Looking at the proposed configuration, we can find the minimum $s$ for that configuration as follows:
notice how the two triangles on the sides overlap. This overlap is a smaller similar triangle (that is, it's equilateral), the larger that triangle is, the smaller $s$ is, so we seek to maximum the size of that triangle. We have the following equalities to work with: The sum of the altitudes of the big triangle in the middle, and the little triangle must be 1.
$A+a=1$
and the length of half a side of the large triangle minus the length of half a side of the small triangle must be 1/2
$\frac{S}{2}-\frac{s}{2}=\frac{1}{2}$.
This latter expression comes from the fact that the triangles on the sides are bisected by the vertical sides of the square.
We appear to have 4 variables, but since $a \propto s$, and $A \propto S$, we really only have two, and can solve this linear system easily enough (I leave the details to the reader).
Now for the trickier part: proving that this configuration is best possible.
To do that, we will pick on the corners of the square. It has 4 corners and any configuration of triangles with covers the square must cover all 4 corners. Therefore we have proven that:
At least one of the triangles in an optimal configuration must cover two of the corners. Observe that the proposed configuration does this.
Now, suppose we have a configuration with a triangle covering the bottom two corners, and one more for each top corner. We are free to move them around, subject to the constraints that they must continue to cover those corners. We concentrate on the triangle covering two corners. Our goal is to use as much of the area of that triangle as possible to cover the square, which is achieved when it's side is flush with the square (100% usage). We can shrink the the triangle all the way down to $s=1$ and still have it covering both corners, but as the linked problem's solution showed, $s=1$ is not sufficient, so $s>1$.
What about the other two triangles? For them, we need that they, together, cover whatever remaining area of the square that the middle triangle (the one covering two corners) doesn't cover. The idea is to position them so that the area of overlap, among the triangles is minimized, this suggests that we would want them to be flush with the middle triangle, but if the length of their altitudes (which are all the same length) is less than 1, having no overlap is impossible by the shape of the remaining area not being two triangles, and as the configuration shows, this can happen. So we will have to have some overlap.
This shows that the shape of the remaining area has the general form of two triangles with quadrilaterals attached to each of them (we can cut it in two, by extending the vertical altitude of the middle triangle, for example). Each of these pieces has a corner of the square as one of its corners, hence the applicable triangle must still cover that corner. Another way of thinking of this shape, is that it is the rectangle given by the the two corners which have a triangle each, and the intersections of the sides of the middle triangle with the vertical edges of the square, with a triangular piece cut out from the bottom (it took all of the bottom edge with it). This is a non-convex pentagonal shape, and we must cover it with just two triangles, hence, certainly the convex rectangle (is this the convex hull of the pentagon?) must have its 4 corners covered by two triangles. So one of the remaining triangles must cover at least two corners of the rectangle, I claim that having it cover three will be less efficient. The reason being that it must cover the diagonal of the rectangle which will have length greater than 1 because the top side is length 1 already (we can find the exact length from knowing where the middle triangle intersects the vertical sides of the square using Pythagoras's theorem). So we have two triangles covering two corners each (left and right sides). The must also must both cover the inside point of the pentagon and all the area above it together. This is the necessary overlap, and it is minimized exactly when the triangles are flush with the edges which have the inside point as an endpoint (or in other words when they are flush with the middle triangle).
The idea from here is show that, as $S$ becomes smaller and smaller, that the triangles have less and less freedom as to how they can positioned until you end up with exactly the configuration pictured in the OP. This may be able to be argued by showing that the pictured configuration always minimizes overlapped area for a fixed $S$.
Best Answer
Here's a solution using "advanced mathematics" that may nevertheless be useful. Our starting point will be the equation $\sin^2(\pi/p)+\sin^2(\pi/q)=\sin^2(\pi/r)$ from David's comment. Part I will be the only part using algebraic number theory; everything else will be "elementary." It's quite plausible that Part I can be circumvented somehow.
Part I. Showing that $p\mid 6\operatorname{lcm}(q,r)$, $q\mid 6\operatorname{lcm}(r,p)$, and $r\mid 6\operatorname{lcm}(p,q)$.
Firstly, for any positive integer $n$, we see $$\mathbb Q\left(\sin^2\frac{\pi}n\right)=\mathbb Q\left(\cos\left(\frac{2\pi}n\right)\right),$$ since $\sin^2 x=1-\cos(2x)$. The field $\mathbb Q(\cos(2\pi/n))$ is the maximal real subfield of $\mathbb Q(\zeta_n)$, where $\zeta_n=e^{2\pi i/n}$ is a primitive $m$th root of unity. Let $(a,b,c)$ be a positive integers with $$\pm\sin^2\frac{\pi}a\pm\sin^2\frac{\pi}b\pm\sin^2\frac{\pi}c=0$$ for some choice of the three $\pm$ signs. Then $$\cos\left(\frac{2\pi}a\right)\in\mathbb Q\left(\sin^2\frac{\pi}a\right)\subset\mathbb Q\left(\sin^2\frac{\pi}b,\sin^2\frac{\pi}c\right)\subset \mathbb Q(\zeta_b,\zeta_c).$$ Since $\mathbb Q(\zeta_b,\zeta_c)=\mathbb Q(\zeta_{\operatorname{lcm}(b,c)})$, we get $$\cos\left(\frac{2\pi}a\right)\in\mathbb Q\left(\zeta_{\operatorname{lcm}(b,c)}\right).$$ This means that $\zeta_a$ is at most degree $2$ over $\mathbb Q\left(\zeta_{\operatorname{lcm}(b,c)}\right)$, and thus that $$\left[\mathbb Q\left(\zeta_{\operatorname{lcm}(a,b,c)}\right):\mathbb Q\left(\zeta_{\operatorname{lcm}(b,c)}\right)\right]\leq 2.$$ Since $\mathbb Q(\zeta_n)$ is degree $\varphi(n)$ over $\mathbb Q$, where $\varphi$ is Euler's totient function, this means $$\frac{\varphi(\operatorname{lcm}(a,b,c))}{\varphi(\operatorname{lcm}(b,c))}\leq 2.$$ Using the multiplicative properties of $\varphi$, we see $$\frac{\varphi(xy)}{\varphi(y)}=x\prod_{\substack{\ell\text{ prime}\\\ell\mid x,\ell\nmid y}}\left(1-\frac1\ell\right)$$ is a product over all primes dividing $x$ of either $\ell$ (if $\ell\mid y$) or $\ell-1$ (if $\ell\nmid y$), so for this to be at most $2$ we must have $x\mid 6$; in particular, $$a\mid \operatorname{lcm}(a,b,c)\mid 6\operatorname{lcm}(b,c),$$ as desired.
Part II. Technical bounding.
Lemma 1. Let $x_1,x_2,x_3$ and $k$ be positive integers so that $x_i\mid k\operatorname{lcm}(x_j,x_k)$ for every permutation $(i,j,k)$ of $(1,2,3)$. Then $$\operatorname{lcm}(x_1,x_2,x_3)^2\mid kx_1x_2x_3.$$
Proof. We'll show for each prime $\ell$ that $\nu_\ell$ of the left side is at most $\nu_\ell$ of the right. Let $y_1,y_2,y_3$ be $\nu_\ell(x_1),\nu_\ell(x_2),\nu_\ell(x_3)$, ordered so that $y_1\geq y_2\geq y_3$, and let $(i,j,k)$ be a permutation of $(1,2,3)$ so that $y_1=\nu_\ell(x_i)$, $y_2=\nu_\ell(x_j)$, and $y_3=\nu_\ell(x_k)$. Then, since $x_1\mid k\operatorname{lcm}(x_2,x_3)$, $y_1\leq \nu_\ell(k)+y_2$. This means that $$\nu_\ell\left(\operatorname{lcm}(x_1,x_2,x_3)^2\right)=2y_1\leq y_1+\nu_\ell(k)+y_2\leq \nu_\ell(kx_1x_2x_3),$$ as desired. $\square$
Lemma 2. If $0<\alpha,\beta,\gamma<\pi/2$ with $\sin^2\alpha+\sin^2\beta=\sin^2\gamma$, then $$0<\gamma^2-\alpha^2-\beta^2\leq \frac{\gamma^4}3-\frac{\gamma^6}{36}.$$ Proof. For the right inequality, we need only use the Taylor-derived inequalities $$\left(x-\frac{x^3}6\right)^2\leq \sin^2 x\leq x^2,$$ valid for all of $\mathbb R$. For the left inequality, consider the function $$f(x)=\arcsin(\sqrt x)^2$$ defined on $[0,1]$. It satisfies $f''(x)>0$, and so $f'$ is increasing, meaning that for $\beta>0$ $$\gamma^2=f(\sin^2\alpha+\sin^2\beta)=f(\sin^2\alpha)+\int_{\sin^2\alpha}^{\sin^2\alpha+\sin^2\beta}f'(x)dx>\alpha^2+\int_0^{\sin^2\beta}f'(x)dx=\alpha^2+\beta^2.$$
Lemma 3. If $r>10$ and $\sin^2(\pi/p)+\sin^2(\pi/q)=\sin^2(\pi/r)$ with $p\geq q\geq r$ integers, then $q\leq (r+0.2)\sqrt 2$ and $p\leq \sqrt{r^3/2}+r/2$.
Proof. By Lemma 2, after dividing out by $\pi^2$, $$0<\frac1{r^2}-\frac1{p^2}-\frac1{q^2}\leq \frac{\pi^2}{3r^4}-\frac{\pi^4}{36r^6}\implies\frac1{p^2}+\frac1{q^2}\geq \left(\frac1r-\frac{\pi^2}{6r^3}\right)^2.$$ On one hand, this implies $$\frac2{q^2}\geq \left(\frac1r-\frac{\pi^2}{6r^3}\right)^2\implies q\leq \sqrt2(r+0.2)$$ for $r>10$. On the other hand, since $q\geq r+1$, this gives $$\frac1{p^2}\geq \frac1{r^2}-\frac{\pi^2}{3r^4}+\frac{\pi^4}{36r^6}-\frac1{(r+1)^2}\implies p\leq \sqrt{\frac{r^3}2}+\frac r2$$ for $r>10$.
Part III. Finishing the proof, almost.
We will show there are no solutions to $\sin^2(\pi/p)+\sin^2(\pi/q)=\sin^2(\pi/r)$ if $r>417$. Note that $$\frac1{r^2}-\frac1{p^2}-\frac1{q^2}$$ is a positive real number (by Lemma 2) and has denominator at most $\operatorname{lcm}(p,q,r)^2\leq 6pqr$ (using Lemma 1), so it is at least $\frac1{6pqr}$. On the other hand, by Lemma 2, $$\frac1{6pqr}\leq \frac1{r^2}-\frac1{p^2}-\frac1{q^2}< \frac{\pi^2}{3r^4},$$ so $r^3\leq 2\pi^2 pq$. The bounds from Lemma 3 give that $$r^3\leq 2\pi^2 (\sqrt 2)(r+0.2)\left(\sqrt{\frac{r^3}2}+\frac r2\right).$$ The right side grows like $r^{5/2}$ while the left side grows like $r^3$, so there is some threshold value $r_0$ for which this inequality fails for $r>r_0$. This turns out to be about $417$.
Part IV.
All that's left is to painstakingly check the possible solutions with $r\leq 417$. I prefer to do this with Python, and here's the code I ran.
This finds all solutions which are close numerically (within $10^{-10}$, which should be large enough that any floating point errors introduced won't be an issue), filters by the number-theoretic condition from part 1, and returns those that pass. The code intentionally passes through more cases than necessary (iterating through both $p$ and $q$ for $r>10$ instead of just $q$) to perform few "messy" computations that might introduce floating point error. The three solutions returned are $(6,4,3)$ (as noticed in the question), $(6,6,4)$, and $(10,6,5)$.