I am currently studying maths at a high school level, I have been given this problem but can’t find any route into it, I have experimented with using Heron’s formula and the expression for the nth term of a Fibonacci sequence.I know that you can express terms in a Fibonacci sequence in terms of the golden ratio and I tried combining this golden ratio identies that I knew to simplify the value for half the perimeter so I could use Heron’s formula, however I wasn’t able to do this, I also tried expressing the sum of terms as one term for the perimeter due to the term to term relationship of the Fibonacci sequence, however I also wasn’t able to do this. I would really appreciate a worked solution so I can understand this problem
Here $F_{(n)}$ is the $n$–th number in the sequence of Fibonacci numbers, where $F_{(0)} = 0$, $F_{(1)} = 1, \ldots.$
The triangle $T_{n}$, (for $n > 0$), has sides
$$F_{(n+2)},$$ $$\sqrt{F_{(2n+1)}},$$ $$\sqrt{F_{(2n+2)}}+F_{(n+1)}$$ (you may assume the triangle exists). If $T_{(n)}$ has an area that is an integer, what is this area?
The first side $a$ is a whole number. The second side $b$ cannot be for any positive integer $n$ (the only odd-numbered term in the Fibonacci sequence that's a square is $F_1=1$, corresponding to $n=0$), but it is the square root of a whole number. That leaves the third side $c$, which may or may not be a whole number.
Let's see what we could do with Heron's Formula if the third side were a whole number:
$(Area)=\frac14\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=\frac14\sqrt{[(a+c)^2-b^2][b^2-(a-c)^2]}$
where $a+c,a-c$ and $b^2$ are all whole numbers if we pick $n$ to make $c$ one. Then the area is the square root of a rational number, and we would check whether that rational radicand perchance hits on a whole-number square.
To make $c$ a whole number and thus get a rational square-root radicand, we need $F_{2n+2}$ to be a square, which occurs for $n=0$ or $n=5$. Trying these two values we find $n=0$ gives a degenerate case whereas $n=5$ gives a proper triangle and Heron's Formula hits with an area of $50$.