Question

The lengths of the sides of the a triangle are integrals, and its area is also integer. One side is $21$ and the perimeter is $48$. Find the shortest side.

Answer 1

Given : a + b + c = 48 $\Rightarrow s=\frac{48}{2}=24$

Also, C = 21, hence a + b = 27 $\Rightarrow b=27-a$. we define a>b, so that

a > 13. Area of $△=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{24(24-a)(24-27+a)(24-21)}$

$\Rightarrow A=\sqrt{24\times 3(24-a)(a-3)}=\sqrt[6]{62(a-3)(24-a)}$

$\Rightarrow {\left(\frac{A}{6}\right)}^2=2(a-3)(24-a)$

In order A to be integer, the RMS should be a square number with condition a<24. By condition that a>13 but <24, a may be any of 14,15,16,17,18,19,20,21,22 or 23. So that b = (27-a) shall be any of 13,12,11,10,9,8,7,6,5 or 4 respectively. C cannot be <b since C=21. On the other hand b<a. Therefore b is the shortest of 3 sides. In order that the area A is an integer, we tabulate the possible set of a,b,a-3,24-a and 2(a-3)(24-a) i.e, the desired square RHS.

a b a-3 24-a 2(a-3)(24-a)

14 13 11 10 220

15 12 12 9 216

16 11 13 8 208

17 10 14 7 $196$

18 9 15 6 180

19 8 16 5 160

20 4 17 4 136

21 6 18 3 108

22 5 19 2 76

23 4 20 1 40

By verification, we find that the RHS shall be square only if it is 196. Correspondingly b = 10 is the only possible length of the shortest side of the triangle.