevator on the second floor, half goes to the first floor, and the other half - the third, 2) buyers coming out of the elevator, less third does so on the third floor. Which floor buyers often went from the first floor to the second or third?

4. Is chocolate in the shape of an equilateral triangle with sides n , divided by grooves on equilateral triangles of side 1. Play two. For the course can break off a triangular piece of chocolate along the grooves, eat it, and the remainder to pass to the enemy. The one who gets the last piece - a triangle with sides 1, - the winner. Those who can not make a move early loses. Who wins in a particular game?

5. circle inscribed in a right triangle ABC with hypotenuse AB . Let K - middle arc Sun not containing the point A ; N - the midpoint of AC , M - the ray KN with a circle. At points A and C draw a tangent to the circle that intersect at point E . Prove that /   EMK = 90 o .

6. put in jail 100 prisoners. The warden said to them: "I will give you a night to talk to each other, and then planted the individual cameras, and chat you can no longer. Sometimes I'll take one of you in the room, in which there is a lamp (initially it is disabled). Leaving room, you can leave the lamp turned on as well as off.

If at any time any of you tell me what you all have been in the room, and will be right, I'll let you all to freedom. And if you wrong - all will feed the crocodiles. And do not worry that someone will forget - if you remain silent, then all will visit the room, and no one no visit to the room will not be the last. "

Come up with a strategy that ensures the release of prisoners.

Solutions

1. A : area of figures equal.

First solution. Let the area of the circle is equal to S env , and the area of an equilateral triangle with sides equal to the diameter of the circle is equal to S tr . Since the triangle O 1 O 2 O 3 is equilateral, its angles are 60 o , except the length of its sides equal to the diameter of the

original circle, so its area is S tr .. Hence, the area of each of the shaded sector in Fig. 20 equals S env / 6, and the area of the shaded part is

Fig. 20 Fig. 21

S tr -3 ( S env / 6) = S tr - ( S env / 2)

Therefore, the area is second tadpole S env + S tr - ( S env / 2) = S env + ( S env / 2). Obviously, this is the same as the second area of the tadpole.

The idea of the second solution . Impose a tadpole to another as shown in Fig. 21. It is easy to see that if cut off from the first tadpole hatched segments, and turn them around points A 1 and B 1 to 180 o , you get exactly the second tadpole.

2. A : for example, 5, 6, 7, 8, 1.

We can construct another example: suppose the first four numbers - two. Obtain an equation for the fifth day of x : 16 x = x -1, where x = -1 / 15.

3. A : From the first to the third floor for the day came less customers than from the first to the second.

First solution. Suppose that for the whole day on the first floor of the elevator entered x buyers, on the second - y , the third - z . Note that the number of customers who came out of the elevator on each floor, is the number of customers that are included on the same floor.

By the condition of customers included on the second floor, half goes down, and half - up. Hence, from the second floor to the third goes y / 2 customers, and the same from the second to the first. The second condition can be written as: z <( x + y + z ) / 3. This is equivalent to 2 z < x + y .

From the first floor to the third was committed z - ( y / 2) travel, as all on the third floor of the elevator came z man, and y / 2 of them came from the second floor. And from the first to the second raised those buyers enter the elevator on the ground floor, who did not go to the third, ie, x - ( z - ( y / 2)). To solve the problem you want to compare these two expressions. But the

inequality z - ( y / 2) < x - ( z - ( y / 2)) is equivalent to already proved inequality 2 z < x + y . This proves that from the first floor to the third that day came less customers than from the first to the second.

The second solution . Denoted by n 12 the number of trips from the first floor to the second. Similarly, we define the number n 13 , n 21 , n 23 , n 31 , n 32 . All trips will be divided into three groups: a trip to the third floor, a trip to the third floor, and travel between the first and second floor. According to the problem statement, the first group is less than a third of all trips. On the other hand, it is quantitatively equal to the second, as the number of customers who have come to the third floor is the number of emigrants from the third floor. Hence, the latter group is more than a third of all trips:

( n 31 + n 32 ) = ( n 13 + n 23 ) <( n 12 + n 21 ).

Substituting the inequality n 21 = n 23 , we obtain n 13 < n 12 : on the first floor often went to the second rather than the third.

4. A : if the number n is prime, then the second player wins, otherwise the first player wins.

After the first turn form an isosceles trapezoid. Let's see what the figures are produced in subsequent passages.

Fig. 22 Fig. 23

Let go at some one of the players got a chocolate bar in the shape of an isosceles trapezoid with a smaller base a large base, and B (the length of the side of a trapezoid is equal to B - a ). If he broke off a triangle whose side is less than B - a , then the other player can break off a triangle with side 1, and that player loses (Fig. 22}.) Therefore, in this situation, he should break off triangle with side B - a . After this move is a parallelogram, the length of whose sides are a and B - a .

Suppose a player is in the form of a chocolate bar parallelogram with sides a and B , and a < B . Then by a similar argument it should break off a triangle with sides a . After this move is an isosceles trapezoid with bases B -a and B .

Thus, if at some point one of the players got a chocolate bar in the shape of a parallelogram with sides a and B ( a < B ), then after two turns he will get a chocolate bar in the shape of a parallelogram with sides a and B - a .

If one of the players got a parallelogram with equal sides (ie, diamond), after his stroke a triangle (Fig. 23).

We now present a winning strategy for the second player with a simple n . Him enough each time to break off a piece of the largest size. We show that this strategy results in a win. Let

the first player to break one triangle with side k . After the move of the second player form a parallelogram with sides k and ( n - k ). These numbers are relatively prime, as n - prime. Furthermore, each move of the second player will be obtained parallelogram (until eventually fail diamond). Lengths of the sides of the parallelogram are prime (if k and n are relatively prime, k and ( NK ) is also relatively prime). Mean length of the sides of the diamond is also relatively simple. But this means that the diamond will side 1! The first will be forced to break off from it a triangle with one side, then the second wins.

If n = 1, the first is already won. Suppose now that n is composite. Denoted by p any prime divisor of n . Let the first break off the first triangle with a side of P , and then breaks off every time the biggest piece. After some time, the second player will get a triangle with sides of p and, as analyzed above, a few moves lose.

Note . Note that this process is nothing else than the Euclidean algorithm.

5. Suppose that O - center of the circle (Fig. 24). Since the triangle ABC is rectangular, point O coincides with the center of the hypotenuse AB . Angle NOK line: indeed, the middle line NO triangle ABC is parallel to the side BC , and the line OK is the height of an equilateral triangle BOC (as it is the bisector of the triangle).

 Fig. 24

It is easy to see that the points E , N and O are collinear. Indeed, right triangles ECO and EAO are on a leg and the hypotenuse, so EO - bisector AEC . On the other hand, the triangle AEC - an isosceles triangle, as a tangent to the circumference of a single point are equal. Since segment EN - the median of the triangle, it is also the bisector. Therefore rays EN and EO coincide as bisector AEC .

The points A , E , C and O lie on a circle, as /   ECO = /   EAO = 90 o . The theorem on intersecting chords of a circle that

AN * NC = EN * NO . (1)

Applying the same theorem to the chords AC and MK original circle, we obtain

AN * NC = MN * NK . (2)

Combining (1) and (2), we find that

MN * NK = EN * NO .

By the theorem, return to the theorem of intersecting chords we find that the point M , K , E and O lie on a circle.

Angles of EMK and EOK equal, as based on one arc. But the angle EOK - direct, hence the angle EMK - straight.

6. Prisoners choose one specific person (let's call it "counter"), which will be considered prisoners of such a system: If you are coming into the room, he finds that the light is on, it adds to the already count the number of prisoners of the unit and turn off the light If the light is off, it do not do anything, is returned to his cell. Each of the remaining prisoners of acts by the following rule: If you are coming into the room, he finds that the light is not on, and before that he had never turned the lights on, it includes it. In other cases, it does not change anything. Counted the number of prisoners becomes 99, "count" says that all prisoners have visited the room.

Indeed, every prisoner, except for "counter", turn on the light in the room more than once. When the "counter" were numbered 99, he can be sure that all the other prisoners had already been in the room at least once, except for the fact he had already been in the room. It turns out that by this time all the prisoners were in the room clearly for once.

It remains to prove that each of the 99 prisoners turn on the light. Suppose that this is not the case - the light will be turned on at least 99 times. Then, starting with some of the day n , the light will not turn on. Since no go in the room will not be the last for the counter, he will visit the room after that date (eg, m -day, m > n ). If the light while burning, he turned it off. Hence, from ( m 1)-th day of the light will be switched off at all times. Consider a prisoner, which light has never lit. Because for him no go in the room did not last, he will visit the room after m -th day. But then he has to turn on the light - a contradiction.

66 Moscow Mathematical Competition. 10 class.

Competition was held March 2, 2003, to complete the task was given to five astronomical clock.

Tasks

1. there exist positive integers a , B , and C , that each of the equations ax 2 + BX + C = 0, AX 2 + BX - C = 0, AX 2 - BX + C = 0, AX 2 - BX - C = 0 the two roots - the whole? 

2. along the edge of a convex polyhedron with 2003 vertices held closed broken line passing through each vertex exactly once. Prove that each of the parts into which it divides the broken surface of the polyhedron, the number of faces with an odd number of sides is odd.

3. Suppose that P ( x ) - a polynomial with leading coefficient 1, and a sequence of integers a 1 , a 2 , ... such that P ( a 1 ) = 0, P ( a 2 ) = a 1 , P ( a 3 ) = a 2 , etc. The numbers in the sequence are not repeated. What degree can have P ( x )?

4. Suppose M - the point of intersection of the medians in the  ABC. On the perpendicular from M to the sides BC , AC and AB , we take the point A 1 , B 1 and C 1 , respectively, and A 1 B 1   |     MC and A 1 C 1   |     MB .Prove that M is the intersection of the medians and A 1 B 1 C 1 .

5. several cities in the country, connected by a one-way and two-way traffic. It is known that from every city in every other can be reached exactly one way, does not pass twice through the same city. Prove that the country can be divided into three provinces so that no road connects the two cities from one province.

6. Given an infinite sequence of polynomials P 1 ( x ), P 2 ( x ), ... . It always has a finite set of functions f 1 ( x ), F 2 ( x ), ..., F N ( x ), the composition of which can record any of them (for example, P 1 ( x ) = F 2 ( F 1 (f 2 ( x ))))?

Solutions

1. A : Yes, for example, a = 1, B = 5, C = 6.

Solution. can assume that a = 1, because if x 1 and x 2 - the roots of ax 2 + BX + C , then B = - ( x 1 + x 2 ) a , C = x 1 x 2 a , t is, all the coefficients can be reduced by a . Note also that the roots of equations, differing only in the BX , the opposite, so it is sufficient only to the roots of equations x 2 + BX + C = 0 and x 2 + BX - C = 0 were intact.

For this to be satisfied, it is necessary that the discriminants of these equations are perfect squares: B 2 -4 C = m 2 , B 2 4 C = n 2 . The same condition is also sufficient - for b 2 -4 C of the same parity as B , and hence ( B 2-4 C ) 1/2 (if the root of the whole) is of the same parity. Thus, x 1.2 = (- B + ( B 2 -4 C ) 1/2 ) / 2 (the roots of x 2 + BX + C = 0) - integers. Similarly, for the second equation.

In fact, these B , C , m , n , there are infinitely many. To solve the problem, it suffices to show one example of B , C , m , n . We show how this example could find. Excluding C , we get n 2 - B 2 = B 2 - m 2 , with that c must be an integer, to B , m and n have the same parity. After going through the squares of the odd numbers from 1 to 9, we find one of the solutions: 7 2 -5 2 = 5 2 -1 = 24; C = 24/4 = 6. We found the b and C , such that the discriminants of both equations - completing the square.

To find an example, we needed to pick up the representation 2 b 2 as the sum of m 2 + n 2 , different from b 2 + B 2 . We have found B , m and n just selection, but in reality the problem of representing a given number as a sum of two squares is well established. It is related to the so-called Gaussian integers - the numbers of the form x + y (-1) 1/2 , where x and y - integers. In particular, we show that a non-trivial representation of 2 b 2 = m 2+ n 2 if and only if the expansion of b into primes is at least one prime number of the form 4 k 1. More

information can be found, for example, in the article by Senderov and A. Spivak, "sums of squares and the number of Gaussian integers" (Quantum, 1999, N 3, pp. 14-22).

2. Choose any of these two parts. Consider the amount of a 1 + a 2 + ... + a n , where a I - the number of parties I -th face.

Each edge of a polyhedron in which broken persists counted in this sum is twice the amount of parity and therefore does not depend on the number of such edges. Each edge through which the broken line is a sum of exactly once. Such edges of 2003, so the full amount is odd.

If the number of faces with an odd number of sides were even, then considered the amount would also be even. So this number is odd.

3. A : The degree of the polynomial can be only one.

Decision. If the degree of the polynomial P ( x ) is 0, then P ( x ) = 1 (the leading coefficient is 1). But then the equation P ( a 1 ) = 0 has no solutions. If the degree of the polynomial is 1, then, for example, the polynomialP ( x ) = x -1 and the sequence 1, 2, 3, ... ( a n = n ) satisfy the conditions of the problem.

We will prove that there are no such polynomials of degree 2 and higher. It can be shown (see below) that there exists a positive constant C (for each polynomial - own) that if | x |> C , then | P ( x ) |> | x |. Now note that all members of the sequence modulo limited C . Indeed, if | a n |> C , then C <| a n | <| P ( a n ) | = | a n -1 |. Continuing, we get:

C <| a n | <| P ( a n ) | = | a n -1 | <| P ( a n -1 ) | <... <| P ( a 2 ) | = | a 1 | <| P ( a 1 ) | = | 0 |,

which is false.

Now we see that the sequence consists of the integers modulo not exceed C, a finite number of them, that is, the sequence will necessarily repeat.

It remains to prove the existence of a constant C . For a polynomial P ( x ) = x n + B n -1 x n -

1 + ... + B 2 x 2 + B 1 x + B 0 can take C = | B n -1 | + ... + | b 1 | + | B 0 | +1. 

Indeed, if | x |> C , then

|P(x)|>|x|n-(|bn-1|*|x|n-1+...+|b1|*|x|1+|b0|)> >|x|n-(|bn-1|*|x|n-1+...+|b1|*|x|n-1+|b0|*|x|n-1)= =|x|n-1*(|x|-(|bn-1|+...+|b0|))>|x|n-1>|x|.

4. First solution. denote

BC = a ,   CA = B ,   AB = C , MA ' = x ,   MB ' = y ,   MC ' = z . (in bold denote vectors) 

Then the condition of the problem is written as follows:

( C , z ) = ( a , x ) = ( B , y ) = 0, (1) (( y - z ), ( B - C )) = (( x - z ), ( a - C )) = 0. (2)

Revealing in (2) brackets, and using (1) we obtain ( B , z ) + ( C , y ) = ( a , z ) + ( C , x ) = 0. Since C = - a - B , we have: ( B , z ) - ( a , y ) = ( a , z ) - ( B , x ) = 0.

We must show that x + y + z = 0, which is enough to ( a , ( x + y + z )) = ( B , ( x + y + z )) = 0. Indeed, ( a , ( x + y + z )) = ( a , y ) + ( a , z ) = ( a , y ) + ((- C - B ), z ) = - ( B , z ) + ( a , y ) = 0 . Similarly, ( B , ( x + y + z )) = ( B , x ) + ( B , z ) = ( B , x ) + (- C - a ), z ) = - ( a , z ) + ( B , x ) = 0.

The second solution. For an equilateral triangle statement of the problem is trivial. We reduce the problem to the case of an equilateral triangle.

The points A ' , B ' and C ' are defined by the following conditions:

MA '   |     BC ,   MB '   |     AC ,   MC '   |     AB ,   A'B '   |     MC ,   A'C '   |     MB .

Rotate the triangle A'B'C ' around the point M at 90 o clockwise. The conditions at point A ' , B ' and C ' will go to the following:

MA ' | | BC ,   MB ' | | AC ,   MC ' | | AB ,   A'B ' | | MC ,   A'C ' | | MB .

The task enters equivalent. Draw an affine transformation taking triangle ABC an equilateral. The problem goes back to an equivalent, as expressed in terms of parallelism, and the middle of the intersection of the lines.

Now rotate the triangle A'B'C ' clockwise by 90 o . The problem goes to the original, but with an equilateral triangle ABC .

5. Suppose that in the country, has not yet been divided into provinces, there was an autonomous region consisting of a single city. This area, firstly, can be divided into three provinces (two blank), and second, it satisfies the condition as an independent country, that is, the removal of all other cities. Will add to the autonomous region of maintaining these two conditions.

 Fig. 25

Suppose that the region does not contain all of the city. Then there is the road leading out of the city area (denoted by X ) to the city, have not owned an autonomous region (call it Y ). Consider the path of self-intersections (a sequence of roads) leading from Y to X (Fig. 25).

We prove that in this way they meet the city area, except that the latter city X . Assume the contrary: in this way there is a city Z , which lies in the autonomous region. Then, from X to get to Z nonselfintersecting two ways: one way is through the Y , and the second one is only for cities in the region (such a path exists because of the condition of the problem, as for the whole country). But, by hypothesis is impossible. Consequently, the path from Y to X does not include other cities in the area, except for X .

Now connect all the cities along the way (including Y ) to the autonomous region and take them one by one in those two provinces, in which X is not included. All roads connecting the two attached town - the roads on the way Y -> X , otherwise it would be between two paths. Similarly, all the roads that connect the city connected with the city, already existing in the area - it is the way of the X to Y and the last road on the way Y -> X .Consequently, the region will be properly divided into provinces.

We prove that the new region problem condition is also satisfied. Two non-self-way from one city to another is not possible, otherwise the whole country did not satisfy the condition. From each of the city can be reached (on the field) to X , and from X - before everyone else, so at least one way from one area to another city is always there. Hence, the new region also satisfies the condition.

At each step of the described area is expanded to at least one city ( Y ). Therefore, sooner or later the city will be connected. At this point, the area will coincide with the whole country, but will be divided into provinces.Thus, the statement of the problem is proved.

6. A : It is possible.

Solution . Suppose, for example, F ( x ) =  + arctg  x , G ( x ) = x +  , H ( x ) - a feature that in the interval (- (  / 2) + n ; (  / 2) + n ) is equal to P n (TG  x ) (Fig. 26). Then 

P n ( x ) = H ( G ( G (... ( G ( F (x))) ...))) (the function g is used ( n -1) times.)

Explanation. associate with each pair ( x , n ), where n - a positive integer and x - a real, a number of A ( x , n ) such that the number of A ( x , n ) do not overlap for any two distinct pairs.

To do this, we divide the real line into intervals (- (  / 2) + k , (  / 2) + k ). Each number n is comparable interval (- (  / 2) + n , (  / 2) + n ), each pair ( x ,  n ) - are the number of interval (- (  / 2) + n , (  / 2) + n ), equal to A ( x , n ) = arctg  x + n .

Functions F , G , H such that F ( x ) = A (1, x ) G ( A ( n , x ) = A (( n one), x ) H ( A ( n , x ) = P n ( x ) (see Fig. 26). 

Then H ( G ( G (... ( G ( F ( x ))) ...))) (the function g is used ( n -1) times) is obviously equal to P n ( x ).

 

Fig. 26

66 Moscow Mathematical Competition. 11 class.

Competition was held March 2, 2003, to complete the task was given to five astronomical clock.

Tasks

1. For positive numbers x , y , z holds

x 2

+

y 2

+

z 2

=

x 2

+

z 2

+

y 2

.

y z x z y x

Prove that at least two of the numbers x , y , z are equal.

2. Given a polynomial P ( x ) degree in 2003 with real coefficients and leading coefficient 1. There is an infinite sequence of integers a 1 , a 2 , ..., such that P ( a 1 ) = 0, P ( a 2 ) = a 1 , P ( a 3 ) = a 2 , etc. Show that not all the numbers in the sequence a 1 , a 2 , ... different.

3. Dan inscribed quadrilateral ABCD . The points P and Q are symmetric point C with respect to lines AB and AD respectively. Prove that the line PQ passes through the orthocenter (intersection of heights) H triangle ABD.

4. the perimeter diameter round cake n /  meters are located n cherries. If at the end of an arc are cherries, the number of the remaining cherries on this arc is less than the arc length in meters. Prove that the cake can be cut into n equal sections so that each piece will be cherry.

5. a convex polyhedron internal dihedral angle at each edge is sharp. How can the faces of the polyhedron?

6. On the shore of a circular island Gdetotam has 20 villages, each live to 20 wrestlers. Tournament was held in which each wrestler met with all the fighters from all the other villages. Village A village is considered a strongB , if at least k matches between fighters from the villages ends winning fighter from the village of A . It was found that each village stronger following it clockwise. What is the greatest value can be k ? (All the fighters have different power, and fight the strongest always wins.)

7. Given the equality ( a m 

1 -1) ... ( a m n -1) = ( a k 

1 one) ... ( a k L 1), 

where a , n , L , and all the exponents - natural numbers, with a > 1. Find all possible values of a .

Solutions

1. Freed from the denominator, we present our equality to the form x 3 z - x 3 y + z 3 y - z 3 x + y 3 x - y 3 z = 0. Expanding the left side of the factorization, we get ( x - y ) ( y - z ) ( z - x ) ( x + y + z ) = 0. Note that for positive x , y , z the last bracket is positive. Thus, if all the numbers are different, then all factors are not zero. Contradiction. 

2. A polynomial P ( x ) - x 2003 and has a leading coefficient of 1. Therefore, when x -> + infinity of its value stremitsyak + infinity . Therefore, if the number a n is large enough, a n -

1 - a n = P ( a n ) - a n > 0, ie, the number a n -1 will be more. Then, as a n -2 > a n -1 , and so on, and we get 0 at any step. So, the numbers in the sequence are bounded above. When x -> - infinity value of the polynomial P ( x ) - x tends to - infinity .Therefore, the numbers in the sequence (as above) are bounded from below, that is, they focus on a finite interval. But these numbers are integers, so among them, only a finite set of different.

Note. Compare with decision third task of the 10th class .

3. Suppose U , V - points symmetric H regarding AB and AD , respectively, X - point of intersection of UC , and AB , Y - the point of intersection of VC and AD . Since the line PH passes through X , and QH - by Y , the statement of the problem is equivalent to the point X , H , Y are collinear.

First, we note that the points U and V lie on the circumcircle of the quadrilateral. It is sufficient to note that, for example, / ( UA , UB ) =  - / ( HA , HB ) = / ( DA , DB ) (corners oriented). As

4. Assume a point of the circle as the reference. Let a I - arc length from the origin to the I -th cherries clockwise. Consider the numbers B I = a I - I . From the condition that | B m - B k | <1 for all m , k (enough to consider the two arcs that are m -i and k -i cherries divide the perimeter of cake). Let B s - the smallest of them, then 0 <   B I - B s <1 for all I . It follows that there is an x , that x < B I < x one for all I . Obviously, cut along the radius at points with coordinates x , x one, ..., x + n -1, - desired.

5. first decision . First, we prove that each vertex of is adjacent to three faces. Consider any trihedral angle. Its dual is called triangular corner edges are perpendicular to the faces of the. Obviously, the amount of any of the dihedral angle of the triangular corner with a corresponding flat angle of the dual is  . Since the sum of the plane angles of the dual angle

less than 2  , the sum of the dihedral angles of the larger  . Since the n -sided corner can be cut into n -2 triangular, then the sum of the dihedral angles greater than  ( n -2), ie, at least one more  (1 - (2 / n )). Since for n > 3, the inequality 1 - (2 / n ) > 1/2, at the apex of the polyhedron can converge more than three ribs.

We now show that if the dihedral angles of the trihedral angle acute, then flat, too sharp. Turning to the dual angle, get the equivalent statement: If all plane angles obtuse, then the dihedral too blunt. Assume that for the trihedral angle OABC it is not, and the angle at the edge of OC is not stupid. Since the angles AOC and BOC blunt, the base of the perpendiculars from A and B to the line OC , lie outside the ray OC . Choose points A andB so that these perpendiculars had a common base D . Then /   ADB <    / 2, AD < AO , BD < BO . Consequently, AB 2 <   AD 2 + BD 2 < AO 2 + BO 2 , and the angle AOB acute - a contradiction.

So, this polyhedron flat angles of all the faces are sharp, mean, all the faces --- triangles. In addition, each vertex is adjacent to three faces. Consider any edge KLM . To each side of its adjacent triangular face, and any two of these faces have a common edge. Consequently, the top third of the faces are the same, and the polyhedron is a tetrahedron.

The second solution. For each of the faces look normal vector, ie a vector perpendicular to this face and directed outside the polyhedron.

1) We prove that the angle between any two external normals blunt or expanded. Let's not true - there were two faces of 1 and 2 , the outer normals to which form an angle of not more than  / 2. Then the face 1and 2 half-planes belong to 1 and 2 , which form a dihedral angle of not less than  / 2.

Take a point P on the edge of 2 . Let P ' - the projection of P on the plane of the face 1 . Point P ' is outside of the polygon 1 , so there is a line that contains an edge of r edge 1 and separating P ' of 1 .Polyhedron lies inside an acute dihedral angle corresponding to the edge R , but P lies outside this dihedral angle. Contradiction.

2) It remains to show that in space there is no more than four vectors, the pairwise angles between them stupid or expanded.

Suppose not, and u 0 , u 1 , u 2 , u 3 , u 4 - five vectors pairwise angles between them stupid or expanded. We introduce a rectangular coordinate system so that the axis Oz was the same direction as u 0 . Denoted by v 1 ,V 2 , V 3 , V 4 projections of u 1 , u 2 , u 3 , u 4 the plane Oxy . One of the angles between v 1 , V 2 , V 3 , V 4 , for example, the angle between v 1 and v 2 , does not exceed  / 2. This means that the inner product ( V1 , V 2 ) is nonnegative. Let u 1 = (x 1 , y 1 , z 1 ), u 2 = (x 2 , y 2 , z 2 ). Since the angle between u 1 and u 0 stupid, z 1 <0, and similarly z 2 <0, so, z 1 z 2 > 0. We have: ( u 1 , u 2 ) = x 1 x 2 + y 1 y 2 + z 1 z 2 = ( V 1 , V 2) + z 1 z 2 > 0. We find that the angle between u 1 and u 2 acute - a contradiction.

6. A: 290.

Solution . We show that for k > 290, this situation is impossible. We order in every village fighters descending power and choose in every village the tenth to force fighter. We show that the village in which he lives the weakest of the selected wrestlers can not be stronger

than the one after. Denote the selected fighters in our village and the next through A and B , respectively. Then in our village 11 fighters are not stronger than A , and the next - 10 fighters at least the same strength as B . All bouts between the fighters will end in favor of the second village, and these fights exactly 110, that is, the fights, which won the champion of our village, no more than 20 * 20-110 = 290.

Here is an example showing that if k < 290 described situation is possible. Suppose that among the wrestlers has 210 newcomers and 190 masters (any newcomer less any master). We number the village against theclockwise direction. Was in the first village one weak freshman and 19 masters weakest in the second - two newcomers, the weakest of the remaining, and 18 masters, the weakest of the other, in the third - the weakest of the remaining three newcomers and 17 masters, the weakest of the other; ...; in the last village we put 20 strongest newcomers. This accommodation is in the villages is shown in Table. 3 (in the "Fighters" numbers dialed roman mean force masters in italics - the power beginners):

Table 3.

Villages

Wrestlers

1 1 , 211-229

2 2.3 , 230-247

3 4.6 , 248-264

4 10.7 , 265-280

5 11-15 , 281-295

6 16-21 , 296-309

7 22-28 , 310-322

8 29-36 , 323-334

9 37-45 , 335-345

10 46-55 , 346-355

11 56-66 , 356-364

12 67-78 , 365-372

13 79-91 , 373-379

14 92-105 , 380-385

15 106-120 , 386-390

16 121-136 , 391-394

17 137-153 , 395-397

18 154-171 , 398-399

19 172-190 , 400

20 191-210

We show that I -I village stronger ( I -1)-th at I > 1. Indeed, in the k -th country is k beginners and 20 - k masters. In this case, the master I -th win everyone in the village ( I -1)-th, and newcomers win novice and wins willI ( I -1) {+ 20 (20 - I ) = I 2 -21 I 400. The vertex of the parabola is at the point I = 10.5, and the branches are directed upwards, so the minimum value of the whole point is reached at I = 10, 11 and equal to 290, that is,I -I village stronger ( I -1)-th for k < 290. In addition, the master of the first village newcomers win the 20th, and the wins will be 20 * 19 = 380> 290, ie, all the conditions are met.

7. A : 2 and 3 (ie, the equalities 2 2 1 = 2 +1, (3-1) 2 = 3 +1).

Solution. Suppose that for some a > 3 required complete

A = ( a m 1 -1) ... ( a m 

n -1) = ( a k 1 1) ... ( a k 

L 1).

We first show that as any m I , and the number a -1 is a power of two. Let  - odd divisor of m I , and P - any prime divisor of a  -1. Then on the right there is a work factor a k 

j 1, is divisible by P . Therefore the number of (( a k 

j )  1) - (( a  ) k j -1) = 2 

divided by P , ie, all the prime factors of a  -1 - two and ( a  -1 + + ... a one) ( a -1) = a  = -1 2 n . Since a > 3, n > 1, and  is odd, so the expression in the first bracket - odd divisor of 2 n . Hence,  = 1 and a = -1 2 n . In particular, a -1 is divisible by 4 (for a -1> 2), so a k 1 has remainder 2 when divided by 4.

Each number on the left side of the original equation can be represented as a 2 D -1 = ( a -1) ( a 1) ( a 2 one) ( a 4 one) ... ( a 2 D -1 one) = 2 n 2 * D * a 0 ... a D -1  , where a I = ( a 2 I one) / 2 - odd. Note that for m > nthe number of a 2 m -1 is divided by a 2 n 1, so gcd ( a m , a n ) < ( a 2 m one) - ( a 2 m -1) = 2, since a I are odd, they are relatively prime. Let q - the maximum power of two, part of the m I or k j . Then A is represented as A = 2 N ( a 0 ) N 

0 ... ( a q ) N 

q , where N > N 0 + ... + N q , since this is true for any representation a 2 D -1 . 

Since any number of the form a k j 1 is divisible by the first power of two, in their

representation of the number A is not less than N . But each of them is divisible by one of the numbers a I : if k j = 2 R s , where s is odd, thena k 

j 1 is divided into a R . Then, since N > N 0 + ... + N q , then for a number of a I share more than N I number a k 

j one, and hence, A is divided by the degree b'olshuyu a I , than N I . This contradicts the fact that a I are odd and pairwise relatively prime.