HMMT 十一月 2021 · 团队赛 · 第 10 题

HMMT November 2021 — Team Round — Problem 10

[60] Three faces X , Y , Z of a unit cube share a common vertex. Suppose the projections of X , Y , Z onto a fixed plane P have areas x , y , z , respectively. If x : y : z = 6 : 10 : 15, then x + y + z can be m written as , where m, n are positive integers and gcd( m, n ) = 1. Find 100 m + n . n

解析

[60] Three faces X , Y , Z of a unit cube share a common vertex. Suppose the projections of X , Y , Z onto a fixed plane P have areas x , y , z , respectively. If x : y : z = 6 : 10 : 15, then x + y + z can be m written as , where m, n are positive integers and gcd( m, n ) = 1. Find 100 m + n . n Proposed by: Sean Li Answer: 3119 Solution: Introduce coordinates so that X , Y , Z are normal to (1 , 0 , 0), (0 , 1 , 0), and (0 , 0 , 1), respec- tively. Also, suppose that P is normal to unit vector ( α, β, γ ) with α, β, γ ≥ 0. Since the area of X is 1, the area of its projection is the absolute value of the cosine of the angle between X and P , which is | (1 , 0 , 0) · ( α, β, γ ) | = α . (For parallelograms it suffices to use trigonometry, but this is also true for any shape projected onto a plane. One way to see this is to split the shape 2 2 2 into small parallelograms.) Similarly, y = β and z = γ . Therefore x + y + z = 1, from which it is not hard to calculate that ( x, y, z ) = (6 / 19 , 10 / 19 , 15 / 19). Therefore x + y + z = 31 / 19.