The equation of one side of an equilateral triangle is $x+y=2$ and one vertex is $(2,-1)$. The length of the side is

  • A
    $\frac{\sqrt{2}}{\sqrt{3}}$
  • B
    $\frac{1}{2\sqrt{3}}$
  • C
    $\frac{\sqrt{3}}{\sqrt{2}}$
  • D
    $\frac{2}{\sqrt{3}}$

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Let $ABC$ be a triangle and $M$ be a point on side $AC$ closer to vertex $C$ than $A$. Let $N$ be a point on side $AB$ such that $MN$ is parallel to $BC$ and let $P$ be a point on side $BC$ such that $MP$ is parallel to $AB$. If the area of the quadrilateral $BNMP$ is equal to $\frac{5}{18}$ of the area of $\triangle ABC$,then the ratio $AM/MC$ equals

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