XIX OBM

Let $A=(F_{n+4},0)$, $B=(0,F_{n+2})$ and $C=(F_{n+3},F_n)$. Notice that $$AB=\sqrt{F_{n+4}^2+F_{n+2}^2}=V_{n+2},BC=\sqrt{F_{n+3}^2+(F_{n+2}-F_n{)}^2}=\sqrt{F_{n+3}^2+F_{n+1}^2}=V_{n+1}$$ and $CA=\sqrt{(F_{n+4}-F_{n+3}{)}^2+F_n^2}=\sqrt{F_{n+2}^2+F_n^2}=V_n$. Its area equals $\frac{|D|}{2}$, where $$\begin{array}{rl}D& =\left|\begin{array}{ccc}{F}_{n+4}& 0& 1\\ 0& F_{n+2}& 1\\ F_{n+3}& F_n& 1\end{array}\right|=F_{n+4}{F}_{n+2}-F_{n+3}{F}_{n+2}-F_n{F}_{n+4}\\ & =F_{n+4}(F_{n+2}-F_n)-F_{n+3}{F}_{n+2}=F_{n+4}{F}_{n+1}-F_{n+3}{F}_{n+2}\\ & =\left|\begin{array}{cc}{F}_{n+1}& F_{n+2}\\ F_{n+3}& F_{n+4}\end{array}\right|\end{array}$$ Let $A_{n+1}=$. An easy induction proves that $A_n=$. $($. Indeed, $A_0=$ and $$\begin{array}{rl}{A}_{n+1}\cdot \left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)& =\left(\begin{array}{cc}{F}_{n+1}& F_{n+2}\\ F_{n+3}& F_{n+4}\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}{F}_{n+2}& F_{n+1}+F_{n+2}\\ F_{n+4}& F_{n+3}+F_{n+4}\end{array}\right)\\ & =\left(\begin{array}{cc}{F}_{n+2}& F_{n+3}\\ F_{n+4}& F_{n+5}\end{array}\right)=A_{n+2}\end{array}$$ Hence $$D=|A_{n+1}|=\left|\begin{array}{cc}0& 1\\ 1& 2\end{array}\right|\cdot {\left|\begin{array}{cc}0& 1\\ 1& 1\end{array}\right|}^{n+1}=(0\cdot 2-1\cdot 1)\cdot (0\cdot 1-1\cdot 1{)}^{n+1}=(-1{)}^{n+2}$$ and area $ABC=\frac{|D|}{2}=\frac{1}{2}$.