HMMT 二月 2005 · 几何 · 第 7 题

7. Let ABCD be a tetrahedron such that edges AB , AC , and AD are mutually perpen- dicular. Let the areas of triangles ABC , ACD , and ADB be denoted by x , y , and z , respectively. In terms of x , y , and z , find the area of triangle BCD . √ 2 2 2 Solution: x + y + z Place A , B , C , and D at (0 , 0 , 0), ( b, 0 , 0), (0 , c, 0), and (0 , 0 , d ) in Cartesian coordinate space, with b , c , and d positive. Then the plane through B , C , and D is given by the y x z equation + + = 1. The distance from the origin to this plane is then b c d 1 bcd bcd √ √ √ = = . 2 2 2 2 2 2 2 2 2 1 1 1 2 x + y + z b c + c d + d b

2 2 2 b c d Then if the area of BCD is K , the volume of the tetrahedron is bcd bcdK √ = , 2 2 2 6 6 x + y + z 2 √ 2 2 2 implying K = x + y + z . Alternative Solution: The area of BCD is also half the length of the cross prod- − − → − − → uct of the vectors BC = (0 , − c, d ) and BD = ( − b, 0 , d ). This cross product is √ 2 2 2 ( − cd, − db, − bc ) = − 2( y, z, x ), which has length 2 x + y + z . Thus the area of √ 2 2 2 BCD is x + y + z .