Đam mê Toán học Mathematical Delight 2004

Đam mê Toán học Mathematical Delight 2004 Sách đề cập nhiều đến các bài toán cổ những năm từ 1950 đến 2000, sách do Hiệp Hội Toán Học Hoa Kỳ xuất bản. Đây là một tài liệu rất hay cho ai đam mê những bài toán cổ và những bài toán rất hay gặp trong trường học.

[...]...To Don Albers Gleanings SECTION From 1 Mathematical Miniatures Mathematical Miniatures, by Titu Andreescu and Svetoslav Savchev (Anneli Lax New Mathematical Library Series MAA, 2003), is a goldmine of elemen­ tary delights 1 Six points are given in space such that the lengths of the 15 segments that join them in pairs are all... and D is far down the highway coming towards them A passed B at 8 A.M and C at 9 A.M., and was the first to encounter D, whom he met at 1 0 A.M D met B at noon, and C at 2 P.M When did B pass C? 3 Mathematical Delights 4 In Figure I, let the position on the highway be plotted along the y -axis against the time along the x-axis Let the time at the origin 0 be 8 o'clock and let the point on the highway... of the triangle Then LPLC = L B = 60° = L C, making PLCM an isosceles trapezoid Hence the diagonals PC and LM are equal Similarly, PA = MN and PB = LN, and �LMN has sides of lengths PA, PB, PC Mathematical Delights 6 B c L FIGURE 2 5 Asterisks are placed in some cells of an m x n matrix, where m < n , so that there is at least one asterisk in each column Prove there is an asterisk such that there... remains, then, to show there exists a pair (p, q) that is "smaller" than the initial pair (a, b) (b) The central relation, a2 + b k - ­2 ab + 1' is easily put into the form a 2 - kab + b2 = k, Mathematical Delights 8 which displays (a b) as an integer solution of the equation x =0 k =x 2 • k (x, y) Now, if x2 - kxy+y 2=k k =y2 , in this equation, then y = 0, integers x x y and if then Since is not... (99) X98 4 98 x i=.J=T, + i) 99 ; ( l - i) � c;) - en + � - - G:) = s (I () + (1 = 2 2 0 2 = This is an identity, and for • we obtain = Now, in polar form, 1 + i=-J2 ( : + i : ) cos sin 11 Mathematical Delights 12 Hence, by the theorem of de Moivre, 991l' 991l' ( 4 4) 3 (24rr 3: ) ] = (h) 99 [ (24rr :) 3.7r 3.7r J ;;:;= (vk) 99 [ ;;:;( 1 + i) 99 = (v2 ) 99 + i sin cos cos + + i sin + +isin cos... integers b; inherit property P: one of the results Removing any one b; leaves 2n o f them which are determined from 2n of the a's Now, these 2n a's can be divided into two n-subsets with equal 15 16 Mathematical Delights sums T , and subtracting a from each of these a; yields two n-subsets of b's with the same sum T na - Accordingly, all the b's must have the same parity, and since one of them is zero,... Then CP = CO = r, and PO = R Also let LPCO = a Let C and CO be the origin and axis of a system of polar coordinates The area A in question is obtained by revolving the circular arc PO about 17 Mathematical Delights 18 CO An element of this arc is rd() and its distance from the axis is Hence an element of the area A is given by r sin e d A = 2rr(r sin ())(rd()), and A = 2rrr 2 loa sin(}d(} = 2rrr 2... ABD, BCE, and CAF are drawn to give tlDEF (Figure 4) triangles From A, B, C, perpendiculars are drawn, respectively, to DF, DE, and EF Prove these perpendiculars are concurrent A D c E FJGURE 4 Mathematical Delights 20 We follow the very clever second published solution Since MBD is isosceles, the circle C 1 with center D and radius DA goes through A and B Similarly, the circle C2 with center E and... so is 1r, a contradiction (Although algebraic numbers are defined in terms of polynomial equations with integer coefficients, it is known that algebraic coefficients also yield algebraic roots.) Mathematical Delights 22 5 (Problem A3 from the 1999 Putnam Competition (Feb., 2000, page 74)) Prove that the series expansion I � + at X + · + an x n + ·· · I- Zx - x 2 = ao · · has the remarkable property... divisible by 1 1? We follow the published solution Every Putnam competitor knows the test for divisibility by ger n 1 1: an inte­ = abed , in decimal notation, is divisible by 1 1 if and only if the Mathematical Delights 24 alternating sum of its digits, a - b + c - d + - · ·· , is divisible by 1 1 Let rn and dn respectively denote the number of digits and the alter­ nating sum of the digits of A n From