China National Olympiad

Let $\triangle ABC$ be an equilateral triangle of side length 1. Let $D, E, F$ be points on $BC, AC, AB$ respectively, such that $\frac{DE}{20} = \frac{EF}{22} = \frac{FD}{38}$. Let $X, Y, Z$ be on lines $BC, CA, AB$ respectively, such that $XY \perp DE$, $YZ \perp EF$, $ZX \perp FD$. We aim to find all possible values of $\frac{1}{[DEF]} + \frac{1}{[XYZ]}$.

Consider the center $K$ of the spiral similarity $\Phi: \triangle DEF \to \triangle XYZ$. By angle chasing, we conclude that $K$ is the Miquel point of $D, E, F$ with respect to $\triangle ABC$. The transformation $\Phi$ rotates the plane by $90^\circ$, making $KF \perp KZ$, etc.

Let $KP \perp BC$, $KQ \perp CA$, $KR \perp AB$, and $\theta := \angle KFB = \angle KDC = \angle KEA$. We have:
$$\frac{1}{[DEF]} + \frac{1}{[XYZ]} = \frac{1}{[DEF]} \left(1 + \left(\frac{KF}{KZ}\right)^2\right) = \frac{1}{[DEF]} \left(1 + \cot^2 \theta\right) = \frac{1}{[DEF] \sin^2 \theta}.$$

Since $\triangle KEF \sim \triangle KQR$ with ratio $1 : \sin \theta$, we get $[KEF] \sin^2 \theta = [KQR]$. Adding up, we find:
$$\frac{1}{[DEF]} + \frac{1}{[XYZ]} = \frac{1}{[PQR]}.$$

Now, $QR = \frac{\sqrt{3}}{2} AK$ but $QR = EF \sin \theta$, so $CK : AK : BK = DE : EF : FD = 10 : 11 : 19$. Let $CK = 10t$.

To find $t$, we use the fact that $AC = \sqrt{291 + 60 \sqrt{6}} t$. Since $AC = 1$, we have $t = \frac{1}{\sqrt{291 + 60 \sqrt{6}}}$.

Next, we calculate $[PQR]$:
$$\begin{align*} [PQR] &= \frac{1}{2} PR \cdot QR \sin(\angle QRK + \angle KRP) \\ &= \frac{1}{2} PR \cdot QR \sin(\angle KBA + \angle KCA) \\ &= \frac{1}{2} \frac{\sqrt{3}}{2} BK \cdot \frac{\sqrt{3}}{2} CK \sin(\angle BKC - 60^\circ) \\ &= \frac{3}{8} \left( \frac{1}{2} BK \cdot CK \sin \angle BKC - \frac{\sqrt{3}}{2} BK \cdot CK \cos \angle BKC \right) \\ &= \frac{3}{8} [BKC] - \frac{3 \sqrt{3}}{32} (BK^2 + CK^2 - BC^2). \end{align*}$$

By rotational symmetry, we have similar expressions for the other segments. Adding them up, we get:
$$\begin{align*} [PQR] &= \frac{1}{8} [ABC] - \frac{\sqrt{3}}{32} (2AK^2 + 2BK^2 + 2CK^2 - AB^2 - BC^2 - CA^2) \\ &= \frac{\sqrt{3}}{32} + \frac{3 \sqrt{3}}{32} - \frac{\sqrt{3}}{32} \cdot 1164 t^2 \\ &= \frac{15 \sqrt{2}}{194 + 40 \sqrt{6}}. \end{align*}$$

Thus, the value is:
$$\boxed{\frac{97 \sqrt{2} + 40 \sqrt{3}}{15}}.$$