60-odd years of moscow mathematical
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Moscow olympiad problems
R =
5 arcsin ³ 1 √ 3 ´ . Figure 124. (Sol. A96) Figure 125. (Sol. A97) 97. Consider a mapping f that converts the map 1 K 0 with a larger scale into the map map K 1 with the smaller scale: each point representing some spot in K 0 (by considering only the part of the town depicted on K 0 we assume that K 0 ⊃ K 1 ) is placed upon the point representing the same spot in K 1 . Denote the image of K 1 under the same mapping by K 2 , see Fig. 125. 1 The assumption that the map of Moscow is continuous was quite unjustified before the time of “glasnost” and probably still is: large portions of Moscow were “classified”, i.e., restricted for “security reasons” and quite a number of scientists were busy producing more or less plausibly distorted maps without “white spots”. But, kidding aside, this assumption is also dubious mathematically if we take into account the fractal theory, see e.g., K. Falconer Fractal Geometry, Wiley, Chichester ea, 1990 and refs therein (esp. on p. xxii). 184 SOLUTIONS Further, set K n = f (K n−1 ), n = 1, 2, . . . . The rectangles K 0 , K 1 , K 2 , . . . , K n , . . . have just one common point, x, since the sizes of the rectangles tend to zero. The point x is precisely the point for pearcing. Indeed, it follows from x ∈ K n−1 that f (x) ∈ K n . Thus the point f (x) also belongs to all rectangles but there is only one such point and so x = f (x). Remark. In general, the following theorem is true: Any continuous map of a rectangle onto itself has a fixed point. So the statement of the problem remains true even if one map is crumpled before being placed on top of the other map and pierced. 98. Since 2222 5555 ≡ 3 5555 ≡ (−4) 5555 ≡ −4 5555 (mod 7) and 5555 2222 ≡ 4 2222 (mod 7), we have 2222 5555 + 5555 2222 ≡ 4 2222 − 4 5555 ≡ 4 2222 (1 − 4 3333 ) ≡ 4 2222 (1 − 64 1111 ) ≡ 4 2222 (1 − 1 1111 ) ≡ 0 (mod 7). Therefore, 2222 5555 + 5555 2222 is a multiple of 7. Another solution. Note that 4 3 = 64 ≡ 1 (mod 7). Hence 2222 5555 ≡ 3 5555 ≡ (−4) 5555 ≡ (−4) 5 ≡ (−4) 2 ≡ −2 (mod 7). Similarly, 5555 2222 ≡ 4 2222 ≡ 4 2 ≡ 2 (mod 7), and we are done. Figure 126. (Sol. A99) 99. A sketch of the required construction with nine threads is shown in Fig. 126. A convenient way of making it is to take 3 pencils. If all threads have the same length l and the rods the same length d, then, for the construction to be rigid, it is necessary to have d = l r 1 + 2 √ 3 . The endpoints of the rods in our construction are vertices of two equilateral triangles arranged in parallel planes which are perpendicular to the straight line connecting the centers of triangles and rotated through an angle relative to one another. The rods themselves lie on straight lines which cross pairwise. The connection of the rods and threads is the same as for an octahedron. It is extremely difficult to prove the sufficiency of the above condition. Remark. This connection was invented in the 1960s by an architect, B. Fuller. Many different designs of this type have appeared since then. HISTORICAL REMARKS 175 In this section we generously borrow from Introduction to [Le], [BY] and from [GT]. 176 HISTORICAL REMARKS Historical remarks H.1. On this book. The book contains problems of all mathematical Olympiads held in Moscow since the first one, in 1935. For the first time all of them are provided with solutions or at least hints and answers. This book is not the first compendium of problems of Moscow Mathematical Olympiads. Selected problems of Moscow Olympiads 1–15 were collected in [SCY]; those from Olympiads 1–27 were compiled almost completely and published in Russian in [Le]. The compiler, A. A. Leman, and all who helped him, did a tremendous job to put through that edition. Written with care and with easy to understand solutions, the collection [Le] has become a rarity long ago. Both [SCY] and [Le], however, only contained solutions to selected problems. A critical review of these collections revealed a number of omissions and errors which we corrected as best as we could 1 . We generously borrowed from Introduction of [Le] and from the book [GT], where most of the problems from Olympiads 1–49 are supplied with at least an answer or a hint. In this book, we offer the reader for the first time all (see footnote) problems of the first 27 Olympiads and all problems of the Olympiads after the 49-th. (Problems from Olympiad 50 can be found in Chinese [GT*]. The book was published in precopyright era without anybody’s consent 2 .) No complete (and correct) solutions of ALL problems had ever been published yet. A new generation of high school students will be able to get acquainted with a great number of interesting and beautiful ideas contained in more than 2 000 problems of the Moscow Mathematical Olympiads and learn the history of these happenings. The problems for the Olympiads were put together and composed by many generations of graduate, undergraduate and post graduate students (mainly) from the Moscow University. The preparatory problems, those of the Olympiads themselves, and the problems told to school pupils at consultation sessions and lectures constitute a very valuable material for study; a sort of the mathematical folklore. Enthusiastic university students pester undergraduates and professors offering their problems, fiercely criticizing others’ problems and demanding to create more and more of new problems to the pool. Sometimes the discussions are very heated; sometimes a problem is discussed in whispers and the speakers look around like conspirators. It means that they discuss a problem that has a chance to be accepted for the Olympiad. More often than not a problem is so transformed during the discussion that the author can hardly recognize his creation. Thus, the preparatory problems and those of the Olympiad are mainly the result of a collective brain storm. Unfortunately, this most valuable folklore is lost to a great extent and partly beyond recover. It is only with great difficulties that we managed to restore some of the problems of Moscow Olympiads and sometimes even complete original solutions that seemed to have been lost. H.2. On necessity of Olympiads. In the mid-1930s many Soviet mathematicians pondered about the need for cooperation with the high school to bring up the next mathematical generation. The training of future mathematicians should begin in their childhood, the earlier the better. Nobody is surprised to see a ballet dancer or a musician starting their career at the age of 8 or 6 or earlier. The explanation is that it is impossible for a teenager to master all intricacies of the dancing art or of music, without specialized training when a child, to develop the ear and the feeling of rhythm, the flexibility of knuckles or agility of 1 Concerning completeness, this goal seem to be out of reach. Apart from Olympiads held during WW II, several problems of one more Olympiad seem to have sunk in Lethe. As I. M. Yaglom writes in Problems, Problems, Problems. History and Download 1.08 Mb. Do'stlaringiz bilan baham: |
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