16th Chinese 2001 Problem A3 There is a bird at each vertex of a regular n-gon. Suddenly the birds all fly away and then return a minute later. After their return there is again one bird at each vertex, but not necessarily the same bird as before. For which n can we always find three birds such that the triangle formed by their initial positions is of the same type (acute, obtuse or right-angled) as the triangle formed by their final positions? Solution
: all n ≥ 3, except n = 5. Let O be the center of the n-gon. The triangle is acute, obtuse or right-angled according as O lies inside, outside or on the triangle. If n is odd, then O cannot lie on the triangle, so the triangle is not right- angled. The basic idea is that for n sufficiently large more than half the triangles are obtuse, so three birds must form an obtuse triangle both times. In fact, more than half the triangles are obtuse for odd n greater than 5 and for even n greater than 10. But it is easier to use a different argument for even n to avoid looking separately at n = 4, 6, 8 and 10. So consider first n even. Suppose birds A and B are directly opposite in the first arrangement. That means that ABC is right-angled for all C (not equal to A or B). If A and B are also directly opposite in the second arrangement, then any ABC is right-angled both times. If not, then suppose C is the bird directly opposite A in the second arrangement. That means ABC is right-angled in both arrangements. So for even n we can always find three birds whose two triangles are of the same type. Now consider n odd. Label the vertices 1, 2, 3, ... , 2n+1. Let us count triangles with 1 as a vertex. A triangle is acute-angled if the number of vertices between each pair of vertices of the triangle is n-1 or less. It is obtuse two vertices of the triangle have n or more vertices between them. So there is just 1 acute-angled triangle with vertices 1 and 2 (namely n+2). There are 2 with vertices 1 and 3 (namely n+2 and n+3), there are 3 with vertices 1 and 4 (namely n+2, n+3 and n+4), ... , and there are n with vertices 1 and n+1 (namely n+2, n+3, ... 2n). Similarly, there are n with vertices 1 and n+2, there are n- 1 with vertices 1 and n+3, ... , and there is 1 with vertices 1 and 2n+1. We have counted each triangle twice, so the total number of acute-angled triangles with 1 as a vertex is n(n+1)/2. The total number of triangles with 1 as a vertex is 2n(2n-1)/2, so there are n(3n-3)/2 obtuse triangles with 1 as a vertex. Thus more than half the triangles are obtuse iff 3n-3 > n+1, or n > 2 or 2n+1 >= 7. It remains to consider n = 3 and n = 5. The first is trivial because there is only one triangle. In the case n = 5 it is easy to check that if the birds initially at 1, 2, 3, 4, 5 move to 1, 3, 5, 2, 4, then all birds forming an acute-angled triangle initially form an obtuse triangle in the second arrangement and vice-versa. 16th Chinese 2001 © John Scholes jscholes@kalva.demon.co.uk 5 June 2002