How many different triangles with integer side lengths are there such that the sum of the lengths of any 2 sides is at least 5 units more than the length of the third side, and that the area is numerically twice the perimeter?
We are going to consider the more general problem where the area is numerically twice the perimeter, without the +5 constraint. We can use Heron and the information in the problem to derive the controlling equality:
64(a + b + c) = ( b + c-a) (a + c-b) (a + b - c)
Clearly, the smallest of the 3 differences, s, of ( b + c-a), (a + c-b), and (a + b - c) must be even. Say s = 2. Then, for
64(a + b + c) = ( b + c-a) (a + c-b) (a + b - c), a + b = c + 2, we have -a^2+ac+2a-17c-17=0, with 20 solutions.
We can do the same for s=4,6,8, and so on, and always be guaranteed a finite number of solutions. There is no need to test beyond 12, since only the triplet {13,14,15} has such a large s, because its members are consecutive.*
We can now list every triple in the natural numbers whose area is twice its perimeter:
1. Twice every equable triangle (there are only 5):
10 8 6, so 12,16,20: area 96 (in fact, 4 times the 3,4,5 triangle) (s=8)
13 12 5, so 10,24,26: area 120 (s=8)
17 10 9, so 18,20,34: area 144
20 15 7, so 14,30,40: area 168
29 25 6, so 12,50,58: area 240
2. Three times the Heronian triples:
8 5 5, so 15,15,24: area 108 (s=6) and
3 25 26, so 9,75,78: area 162 (s=6)
3. All others:
15 14 13: area 84 (s=12)
30 25 11: area 132 (s=6)
37 26 15: area 156
39 35 10: area 168 (s=6)
41 40 9: area 180 (s=8)
65 34 33: area 264
74 51 25: area 300
97 90 11: area 396
21 85 104: area 420
19 153 170: area 684
18 289 305: area 1224
Those with s>5 are so marked, 8 in all, as Larry correctly observed.
*Note: -a^3-3a^2+193a+195 = 0, with 13 as the sole solution in the positive integers.
Edited on November 22, 2019, 1:53 am
|
|
Posted by broll
on 2019-11-22 01:42:22 |
Full Solution
