<s>[INST] Geometry question on square $ABCD$ is a square and $AB$ = 1. Equilateral triangles $AYB$ and $CXD$ are drawn such that $X$ and $Y$ are inside the square.How can I find the length of $XY$ ?
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You can use also this approach if you like,also if, as already said, you can really solve it by simmetry arguments.
Area of $\Delta DPR = \cfrac {DP ^2 \cdot \sin P \cdot \sin D}{2 \sin R}=\cfrac {1}{4} \cdot \cfrac{ \sin 90 \cdot \sin 30 }{2 \sin60} =\cfrac { \sqrt {3}}{24}$. 
( ----> Area of any triangle $\Delta ABC$ given side $a$ and two angles is : $\cfrac{a^2 \sin B \sin C}{2 \sin A}$)
So we have that the area of $DROY' =\cfrac {1}{4} - \cfrac { \sqrt {3}}{24}$.
This is the equivalent to say that $DROY'=\cfrac {(DY'+RO) \cdot OY'}{2}= \cfrac {1}{4} - \cfrac { \sqrt {3} }{24}$, now since $DY=OY'=\cfrac {1}{2}$ , we find that $RO= \cfrac {3- \sqrt{3}}{6} $.
Now note that $OX=\sqrt{3} \cdot OR$  (30-60-90 triangle properties) ,so we find that $OX=\cfrac {3- \sqrt{3}}{6} \cdot \sqrt{3} =\cfrac {\sqrt {3}}{2} -\cfrac {1}{2} $.
Now since $XY=2OX$ we find $XY=\sqrt{3} -1$
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