Consider the center $K$ of the spiral similarity $\Phi:\triangle DEF\to\triangle XYZ$. Simple angle chasing concludes $K$ is in fact the Miquel point of $D,E,F$ wrt $\triangle ABC$. $\Phi$ rotates the whole plane by $90^{\circ}$ so $KF\perp KZ$ etc. Let $KP\perp BC,KQ\perp CA,KR\perp AB$ and $\theta:=\angle KFB=\angle KDC=\angle KEA$. $\frac{1}{[DEF]} + \frac{1}{[XYZ]}=\frac{1}{[DEF]}\left(1+\left(\dfrac{KF}{KZ}\right)^2\right)=\frac{1}{[DEF]}\left(1+\cot^2\theta\right)=\frac{1}{[DEF]\sin^2\theta}$ The cautious reader might notice $\triangle KEF\sim\triangle KQR$ with ratio $1:\sin\theta$ so $[KEF]\sin^2\theta=[KQR]$, by adding up we get $\frac{1}{[DEF]} + \frac{1}{[XYZ]}=\frac{1}{[PQR]}$ Now $QR=\frac{\sqrt{3}}{2}AK$ but $QR=EF\sin\theta $, so $CK:AK:BK=DE:EF:FD=10:11:19$. Let $CK=10t$

It should be easy

to get $AC=\sqrt{291+60\sqrt{6}}t$. But then $AC=1$, so $t=\frac{1}{\sqrt{291+60\sqrt{6}}}$. The path is clear. \begin{align*}[PQR]=&\frac12 PR\cdot QR\sin(\angle QRK+\angle KRP)\\=&\frac12 PR\cdot QR\sin(\angle KBA+\angle KCA)\\=&\frac12\frac{\sqrt{3}}{2}BK\cdot\frac{\sqrt{3}}{2}CK\sin(\angle BKC-60^{\circ})\\=&\frac38 (\frac12 BK\cdot CK\sin\angle BKC-\frac{\sqrt{3}}2 BK\cdot CK \cos\angle BKC)\\=&\frac38[BKC]-\frac{3\sqrt3}{32}(BK^2+CK^2-BC^2) \end{align*}There are two other expressions like that (rotational symmetry) and by adding them up: \begin{align*}[PQR]=&\frac18[ABC]-\frac{\sqrt3}{32}(2AK^2+2BK^2+2CK^2-AB^2-BC^2-CA^2)\\ =&\frac{\sqrt3}{32}+\frac{3\sqrt3}{32}-\frac{\sqrt3}{32}\cdot 1164t^2\\=&\frac{15\sqrt2}{194+40\sqrt{6}} \end{align*}So $$\boxed{\frac{1}{[DEF]} + \frac{1}{[XYZ]}=\frac{1}{[PQR]}=\frac{97\sqrt2+40\sqrt3}{15}}$$

Another proof to the uniqueness of the value. The calculation part is trivial, and I'll omit it. Lemma Given a triangle $DEF$ on the plane. Construct equilateral triangle $ABC$ s.t. $D \in BC, E\in CA, F\in AB$, the line $l_1$ passing through $D$ perpendicular to $BC$, the line $l_2$ passing through $E$ perpendicular to $CA$ and the line $l_3$ passing through $F$ perpendicular to $BC$ intersects each other at $U, V, W$ $U \equiv l_2 \cap l_3$, etc. Then $[UVW] + [ABC]$ is a constant independent of $A, B, C$. proof. Denote the circumcenter of $\triangle EFA, \triangle FDB, \triangle DEC$ by $O_1, O_2, O_3$, respectively. (1st Napoleon triangle of $\triangle DEF$) Note that $\triangle DCA \sim \triangle DWU \sim \triangle AO_3O_2$. Hence (Pythagorean theorem in ${\rm Rt}\triangle CDW$, for the last equality) $$[UVW]+[ABC] = \frac{\sqrt{3}}{4} (UW^2+AC^2) = \frac{\sqrt{3}}{4} \cdot \frac{O_1O_3^2}{DO_3^2}(DC^2+DW^2) = \sqrt{3}O_1O_3^2\;(\text{constant})\Box$$ Back to the main problem. Notice that $XYZABC \sim DEFUVW$ (notations are the same as those in Lemma). Hence $$\frac{1}{[DEF]} + \frac{1}{[XYZ]} = \frac{4}{\sqrt{3}}(\frac{[ABC]}{[DEF]} + \frac{[ABC]}{[XYZ]}) = \frac{4}{\sqrt{3}}(\frac{[ABC]}{[DEF]} + \frac{[UVW]}{[DEF]}) = \frac{4O_1O_3^2}{[DEF]}$$Suppose that $DE=10, EF=11, FD=19$ and from somehow trivial calculation we can get the answer. Below is the figure of the Lemma.

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Let $O$ be the Miquel point of $D,E,F$ wrt $\triangle ABC$. Assume that $O$ is inside $\triangle ABC$, then note that $\angle DEF=\angle DEO+\angle OEF=\angle OCB+\angle OAB$ whose value is only related to the location $O,A,B,C$.(The claim is true when $O$ is outside $\triangle ABC$) This means that for some fixed $\triangle ABC$ and $O$, we can determine the three angles of any $\triangle DEF$ s.t. $O$ is the Miquel point of $D,E,F$ wrt $\triangle ABC$. Also we certainly have $\triangle DEF\sim\triangle XYZ$, thus $\triangle DEF$ share a Miquel point $O$ with $\triangle XYZ$. Let $OP\perp BC,OQ\perp CA,OR\perp AB$. Note that $O$ is also the Miquel point of $P,Q,R$ wrt $\triangle ABC$, so $\triangle PQR\sim\triangle DEF$. And $OP$ is the height of $Rt\triangle DOX$, so $\frac{1}{OP^2}=\frac{DO^2+OX^2}{DO^2\cdot OX^2}=\frac{1}{DO^2}+\frac{1}{OX^2}$, from which we can get that the desired $\frac{1}{[DEF]} + \frac{1}{[XYZ]}$ is actually equal to $\frac{1}{[PQR]}$. But $\triangle ABC$ is an equilateral triangle, so $O$ is Fermat point of $\triangle PQR$. We need to calculate $[PQR]$ with $\frac{PQ}{20} = \frac{QR}{22} = \frac{RP}{38}$. Let $PQ=10x,QR=11x,RP=19x$, then $\cos \angle PQR=-\frac{7}{11}$ which means that $O$ is outside $\triangle PQR$ so $OP+OR-OQ=\frac{\sqrt3}{2}$. Then $\frac 34=(OP-OQ+OR)^2=(OP+Q'R'-OQ')^2=PR^2=PQ^2+QR^2-2PQ\cdot QR\cos (300^{\circ}-\angle PQR)$ Plug in $\cos (300^{\circ}-\angle PQR)=-\frac{7}{11}\times \frac 12-\sqrt{1-\frac{7^2}{11^2}}\times\frac{\sqrt3}{2}=-\frac{7+6\sqrt6}{22}$, and we can get $ \frac 34=221x^2-220x^2\cdot (-\frac{7+6\sqrt6}{22})=(291+60\sqrt 6)x^2$. On the other hand, $\frac{1}{[PQR]}=\frac{1}{\frac 12\cdot 10x\cdot 11x\cdot\sqrt{1-\frac{7^2}{11^2}}}=\frac{1}{30\sqrt2 x^2}=\frac{\frac 43\cdot(291+60\sqrt6)}{30\sqrt2}=\boxed{\frac{97\sqrt2+40\sqrt3}{15}}$, which is the desired answer.

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A very beautiful solution without Miquel Theorem solved by Zhao Jiyun. Let perpendicular to $AC$ through $E$, perpendicular to $BC$ through $D$ and perpendicular to $AB$ through $F$ intersect at $R,S,T$. Then $\triangle XYZ\sim \triangle DEF$. Let the midpoint of $\overset{\LARGE{\frown}}{FD}$ be $M,N$, the center of $\odot{FDB}$ be $O$. Then $S,O,B$ is collinear. \begin{align*}[DFB]-[DSF]=&\frac12DF\cdot ON\\=&\frac{OF^2}{2\sqrt3}. \end{align*}So we have \begin{align*} \frac{1}{[DEF]} + \frac{1}{[XYZ]}=&\frac{4}{\sqrt3}(\frac{[ABC]}{[DEF]} + \frac{[ABC]}{[XYZ]})\\=& \frac{4}{\sqrt3} (\frac{[ABC]}{[DEF]} + \frac{[RST]}{[DEF]})\\=& \frac{4}{\sqrt3} (1+1+ \sum \frac{[DBF]-[DSF]}{[DEF]}) \\=& \frac{4}{\sqrt3} (2+\frac{DF^2+DE^2+FE^2}{2\sqrt3 \cdot [DEF]})\\=& \frac{4}{\sqrt3} (2+\frac{582}{2\sqrt3 \cdot 30\sqrt2})\\=& \frac{97\sqrt2+40\sqrt3}{15}. \end{align*}

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What A STRANGE geometry problem.

这道问题就是纯粹的狗屎

Invincible-Tyrannosaurus wrote: 这道问题就是纯粹的狗屎 I think the geometry part of this problem is nice. But yyj is exactly like what you said

Newmaths wrote: Invincible-Tyrannosaurus wrote: 这道问题就是纯粹的狗屎 I think the geometry part of this problem is nice. But yyj is exactly like what you said 姚一隽稳定发挥（

I agree with you

Actually geometry part of this problem is really elegant.