IMO 1971/B1 solution

IMO 1971

Problem B1

All faces of the tetrahedron ABCD are acute-angled. Take a point X in the interior of the segment AB, and similarly Y in BC, Z in CD and T in AD.

(a) If ∠DAB + ∠BCD ≠ ∠CDA + ∠ABC, then prove that none of the closed paths XYZTX has minimal length;

(b) If ∠DAB + ∠BCD = ∠CDA + ∠ABC, then there are infinitely many shortest paths XYZTX, each with length 2 AC sin k, where 2k = ∠BAC + ∠CAD + ∠DAB.

Solution

The key is to pretend the tetrahedron is made of cardboard, cut it along three edges and unfold it. Suppose we do this to get the hexagon CAC'BDB'. Now the path is a line joining Y on B'C to Y' on the opposite side BC' of the hexagon. Clearly this line must be straight for a minimal path. If B'C and BC' are parallel, then we can take Y anywhere on the side and the minimal path length is the expression given.

But if they are not parallel, then the minimal path will come from an extreme position. Suppose CC' < BB'. If the interior angle CAC' is less than 180^o, then the minimal path is obtained by taking Y at C. But this does not meet the requirement that Y be an interior point of the edge, so there is no minimal path in the permitted set. If the interior angle CAC' is greater than 180, then the minimal path is obtained by taking X and T at A. Again this is not permitted.

The problem therefore reduces to finding the condition for B'C and BC' to be parallel. This is evidently angles BCD + DCA + CAD + BAD + BAC + ACB = 360^o. But DCA + CAD = 180^o - ADC, and BAC + ACB = 180^o - ABC, so we obtain the condition given.

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.

13th IMO 1971