PUMaC 2020 · 团队赛 · 第 1 题

PUMaC 2020 — Team Round — Problem 1

Consider a 2021-by-2021 board of unit squares. For some integer k, we say the board is tiled by k -by- k squares if it is completely covered by (possibly overlapping) k -by- k squares with their corners on the corners of the unit squares. What is the largest integer k such that the minimum number of k -by- k squares needed to tile the 2021-by-2021 board is exactly equal to 100?

解析

If Q is a point on the circle of radius 2020 centered at the origin such that the line P Q is tangent to the circle at Q, then P Q has integral length. 2) The x -coordinate of P is 38. Proposed by: Ollie Thakar Answer: 16 2 2 Notice that 38 + 24 = 2020 . Then, let P have coordinates (38 , y ) , and label the length of P Q as T. For now, we will only deal with positive y. We know from power of a point theorem that 2 2 6 2 ( y + 24)( y − 24) = T . Re-arranging this expression gives us ( y + T )( y − T ) = 24 = 2 · 3 . 6 2 Now, we know that y + T and y − T must be integer factors of 2 · 3 . There are (6+1)(2+1) = 21 6 2 factors of 2 · 3 , of which 20 come in pairs and 1 is a perfect square. Thus, there are 11 pairs 6 2 of factors multiplying to 2 · 3 . 2 Of those pairs of factors, 3 pairs have 1 odd factor and 1 even factor, while the remaining pairs 6 2 have 2 even factors. ( y + T )( y − T ) = 2 · 3 means that the only the pairs with 2 even factors will lead to integer values of T and y. Each factor pair leads to a unique solution pair of y and T. Thus, there are 8 possibilities for y, when y > 0 . Then, there are 8 more possibilities for y that are negative, so the total is 16. Note: We also accepted the answer of 14 since it isn’t clear that P is allowed to be taken on the circle and still yield a valid configuration.