60-odd years of moscow mathematical
Download 1.08 Mb. Pdf ko'rish
|
Moscow olympiad problems
Grade 8
15.2.8.1. Calculate √ 0.9...9 (60 nines) to 60 decimal places. 15.2.8.2. From a point C, tangents CA and CB are drawn to a circle O. From an arbitrary point N on the circle, perpendiculars N D, N E, N F are dropped to AB, CA and CB, respectively. Prove that the length of N D is the mean proportional of the lengths of N E and N F . (See Fig. 15). OLYMPIAD 15 (1952) 41 15.2.8.3. Seven chips are numbered 1, 2, 3, 4, 5, 6, 7. Prove that none of the seven-digit numbers formed by these chips is divisible by any other of these seven-digit numbers. 15.2.8.4. 99 straight lines divide a plane into n parts. Find all possible values of n less than 199. Grade 9 15.2.9.1. Solve the system of equations 1 − x 1 x 2 = 0, 1 − x 2 x 3 = 0, . . . . . . . . . . . . . . . . 1 − x n−1 x n = 0, 1 − x n x 1 = 0. How does the solution vary for distinct values of n? 15.2.9.2. How to arrange three right circular cylinders of diameter a 2 and height a into an empty cube with side a so that the cilinders could not change position inside the cube? Each cylinder can, however, rotate about its axis of symmetry. 15.2.9.3. See Problem 15.2.8.3. 15.2.9.4. In an isosceles triangle 4ABC, ∠ABC = 20 ◦ and BC = AB. Points P and Q are chosen on sides BC and AB, respectively, so that ∠P AC = 50 ◦ and ∠QCA = 60 ◦ . Prove that ∠P QC = 30 ◦ . (See Fig. 16). Figure 16. (Probl. 15.2.9.4) Figure 17. (Probl. 16.1.8.1) 15.2.9.5. 200 soldiers occupy in a rectangle (military call it a square and educated military a carr´ee): 20 men (per row) times 10 men (per column). In each row, we consider the tallest man (if some are of equal height, choose any of them) and of the 10 men considered we select the shortest (if some are of equal height, choose any of them). Call him A. Next the soldiers assume their initial positions and in each column the shortest soldier is selected; of these 20, the tallest is chosen. Call him B. Two colonels bet on which of the two soldiers chosen by these two distinct procedures is taller: A or B. Which colonel wins the bet? Grade 10 15.2.10.1. Prove that for arbitrary fixed a 1 , a 2 , . . . , a 31 the sum cos 32x + a 31 cos 31x + · · · + a 2 cos 2x + a 1 cos x can take both positive and negative values as x varies. 15.2.10.2. See Problem 15.2.9.2. 15.2.10.3. Prove that for any integer a the polynomial 3x 2n +ax n +2 cannot be divided by 2x 2m +ax m +3 without a remainder. 15.2.10.4. See Problem 15.2.9.4. 15.2.10.5. See Problem 15.2.9.5. 42 MOSCOW MATHEMATICAL OLYMPIADS 1 – 59 Olympiad 16 (1953) Tour 16.1 Grade 7 16.1.7.1. Prove that the sum of angles at the longer base of a trapezoid is less than the sum of angles at the shorter base. 16.1.7.2. Find the smallest number of the form 1...1 in its decimal expression which is divisible by 3...3 (100 three’s). 16.1.7.3. Divide a segment in halves using a right triangle. (With a right triangle one can draw straight lines and erect perpendiculars but cannot drop perpendiculars.) 16.1.7.4. Prove that n 2 + 8n + 15 is not divisible by n + 4 for any positive integer n. Grade 8 16.1.8.1. Three circles are pair-wise tangent to each other. Prove that the circle passing through the three tangent points is perpendicular to each of the initial three circles; see Fig. 17. 16.1.8.2. Prove that if in the following fraction we have n radicals in the numerator and n − 1 in the denominator, then 2 − q 2 + p 2 + · · · + Download 1.08 Mb. Do'stlaringiz bilan baham: |
Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling
ma'muriyatiga murojaat qiling