30th IMO shortlist 1989/17 solution

30th IMO 1989 shortlist

Problem 17

Show that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle, but that the obtuse angle cannot exceed 120^o.

Solution

We may take one vertex of the tetrahedron to be the origin, and one edge to lie along the x-axis, another along the vector (1, √3, 0) and the third along the vector (1, 1/√3, √(8/3) ). There is no loss of generality in assuming that one vertex of the triangle lies on each of these edges, and indeed we may take one vertex to be P = (2, 0, 0). So the others are Q = h (1, √3, 0) and R = k (1, 1/√3, √(8/3) ) with h, k > 0.

So PQ² = (2 - h)² + 3h² = 4 - 4h + 4h², PR² = (2 - k)² + k²/3 + 8k²/3 = 4 - 4k + 4k², QR² = (h - k)² + (h√3 - k/√3)² + 8k²/3 = 4h²/3 - 4hk + 4k².

Note also that OP² = 4, OQ² = 4h², OR² = 4k². So if h = k = 1, then all of OP, OQ, OR, PQ, PR, QR have length 2, which proves that the vectors above were correctly chosen.

The cosine formula gives cos P = (PQ² + PR² - QR²)/(2 PQ.PR) = (1 - h + h² + 1 - k + k² - h² + hk - k²)/(2 √(1 - h + h²)√(1 - k + k² ) = (1 + (h - 1)(k - 1) )/(2 √(1 - h + h²)√(1 - k + k² ).

For an obtuse angle we require cos P < 0, so one of h, k is greater than 1 and the other is less than 1. Without loss of generality h < 1, k > 1. Clearly, we can get cos P < 0. For example, h = 1/3, k = 3 gives cos P = -1/7.

(h - 1) > -1, so for cos P < 0 we need k > 2. As P increases from 90^o to 180^o, cos P decreases from 0 to -1, and cos 120^o = -1/2, so it is sufficient to show that (h + k - hk - 2)² < (1 - h + h²)(1 - k + k²) or that hk² + (h² - 5h + 3)k - 3 + 3h > 0 for 0 < h < 1 and k > 2. We can regard this as a quadratic in k. By completing the square, we see that its minimum value is when k = - (h² - 5h + 3)/2h. Certainly this is less than 2, because - (h² - 5h +3 ) < 4h or h² - h + 3 > 0 for 0 < h < 1. So the quadratic is an increasing function of k for k >= 2. Hence is value for k > 2 is greater than its value at k = 2 which is 4h + 2h² - 10h + 6 - 3 + 3h = 3 - 3h + 2h² > 3(1 - h) > 0.

Note that if we take h small and k large, then cos P is roughly -(k-2)/2k, so we can get close to 120^o.

30th IMO shortlist 1989