PUMaC 2022 · 团队赛 · 第 12 题
PUMaC 2022 — Team Round — Problem 12
题目详情
- Observe the set S = { ( x, y ) ∈ Z : | x | ≤ 5 and − 10 ≤ y ≤ 0 } . Find the number of points P in 2 S such that there exists a tangent line from P to the parabola y = x + 1 that can be written in the form y = mx + b, where m and b are integers.
解析
- Observe the set S = { ( x, y ) ∈ Z : | x | ≤ 5 and − 10 ≤ y ≤ 0 } . Find the number of points P in 2 S such that there exists a tangent line from P to the parabola y = x + 1 that can be written in the form y = mx + b, where m and b are integers. Proposed by Frank Lu Answer: 15 2 First, suppose that the line y = mx + b is tangent to the parabola. Then, it follows that x +1 = 2 mx + b has exactly one solution, which in particular requires us to have that x − mx +1 − b = 0 2 m to have one solution. But from completing the square, this is only possible if 1 − b = , or 4 √ 2 that m = 2 1 − b. For m, b to be integers, notice that we must have b of the form 1 − k , so m = 2 k ; if m were odd, then 1 − b, ergo b, would not be an integer. 2 Thus, our lines are of the form y = 2 kx + (1 − k ) for some integer k ∈ Z . We now seek to classify the points ( x, y ) that lie on a line of this form. Given such a point P in our set, 2 we solve for k. Notice that solving for k here yields us with k − 1 − 2 kx + y = 0 , or that 6 p 2 k = x ± x + 1 − y. We require this to be an integer, and we are picking x, y to also be 2 2 integers. Therefore, we must have that y = x + 1 − l for some integer l, whereby we have that k = x ± l is an integer, given x is an integer. 2 To count these points: notice that x + 1 takes on the values 1 , 2 , 5 , 10 , 17 , 26 , and that the negative squares are 0 , − 1 , − 4 , − 9 , − 16 , − 25 , − 36 . We now wish to count how many pairs ( x, l ) 2 will yield a y that lies between − 10 and 0 . For x + 1 = 1 , these are − 1 , − 4 , − 9 , so there are 3 . Repeating this procedure, we find that for 2 , 5 , 10 , 17 , 26 that there are 2 , 1 , 1 , 1 , 1 , respectively. So the number of pairs ( x, y ) is thus 3 + 2 · (2 + 1 + 1 + 1 + 1) = 3 + 12 = 15 .