23rd CanMO 1991 Problem 5 An equilateral triangle side n is divided into n 2 equilateral triangles side 1 by lines parallel to its sides. How many parallelograms can be formed from the small triangles? [For example, if n = 3, there are 15, nine composed of two small triangles and six of four.] Solution
: (n+2)(n+1)n(n-1)/8 = 3 (n+2)C4. The parallelogram must have angles 60 o , 120 o , 60 o , 120 o . The 60 o angles can be aligned in three ways (with one of the vertices of the triangle) and the number of parallelograms for each way is obviously equal. Suppose it is a n (so that the total number of parallelograms is 3 a n ). Let the big triangle be ABC. Let b n be the number of parallelograms where one vertex of the parallelogram is A. Then we evidently have a n = b n + 2 b n-1 + 3 b n-2 + ... + (n-1) b 2 . But it is easy to calculate b n directly. Such a parallelogram is uniquely defined by the position of the vertex opposite to A. But that can have any position in the rows of vertices (of the small triangles) provided it is at least two rows below A and not at either end. Thus there are 1 + 2 + ... + n-1 = n(n-1)/2 possible positions for it. So b n = n(n-1)/2 and b n+1 - b n = n. Hence a n+1 - a n = 1.n + 2(n-1) + 3(n-2) + ... + n.1 = n(1 + 2 + ... + n) - (1.2 + 2.3 + ... + (n-1)n) = n 2 (n+1)/2 - (n-1)n(n+1)/3 = n(n+1)(n+2)/6 = (n+2)C3. Now we claim that a n = (n+2)C4. It is true for n = 2 (obviously a 2 = 1). So the result is a trivial induction, since (n+2)C4 + (n+2)C3 = (n+3)C4. Hence, finally, the total number of parallelograms is 3a n = 3 (n+2)C4. 23rd CanMO 1991 © John Scholes jscholes@kalva.demon.co.uk 29 Aug 2002