(1) First, consider the rotation of $90^\circ$ (clockwise or counterclockwise), centered at an arbitrary point on the plane. Then the images $X'$, $Y'$, and $Z'$ of the points $X$, $Y$, and $Z$, respectively, satisfy
$$X'Y' \parallel DE, \quad Y'Z' \parallel EF, \quad Z'X' \parallel FD.$$
In this case, $\triangle X'Y'Z'$ and $\triangle DEF$ are directly similar. So, every good triangle pair is directly similar.

(2) **Rotate $\triangle XYZ$ (together with the equilateral triangle $ABC$) $90^\circ$, and properly rescale and translate the picture so that the image of $\triangle XYZ$ coincides with $\triangle DEF$. Under this transformation, the points $A$, $B$, and $C$ are mapped to points $A_1$, $B_1$, and $C_1$. So there are three points $A_1$, $B_1$, and $C_1$ on the plane satisfying:**
* $\triangle A_1B_1C_1$ is an equilateral triangle;
* the points $D$, $E$, and $F$ lie on the lines $B_1C_1$, $C_1A_1$, and $A_1B_1$; and
* $A_1B_1 \perp AB$, $B_1C_1 \perp BC$, and $C_1A_1 \perp CA$.
From this, we see that the points $A_1$, $B_1$, and $C_1$ lie on the circumcircles of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$, respectively. Moreover, $A_1$, $B_1$, and $C_1$ are the antipodes of $A$, $B$, and $C$ in the corresponding circles; see the picture below.

(3) Note that $S_{\triangle ABC} : S_{\triangle XYZ} = S_{\triangle A_1B_1C_1} : S_{\triangle DEF}$. So it suffices to compute the ratio of $S_{\triangle ABC} + S_{\triangle A_1B_1C_1}$ to $S_{\triangle DEF}$. We have
$$S_{\triangle ABC} = S_{\triangle DEF} + S_{\triangle EAF} + S_{\triangle FBD} + S_{\triangle DCE}, \\ S_{\triangle A_1B_1C_1} = S_{\triangle DEF} + S_{\triangle EA_1F} + S_{\triangle FB_1D} + S_{\triangle DC_1E},$$
Here the right hand sides are expressed in terms of oriented areas. Since the two equilateral triangles on the left hand side are directly similar, the signs on the areas are the same. Using the properties of antipodes, if we denote the circumcenters of $\triangle AEF$, $\triangle BFD$, and $\triangle CDE$ by $O_1$, $O_2$, and $O_3$, respectively, then the condition "$D$, $E$, and $F$ lie on the interior of three sides" ensures that the three circumcenters $O_1$, $O_2$, and $O_3$ lie outside of $\triangle DEF$. So we have the following equality of areas:
$$S_{\triangle ABC} + S_{\triangle A_1B_1C_1} = 2(S_{\triangle DEF} + S_{\triangle EO_1F} + S_{\triangle FO_2D} + S_{\triangle DO_3E}).$$
In fact, $\triangle O_1O_2O_3$ is the outer Napoleon triangle of $\triangle DEF$; its area is exactly the half of sum of the areas in the parentheses.

(4) So we need to compute, for a triangle with side length ratio $20 : 22 : 38$, the ratio of the area of the hexagon formed by the vertices of the triangle and the vertices of the outer Napoleon triangle, to the area of the original triangle.
By Heron formula, the area of a triangle with side lengths $20$, $22$, $38$ is
$$\sqrt{40 \cdot (40 - 20) \cdot (40 - 22) \cdot (40 - 38)} = 120\sqrt{2}.$$
On the other hand,
$$S_{\triangle EO_1F} + S_{\triangle FO_2D} + S_{\triangle DO_3E} = \frac{\sqrt{3}}{3} \cdot (11^2 + 10^2 + 19^2) = 194\sqrt{3}.$$
So
$$\begin{align*} & \frac{1}{S_{\triangle DEF}} + \frac{1}{S_{\triangle XYZ}} \\ = & \frac{1}{S_{\triangle ABC}} \left( \frac{S_{\triangle ABC}}{S_{\triangle DEF}} + \frac{S_{\triangle ABC}}{S_{\triangle XYZ}} \right) \\ = & \frac{1}{S_{\triangle ABC}} \frac{S_{\triangle ABC} + S_{\triangle A_1B_1C_1}}{S_{\triangle DEF}} \\ = & \frac{4}{\sqrt{3}} \frac{2(120\sqrt{2} + 194\sqrt{3})}{120\sqrt{2}} \\ = & \frac{97\sqrt{2} + 40\sqrt{3}}{15} \end{align*}$$