Throughout the solution, triangles $A A_{1} D_{1}, B B_{1} A_{1}, C C_{1} B_{1}$, and $D D_{1} C_{1}$ will be referred to as border triangles. We will denote by $[\mathcal{R}]$ the area of a region $\mathcal{R}$. First, we show that $k \geq 1$. Consider a triangle $A B C$ with unit area; let $A_{1}, B_{1}, K$ be the midpoints of its sides $A B, B C, A C$, respectively. Choose a point $D$ on the extension of $B K$, close to $K$. Take points $C_{1}$ and $D_{1}$ on sides $C D$ and $D A$ close to $D$ (see Figure 1). We have $\left[B B_{1} A_{1}\right]=\frac{1}{4}$. Moreover, as $C_{1}, D_{1}, D \rightarrow K$, we get $\left[A_{1} B_{1} C_{1} D_{1}\right] \rightarrow\left[A_{1} B_{1} K\right]=\frac{1}{4}$, $\left[A A_{1} D_{1}\right] \rightarrow\left[A A_{1} K\right]=\frac{1}{4},\left[C C_{1} B_{1}\right] \rightarrow\left[C K B_{1}\right]=\frac{1}{4}$ and $\left[D D_{1} C_{1}\right] \rightarrow 0$. Hence, the sum of the two smallest areas of border triangles tends to $\frac{1}{4}$, as well as $\left[A_{1} B_{1} C_{1} D_{1}\right]$; therefore, their ratio tends to 1, and $k \geq 1$.

We are left to prove that $k=1$ satisfies the desired property.

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Figure 1

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Figure 2

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Figure 3

Lemma. Let points $A_{1}, B_{1}, C_{1}$ lie respectively on sides $B C, C A, A B$ of a triangle $A B C$. Then $\left[A_{1} B_{1} C_{1}\right] \geq \min \left\{\left[A C_{1} B_{1}\right],\left[B A_{1} C_{1}\right],\left[C B_{1} A_{1}\right]\right\}$.

Proof. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the midpoints of sides $B C, C A$ and $A B$, respectively.

Suppose that two of points $A_{1}, B_{1}, C_{1}$ lie in one of triangles $A C^{\prime} B^{\prime}, B A^{\prime} C^{\prime}$ and $C B^{\prime} A^{\prime}$ (for convenience, let points $B_{1}$ and $C_{1}$ lie in triangle $A C^{\prime} B^{\prime}$; see Figure 2). Let segments $B_{1} C_{1}$ and $A A_{1}$ intersect at point $X$. Then $X$ also lies in triangle $A C^{\prime} B^{\prime}$. Hence $A_{1} X \geq A X$, and we have
$$\frac{\left[A_{1} B_{1} C_{1}\right]}{\left[A C_{1} B_{1}\right]}=\frac{\frac{1}{2} A_{1} X \cdot B_{1} C_{1} \cdot \sin \angle A_{1} X C_{1}}{\frac{1}{2} A X \cdot B_{1} C_{1} \cdot \sin \angle A X B_{1}}=\frac{A_{1} X}{A X} \geq 1$$
as required.

Otherwise, each one of triangles $A C^{\prime} B^{\prime}, B A^{\prime} C^{\prime}, C B^{\prime} A^{\prime}$ contains exactly one of points $A_{1}$, $B_{1}, C_{1}$, and we can assume that $B A_{1}1$; also, lines $A_{1} C^{\prime}$ and $C A$ intersect at a point $Z$ on the extension of $C A$ beyond point $A$, hence $\frac{\left[A_{1} B_{1} C^{\prime}\right]}{\left[A_{1} B^{\prime} C^{\prime}\right]}=\frac{B_{1} Z}{B^{\prime} Z}>1$. Finally, since $A_{1} A^{\prime} \| B^{\prime} C^{\prime}$, we have $\left[A_{1} B_{1} C_{1}\right]>\left[A_{1} B_{1} C^{\prime}\right]>\left[A_{1} B^{\prime} C^{\prime}\right]=\left[A^{\prime} B^{\prime} C^{\prime}\right]=\frac{1}{4}[A B C]$.

Now, from $\left[A_{1} B_{1} C_{1}\right]+\left[A C_{1} B_{1}\right]+\left[B A_{1} C_{1}\right]+\left[C B_{1} A_{1}\right]=[A B C]$ we obtain that one of the remaining triangles $A C_{1} B_{1}, B A_{1} C_{1}, C B_{1} A_{1}$ has an area less than $\frac{1}{4}[A B C]$, so it is less than $\left[A_{1} B_{1} C_{1}\right]$.

Now we return to the problem. We say that triangle $A_{1} B_{1} C_{1}$ is small if $\left[A_{1} B_{1} C_{1}\right]$ is less than each of $\left[B B_{1} A_{1}\right]$ and $\left[C C_{1} B_{1}\right]$; otherwise this triangle is big (the similar notion is introduced for triangles $B_{1} C_{1} D_{1}, C_{1} D_{1} A_{1}, D_{1} A_{1} B_{1}$ ). If both triangles $A_{1} B_{1} C_{1}$ and $C_{1} D_{1} A_{1}$ are big, then $\left[A_{1} B_{1} C_{1}\right]$ is not less than the area of some border triangle, and $\left[C_{1} D_{1} A_{1}\right]$ is not less than the area of another one; hence, $S_{1}=\left[A_{1} B_{1} C_{1}\right]+\left[C_{1} D_{1} A_{1}\right] \geq S$. The same is valid for the pair of $B_{1} C_{1} D_{1}$ and $D_{1} A_{1} B_{1}$. So it is sufficient to prove that in one of these pairs both triangles are big.

Suppose the contrary. Then there is a small triangle in each pair. Without loss of generality, assume that triangles $A_{1} B_{1} C_{1}$ and $D_{1} A_{1} B_{1}$ are small. We can assume also that $\left[A_{1} B_{1} C_{1}\right] \leq$ $\left[D_{1} A_{1} B_{1}\right]$. Note that in this case ray $D_{1} C_{1}$ intersects line $B C$.

Consider two cases.

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Figure 4

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Figure 5

Case 1. Ray $C_{1} D_{1}$ intersects line $A B$ at some point $K$. Let ray $D_{1} C_{1}$ intersect line $B C$ at point $L$ (see Figure 4). Then we have $\left[A_{1} B_{1} C_{1}\right]\left[A_{1} B_{1} C_{1}\right]$, and that ray $K C_{1}$ intersects line $B C$ at some point $L$ (see Figure 5). Since ray $C_{1} D_{1}$ does not intersect line $A B$, the points $A$ and $D_{1}$ are on different sides of $K L$; then $A$ and $D$ are also on different sides, and $C$ is on the same side as $A$ and $B$. Then analogously we have $\left[A_{1} B_{1} C_{1}\right]<\left[C C_{1} B_{1}\right]<\left[L C_{1} B_{1}\right]$ and $\left[A_{1} B_{1} C_{1}\right]<\left[B B_{1} A_{1}\right]$ since triangle $A_{1} B_{1} C_{1}$ is small. This (together with $\left[A_{1} B_{1} C_{1}\right]<\left[K A_{1} C_{1}\right]$ ) contradicts the Lemma again.