For Mark

The incentre $I$ of the triangle has barycentric coordinates $a:b:c$ . The orthocentre $H$ of the triangle has barycentric coordinates $\sec A:\sec B :\sec C$ , or $\tfrac{1}{b^2+c^2-a^2}:\tfrac{1}{a^2+c^2-b^2}:\tfrac{1}{a^2+b^2-c^2}$ . Of course, these coordinates are singular if the triangle is right-angled, but the normalized version of these coordinates is not singular when the triangle is right-angled, so there is no problem here.

The Fletcher point $Fl$ has barycentric coordinates $a\left(2 - \tfrac{\cos^2\frac12B + \cos^2\frac12C}{\cos^2\frac12A}\right)\,:\,b\left(2 - \tfrac{\cos^2\frac12A + \cos^2\frac12C}{\cos^2\frac12B}\right) \;:\; c\left(2 - \tfrac{\cos^2\frac12A + \cos^2\frac12B}{\cos^2\frac12C}\right)$ After some substitution, these coordinates cam be written $\tfrac{(b-c)^2 + (b+c)a - 2a^2}{b+c-a} \;:\; \tfrac{(a-c)^2 + (a+c)b - 2b^2}{a+c-b} \;:\; \tfrac{(a-b)^2 + (a+b)c - 2c^2}{a+b-c}$ The area of $HIFl$ is equal to the modulus of $\mathrm{det}M\,|ABC|$ , where $|ABC|$ is the area of $ABC$ and $\mathrm{det}M$ is the determinant of the $3 \times 3$ matrix whose rows are the normalized barycentric coordinates of $H,\,I,\,Fl$ . Thus $|HIFl| \; = \; \left| \frac{(a-b)(a-c)(b-c)(a^2+b^2+c^2-2ab-2ac-2bc)}{2(a+b+c)\big(a^4+b^4+c^4-(a^3b+ab^3+a^3c+ac^3+b^3c+bc^3)+abc(a+b+c)\big)}\right|\,|ABC|$ If we write $a=v+w$ , $b=u+w$ , $c = u+v$ , then we have $|HIFl| \; = \; \left|\frac{(u-v)(u-w)(v-w)(uv + uw + vw)}{2(u+v+w)(u^2(v-w)^2 + v^2(u-w)^2 + w^2(u-v)^2)}\right|\,|ABC|$ and hence $HIFL$ is nondegenerate provided that $u,v,w$ are distinct, and hence provided that $a,b,c$ are distinct.

Let us call the triangle $HIFl$ the FHI-triangle of $ABC$ . If $a,b,c$ are integers and $|HIFl|$ is an integer, then $|ABC|$ is rational, and so we can find $P \in\mathbb{N}$ such that $P^2|ABC|$ is an integer, and hence the triangle with sides $Pa,Pb,Pc$ is Heronian. Thus we can find $Q \in \mathbb{N}$ such that the triangle $\Delta_0$ with sides $\tfrac{P}{Q}a,\tfrac{P}{Q}b,\tfrac{P}{Q}c$ is a primitive Heronian triangle. If we write $\tfrac{P}{Q} = \tfrac{p}{q}$ as a fraction in lowest terms, we see that any prime factor of $p$ will be a common factor of the sides of $\Delta_0$ . Since $\Delta_0$ is primitive, we deduce that $p$ has no prime factors, and hence that $p=1$ . If we let $\Gamma_0$ FHI-triangle of the Heronian triangle $\Delta_0$ , then $|HIFl| = q^2|\Gamma_0|$ and we are interested in choosing $q$ to be as small as possible, since we want to minimize $|HIFl|$ . Suppose that $|\Gamma_0| = \tfrac{m}{n}$ , where $m,n \in \mathbb{N}$ are coprime. If the prime factorisation of $n$ is $p_1^{r_1}p_2^{r_2} \cdots p_N^{r_N}$ , where $p_1,p_2,\ldots,p_N$ are distinct primes and $r_1,r_2,\ldots,r_N \in \mathbb{N}$ , then $q$ must be equal to $p_1^{s_1}p_2^{s_2} \cdots p_N^{s_N}$ , where $s_j \; =\; \left\{ \begin{array}{lll} \frac12r_j & \hspace{1cm} & r_j\;\mbox{even} \\ \frac12(r_j+1) & & r_j\; \mbox{odd} \end{array}\right.$ and hence we deduce that $|HIFl| = \xi(n)m$ , where $\xi(n)$ , sometimes called the core of $n$ , is the product of the primes that occur with odd multiplicity in the prime factorization of $n$ .

Thus, for any nondegenerate primitive Heronian triangle $\Delta$ , calculate the area $|\Gamma|$ of its FHI-triangle $\Gamma$ , which will be a rational $\tfrac{m}{n}$ in lowest terms. Then $\Xi(\Delta) = \xi(n)m$ will be the smallest possible area of the FHI-triangle of a triangle with integer sides, similar to $\Delta$ , whose FHI-triangle has integral area.

The Mathematica code

amax = 1000;
data = {};
For[a = 3, a <= amax, a++,
    For[b = Floor[a/2] + 1, b <  a, b++,
        For[c = a - b + 1, c < b, c++,
            If[GCD[GCD[a,b],c] > 1,,
                If[IntegerQ[Sqrt[(a + b + c) (a + b - c) (a + c - b) (b + c - a)/16]],AppendTo[data, {a, b, c}],
                           ]]]]]

calculates the $5853$ nondegenerate primitive Heronian triangles with largest side of $1000$ or less, and it is easy to apply the above theory to find the smallest triangle of the desired type that is similar to each of these. Of all these triangles, the smallest area is $\boxed{66}$ , coming from the triangle with sides $29,65,68$ . The next smallest area is $286$ (coming from a triangle similar to the $3,4,5$ triangle), and the one after that is $2160$ .

In a separate calculation, I also checked all triangles with $u,v,w \le 600$ (where either $u,v,w$ are all integers or all half-integers) and found no smaller area than $66$ either.

The diagram shows the optimal solution. The Soddy line through the incentre I and the Gergonne point is shown in green, and the Gergonne line is shown in light blue (only two of the Nobbs points are visible on the diagram, but two are enough to define the line). The Fletcher point is the intersection of these two perpendicular lines, and the FHI-triangle is shown in orange.