Problem B3
ABCD is a tetrahedron and D0 is the centroid of ABC. Lines parallel to DD0 are drawn through A, B and C and meet the planes BCD, CAD and ABD in A0, B0, and C0 respectively. Prove that the volume of ABCD is one-third of the volume of A0B0C0D0. Is the result true if D0 is an arbitrary point inside ABC?
Solution
Yes, indeed it is true for an arbitrary point in the plane of ABC not on any of the lines AB, BC, CA
Take D as the origin. Let A, B, C be the points a, b, c respectively. Then D0 is pa + qb + rc with p + q + r = 1 and p, q, r > 0. So a point on the line parallel to DD0 through A is a + s(pa + qb + rc. It is also in the plane DBC if s = -1/p, so A0 is the point - q/p b - r/p c. Similarly, B0 is - p/q a - r/q c, and C0 is - p/r a - q/r b.
The volume of ABCD is 1/6 |a x b.c| and the volume of A0B0C0D0 is 1/6 |(pa + (q + q/p)b + (r + r/p)c) x ((p + p/q)a + qb + (r + r/q)c).((p + p/r)a + (q + q/r)b + rc)|
Thus vol A0B0C0D0/vol ABCD = abs value of the determinant:
| p q + q/p r + r/p | | p + p/q q r + r/q | | p + p/r q + q/r r |which is easily found to be 2 + p + q + r = 3.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
25 Sep 1998
Last updated/corrected 24 Sep 2003