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if the perimeter of triangle is given , then, how to find the shortest area of triangle with integral sides. Let, $P$ be perimeter and $s$ be semi-perimeter. we known,$s=P/2$

area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$\implies $ area of triangle = $\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{4}$

this equation contains four variables $a,b,c,P$. by using relation $P=a+b+c$ we can make the equation contains three independent variables.

area of triangle = $\frac{\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{4}$

then how to I continue?

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  • $\begingroup$ Minimizing $A$ is equivalent to minimizing $A^2$, which is a polynomial. That simplifies things. $\endgroup$ – mr_e_man Jun 3 '18 at 7:20
  • $\begingroup$ This looks like a problem to be solved by Lagrange multipliers, but that works with Real numbers, not integers. en.wikipedia.org/wiki/Lagrange_multiplier $\endgroup$ – mr_e_man Jun 3 '18 at 7:23
  • $\begingroup$ We know that the locus of a point the sum of its distance to two fixed points, constant is ellipse. suppose a(front side to angle A) coincides on the line $F_1 F_2$ , connecting foci. then area of triangle is zero when A locates on vertex on major axis and it is maximum when it locates on co-vertex. How can we estimate minimum area? $\endgroup$ – sirous Jun 3 '18 at 14:24
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    $\begingroup$ Hint: Define (non-negative) integers $$u:=-a+b+c \qquad v := a- b + c \qquad w := a+b-c$$ and note that $u+v+w = a+b+c=p$. Since $$16\cdot\text{area}^2 = puvw$$ your task reduces to minimizing $uvw$ subject to $u+v+w=p$. $\endgroup$ – Blue Jun 3 '18 at 17:56

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