Triangle PerimetersDate: 12/15/98 at 21:36:24 From: Eric Olmsted Subject: Triangle perimeter computation How many triangles with integer sides have a given perimeter? I am trying to use the law of inequality, but I am having problems applying this law to the question. Do you have any suggestions? Date: 12/17/98 at 10:28:59 From: Doctor Rob Subject: Re: Triangle perimeter computation Thanks for writing to Ask Dr. Math! This problem turns out to be quite complicated, but possible. Let the sides be a, b, and c, such that: 0 < a <= b <= c Then: p = a + b + c By the Triangle Inequality: c < a + b (The other two inequalities are automatically true.) Now you have to figure out how to put these together. For starters: p = a + b + c <= 3*c p/3 <= c < a + b p/3 <= p - a - b < a + b p/3 + a + b <= p < 2a + 2b p/2 < a + b <= 2*p/3 p/2 - a < b <= 2*p/3 - a (p+1)/2 - a <= b <= 2*p/3 - a Furthermore, 0 < a <= b <= p - a - b 0 < a <= b <= (p-a)/2 Putting these together, we get: max(a,[(p+1)/2]-a) <= b <= min([2*p/3]-a,[(p-a)/2]) where [x] is the greatest integer less than or equal to x. Further we see that: 3*a <= a + b + c = p 1 <= a <= p/3 1 <= a <= [p/3] Now a <= [(p+1)/2] - a if and only if a <= [(p+1)/2]/2 if and only if a <= [(p+1)/4]. Furthermore, [2*p/3] - a <= [(p-a)/2] if and only if a >= p - 2*[(p+2)/3] (this is not obvious, but true). That means that we should break up the range of a into three parts: 1: 1 <= a <= [(p+1)/4] [(p+1)/2]-a <= b <= [(p-a)/2] 2: [(p+1)/4] + 1 <= a <= p - 2*[(p+2)/3] a <= b <= [(p-a)/2] 3: p - 2*[(p+2)/3] + 1 <= a <= [p/3] a <= b <= [2*p/3]-a The third part is empty (there are no values of a satisfying the conditions) unless (p+2)/3 is an integer, in which case it contains the single value a = (p-1)/3, and then b = (p-1)/3, too. Then the number of possibilities for the pairs (a,b) is: [p/3] min([2*p/3]-a,[(p-a)/2]) N(p) = SUM SUM 1 a=1 b=max(a,[(p+1)/2]-a) which can be evaluated by splitting the outer sum into three parts according to the above scheme, and summing each part separately. I leave the details of this to you. Since the denominators inside [.] are 2, 3, and 4, whose LCM is 12, 12 cases can be considered separately, each corresponding to a remainder obtained when dividing p by 12. For example, when p leaves remainder 0 on division by 12, the number N(p) = p^2/48, and when the remainder is 1, N(p) = (p^2+6*p-7)/48, and so on. In all cases, the highest degree term is p^2/48. The numbers look like this, starting with p = 0 and reading left to right: 0 0 0 1 0 1 1 2 1 3 2 4 3 5 4 7 5 8 7 10 8 12 10 14 12 16 14 19 16 21 19 24 21 27 24 30 27 33 30 37 33 40 37 44 40 48 44 52 48 56 52 61 56 65 61 70 65 75 70 80 75 85 80 91 85 96 91 102 96 108 102 114 108 120 114 127 120 133 127 140 133 147 140 154 147 161 154 169 161 176 169 184 176 192 184 200 192 208 200 217 208 ... - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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