By Andreescu T.
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Additional resources for Mathematical Olympiads 1996-1997: Problems and Solutions from around the World
Sample text
Xs ≤ y1 + . . + ys = (y1 + . . + yt ) + (yt+1 + . . + ys ) and ys ≤ . . ≤ yt+1 ≤ x. 8. Can a 5 × 7 checkerboard be covered by L’s (figures formed from a 2×2 square by removing one of its four 1×1 corners), not crossing its borders, in several layers so that each square of the board is covered by the same number of L’s? First Solution: No such covering exists. Suppose we are given a covering of a 5 × 7 checkerboard with L’s, such that every cell is covered by exactly k L’s. Number the rows 1, .
Let M be a set of n points in the plane, no three collinear, and let f : M → {0, 1, . . , p − 1} be a function such that: (i) only one point of M maps to 0, and (ii) if A, B, C are distinct points in M and k is the circumcircle of 52 the triangle ABC, then f (P ) ≡ 0 (mod p). P ∈M ∩k Show that all of the points of M lie on a circle. Solution: Let X be the point mapping to 0. We first show that if every circle through X and two points of M contains a third point of M , then all of the points of M lie on a circle.
We first have 23pq ≡ 2 (mod 3), which means p and q are odd. In addition, if α is a primitive root mod p, then α3pq−1 ≡ 1 (mod p) implies that p − 1 divides 3pq − 1 as well as 3pq − 1 − 3q(p − 1) = 3q − 1, and conversely that q − 1 divides 3p − 1. If p = q, we now deduce p = q = 3, but 427 ≡ 1 (mod 27), so this fails. Hence p < q. Since p and q are odd primes, q ≥ p + 2, so (3p − 1)/(q − 1) < 3. Since this quantity is an integer, and it is clearly greater than 1, it must be 2. That is, 2q = 3p + 1.
Mathematical Olympiads 1996-1997: Problems and Solutions from around the World by Andreescu T.
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