Chinese Mathematical Olympiad

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



(2) Rotate $△XYZ$ (together with the equilateral triangle $ABC$) $90^{\circ }$, and properly rescale and translate the picture so that the image of $△XYZ$ coincides with $△DEF$. 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: $△A_1{B}_1{C}_1$ is an equilateral triangle; the points $D$, $E$, and $F$ lie on the lines $B_1{C}_1$, $C_1{A}_1$, and $A_1{B}_1$; and * $A_1{B}_1\perp AB$, $B_1{C}_1\perp BC$, and $C_1{A}_1\perp CA$. From this, we see that the points $A_1$, $B_1$, and $C_1$ lie on the circumcircles of $△AEF$, $△BFD$, and $△CDE$, 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 $S_{△ABC}:S_{△XYZ}=S_{△A_1{B}_1{C}_1}:S_{△DEF}$. So it suffices to compute the ratio of $S_{△ABC}+S_{△A_1{B}_1{C}_1}$ to $S_{△DEF}$. We have $$\begin{gathered} S_{△ABC}=S_{△DEF}+S_{△EAF}+S_{△FBD}+S_{△DCE}, \\ {S}_{△A_1{B}_1{C}_1}=S_{△DEF}+S_{△E{A}_{1}F}+S_{△F{B}_{1}D}+S_{△D{C}_{1}E}, \end{gathered}$$ 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 $△AEF$, $△BFD$, and $△CDE$ 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 $△DEF$. So we have the following equality of areas: $$S_{△ABC}+S_{△A_1{B}_1{C}_1}=2(S_{△DEF}+S_{△E{O}_{1}F}+S_{△F{O}_{2}D}+S_{△D{O}_{3}E}).$$ In fact, $△O_1{O}_2{O}_3$ is the outer Napoleon triangle of $△DEF$; 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, $$S_{△E{O}_{1}F}+S_{△F{O}_{2}D}+S_{△D{O}_{3}E}=\frac{\sqrt{3}}{3}\cdot (11^2+10^2+19^2)=194\sqrt{3}.$$ So $$\begin{array}{rl}& \frac{1}{S_{△DEF}}+\frac{1}{S_{△XYZ}}\\ =& \frac{1}{S_{△ABC}}\left(\frac{S_{△ABC}}{S_{△DEF}}+\frac{S_{△ABC}}{S_{△XYZ}}\right)\\ =& \frac{1}{S_{△ABC}}\frac{S_{△ABC}+S_{△A_1{B}_1{C}_1}}{S_{△DEF}}\\ =& \frac{4}{\sqrt{3}}\frac{2(120\sqrt{2}+194\sqrt{3})}{120\sqrt{2}}\\ =& \frac{97\sqrt{2}+40\sqrt{3}}{15}\end{array}$$