There are three parallel lines with given distances $\displaystyle x ~,~ y ~, ~ x+y $ .
Find the length of the side of the equilateral triangle whose three vertices lie on the parallel lines respectively .
Dear simplependulum,
Please refer the attached drawing.
$\displaystyle If~B\hat{D}F=C\hat{D}E=\alpha~and~BF=x,~CE=y$
The length of a side of the traingle ABC= $\displaystyle \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha} }\right)$
Consider the traingle AEC where, $\displaystyle C\hat{A}E=\alpha-60$
Hence, $\displaystyle \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha} }\right)\sin{(\alpha-60)}=y$
By simplification we could obtain, $\displaystyle \tan{\alpha}=\frac{\sqrt{3}(x+y)}{x-y}\Rightarrow{\sin{\alpha}}=\frac{\sqrt{3}(x+y)}{2 \sqrt{x^2+xy+y^2}}$
Therefore the length of a side of the traingle ABC= $\displaystyle \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha} }\right)=\frac{2\sqrt{x^2+xy+y^2}}{\sqrt{3}}$