Solving Triangles - Part 10

The sides $a,b,c$ of a triangle can be written neatly as $a \; = \; u+v \hspace{1cm} b \; = \; u + w \hspace{1cm} c \; = \; v + w$ where $u,v,w > 0$ . Then the semiperimeter of the triangle is $s = u+v+w$ , and so Heron's formula gives the area of the triangle as $\Delta \; = \; \sqrt{uvw(u+v+w)}$ The condition $a^2 + bc = 6$ gives $\begin{aligned} (v+w)^2 + (u+w)(u+v) & = \; 6 \\ u(u+v+w) & = \; 6 - 2(v+w)^2 - vw \end{aligned}$ and hence $\Delta^2 \; = \; vw(6 - 2v^2 - 5vw - 2w^2) \; = \; 1 - (3vw-1)^2 - 2vw(v-w)^2$ and so $\Delta$ takes the maximum value of $1$ when $3vw=1$ and $v=w$ , namely when $v=w=\tfrac{1}{\sqrt{3}}$ , which is an acceptable solution, since then $u = \tfrac{\sqrt{10}-1}{\sqrt{3}}$ . Thus the maximum area of $ABC$ is $\boxed{1}$ .