PQ=FM ,2 equilaterals inscribed in square 2019 Armenia NMO 8.6

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PQ=FM ,2 equilaterals inscribed in square 2019 Armenia NMO 8.6

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#1 h Aug 20, 2021, 9:35 AM

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On the sides $AB$ , $DC$ of the square $ABCD$ are taken points $E , N$ , respectively and on the side $BC$ are taken points $M , F$ such that the triangles $AMN$ and $DEF$ are equilateral. Let $P$ be the intersection point of the segments $AN$ and $DE$ . Let Q be the intersection point of the segments $AM$ and $FE$ . Prove that $PQ=FM$ .

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#2 h Aug 28, 2021, 5:46 PM

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Let $A(0,0),B(b,0),C(b,b),D(0,b)$ .

Points $E(\ b(2-\sqrt{3}),0),N(\ b(2-\sqrt{3}),b),F(b,b(\sqrt{3}-1)\ ),M(b,b(2-\sqrt{3})\ )$ .

Points $P(\frac{b(2-\sqrt{3})}{2},\frac{b}{2}),Q(\frac{b(\sqrt{3}-1)}{2},\frac{b(3\sqrt{3}-5)}{2})$ .

Distance $PQ=FM=b(2\sqrt{3}-2)$ .

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