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 .

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- Jun 19th 2010, 09:14 PMsimplependulumA nice equilateral triangle problem
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 . - Jun 19th 2010, 11:32 PMSudharaka
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= $\displaystyle \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha} }\right)=\frac{2\sqrt{x^2+xy+y^2}}{\sqrt{3}}$ - Jun 20th 2010, 12:47 AMundefined
- Jun 20th 2010, 07:06 AMSudharaka
- Jun 20th 2010, 07:19 PMsimplependulum
Excellent , the key is making two equations with unknowns , the side and the included angle but we have to eliminate the side we are looking for first , that's a trick .