PUMaC 2020 · 团队赛 · 第 11 题
PUMaC 2020 — Team Round — Problem 11
题目详情
- Three (not necessarily distinct) points in the plane which have integer coordinates between 1 and 2020, inclusive, are chosen uniformly at random. The probability that the area of the a triangle with these three vertices is an integer is in lowest terms. If the three points are b collinear, the area of the degenerate triangle is 0. Find a + b .
解析
- Three (not necessarily distinct) points in the plane which have integer coordinates between 1 and 2020, inclusive, are chosen uniformly at random. The probability that the area of the 6 a triangle with these three vertices is an integer is in lowest terms. If the three points are b collinear, the area of the degenerate triangle is 0. Find a + b . Proposed by: Daniel Carter Answer: 13 Let the three points be ( x , y ) for i ∈ { 1 , 2 , 3 } . By the shoelace area formula, the area of the i i triangle is | x y + x y + x y − x y − x y − x y | / 2, so it is an integer if the numerator is 1 2 2 3 3 1 2 1 3 2 1 3 even. Considering the numerator mod 2, shifting any of the x or y by 2 at a time preserves i i the parity of the numerator. Add or subtract an even number from each of x , y , x , and y 2 2 3 3 ′ ′ ′ ′ ′ ′ to make x , y , etc. so that x and x are either x or x + 1 and y and y are either y or 1 1 1 2 2 2 3 2 3 y + 1. If the resulting triangle has integer area, so did the original. 1 Note that there are an equal number of even numbers as odd numbers between 1 and 2020 ′ inclusive. Thus the probability that x = x is 1 / 2, and likewise for the other coordinates 1 2 ′ ′ ′ ′ and possibilities. Out of the sixteen possibilities for x , y , x , and y , six of them form a 2 2 3 3 right triangle with area 1 / 2, and ten of them form a degenerate triangle with area 0. Thus the probability the original triangle had integer area is 10 / 16 = 5 / 8, so the answer is 5 + 8 = 13.