ĐỀ THI TOÁN APMO (CHÂU Á THÁI BÌNH DƯƠNG) ĐỀ 27 doc

Ngày tải lên: 05/08/2014, 10:21

... with sides s −a, s −b, and s −c. This process is repeated until a triangle can no longer be constructed with the side lengths given. For which original triangles can this process be repeated indefinitely? Question

Ngày tải lên: 05/08/2014, 10:21

Ngày tải lên: 05/08/2014, 10:21

Ngày tải lên: 05/08/2014, 10:21

... indexed by (k, ) if |i − k| + |j −| = 1. (Solution) At most n 2 + c 2 −nc −c houses can be saved. This can be achieved under the following order of defending: (2, c), (2, c + 1); (3, c − 1), (3, ... − 2), (4, c + 3); . . . (c + 1, 1), (c + 1, 2c); (c + 1, 2c + 1), . . . , (c + 1, n). (6) Under this strategy, there are 2 columns (column numbers c, c + 1) at which n − 1 houses are saved 2 col...

Ngày tải lên: 05/08/2014, 10:22

... (n, 1) ∈ G and (n − 1, n) ∈ G (because (n, n − 1) ∈ G) and this implies that (n, n − 2) ∈ G and (n − 2, n) ∈ G. If we keep doing this process, we obtain (1, n) ∈ G, which is a contradiction. ... three boxes, occupied by at least one nonzero N(i, j), whose rows and columns are all distinct. This implies (4).) Problem 5. A regular (5 × 5)-array of lights is defective, so that toggling the .....

Ngày tải lên: 05/08/2014, 10:22

... displacement vector for the first section again b e (x, y). This time the car has rotated 90 ◦ clockwise. We can see that the displacements for the second, third and fourth section will be (y, −x), (−x, −y) ... r−1 ∑ k=0 (−i) a k i b k = ( ℓ−1 ∑ k=0 (−i) a k )( r−1 ∑ k=0 i b k ) = 1 ×1 = 1 as required because ℓ ≡ r ≡ 1 (mod 4). Case 4b: ℓ ≡ r ≡ 3 (mod 4) . In this case, we get m k = i a k (−i)...

Ngày tải lên: 05/08/2014, 10:22

Ngày tải lên: 05/08/2014, 10:22

Ngày tải lên: 05/08/2014, 10:22

Ngày tải lên: 05/08/2014, 10:21