HMMT 二月 2005 · 几何 · 第 7 题

英文解析

7. Let ABCD be a tetrahedron such that edges AB , AC , and AD are mutually perpendicular. 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 coordinatespace, with b , c , and d positive. Then the plane through B , C , and D is given by thex zyequation + + = 1. The distance from the origin to this plane is thenb c d
   1 bcd bcd
   √
   √ √
   = = .
   2 2 2
   2 2 2 2 2 2
   1 1 1
   2 x + y + zb c + c d + d b + +
   2 2 2
   b c d
   Then if the area of BCD is K , the volume of the tetrahedron isbcd bcd k
   √
   = ,
   2 2 2
   6 x + y + z 6
   √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 .