Triangle Complex Strike Again

For every positive $c$ the function $f(x)=x^3+cx+1$ is increasing, so it has exactly one real zero. Because $f(0)=1>0,$ it is negative, let's call it $-u,$ for some positive real $u$ . Then the other two zeroes of $f(x)$ are complex conjugates of each other: $v+wi$ and $v-wi,$ where $v$ and $w$ are real numbers, $w>0.$

By the Vieta's formula, the product of these three roots is $-1,$ therefore $(v+wi)\cdot (v-wi)\cdot (-u) =-1,$ thus $v^2+w^2=\frac{1}{u}.$ By the Vieta's formula, the sum of the roots is zero, so $-u+2v=0,$ thus $v=\frac{1}{2}u.$ Therefore, $w=\sqrt{\frac{1}{u}-\frac{u^2}{4}}=\frac{\sqrt{4u-u^4}}{2u}$

Note that since $c>0,$ $f(-1)=-c<0,$ so $0<u<1.$

The area of the triangle formed by the roots equals $(u+v)\cdot w.$ This simplifies to $\frac{3}{4}\sqrt{4u-u^4}.$ So, $\frac{3}{4}\sqrt{4u-u^4}=\frac{3}{25}\sqrt{109},$ therefore $4u-u^4=\frac{16\cdot 109}{5^4}.$ Using the Rational Root theorem, we can guess $u=\frac{4}{5}.$ This is the only solution for $0<u<1,$ because $4u-u^4$ is increasing on this interval (this is easy to check by its derivative).

Finally, by the Vieta's formula $c=(-u)\cdot 2v + (v^2+w^2)=-u^2+\frac{1}{u}=-\frac{16}{25}+\frac{5}{4}=\frac{61}{100}$ Therefore, $a+b=61+100=161.$