60-odd years of moscow mathematical
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Moscow olympiad problems
≥ x
3 ≥ x 2 . 16.2.9.4. Given a 101 × 200 sheet of graph paper, we start moving from a corner square in the direction of the square’s diagonal (not the sheet’s diagonal) to the border of the sheet, then change direction obeying the laws of light’s reflection. Will we ever reach a corner square? 16.2.9.5. Divide a cube into three equal pyramids. Grade 10 16.2.10.1. Find roots of the equation 1 − x 1 + x(x − 1) 1 · 2 − · · · + (−1) n x(x − 1) . . . (x − n + 1) n! = 0. 16.2.10.2. See Problem 16.2.9.2. 16.2.10.3. Let x 0 = 10 9 , x n = x 2 n−1 +2 2x n−1 for n > 0. Prove that 0 < x 36 − √ 2 < 10 −9 . 16.2.10.4. See Problem 16.2.9.5. 16.2.10.5. A knight stands on an infinite chess board. Find all places it can reach in exactly 2n moves. Olympiad 17 (1954) Tour 17.1 Grade 7 17.1.7.1. A regular star-shaped hexagon is split into 4 parts. Construct from them a convex polygon. 17.1.7.2. Given two convex polygons, A 1 A 2 . . . A n and B 1 B 2 . . . B n such that A 1 A 2 = B 1 B 2 , A 2 A 3 = B 2 B 3 , . . . , A n A 1 = B n B 1 and n − 3 angles of one polygon are equal to the respective angles of the other. Find whether these polygons are equal. 17.1.7.3. Find a four-digit number whose division by two given distinct one-digit numbers goes along the following lines: Figure 19. (Probl. 17.1.7.3) 17.1.7.4. Are there integers m and n such that m 2 + 1954 = n 2 ? 17.1.7.5. Define the maximal value of the ratio of a three-digit number to the sum of its digits. Grade 8 17.1.8.1*. Cut out of a 3 × 3 square an unfolding of the cube with edge 1. 17.1.8.2. From an arbitrary point O inside a convex n-gon, perpendiculars are dropped to the (exten- sions of the) sides of the n-gon. Along each perpendicular a vector is constructed, starting from O, directed towards the side onto which the perpendicular is dropped, and of length equal to half the length of the corresponding side; see Fig. 20. Find the sum of these vectors. 17.1.8.3. See Problem 17.1.7.3. 17.1.8.4. Find all solutions of the system consisting of 3 equations: x ³ 1 − 1 2n ´ + y ³ 1 − 1 2n + 1 ´ + z ³ 1 − 1 2n + 2 ´ = 0 for n = 1, 2, 3. OLYMPIAD 17 (1954) 45 Figure 20. (Probl. 17.1.8.2) 17.1.8.5. See Problem 17.1.7.4. Grade 9 17.1.9.1. Prove that if x 4 0 + a 1 x 3 0 + a 2 x 2 0 + a 3 x 0 + a 4 = 0 and 4x 3 0 + 3a 1 x 2 0 + 2a 2 Download 1.08 Mb. Do'stlaringiz bilan baham: |
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