A nice equilateral triangle problem

• Jun 19th 2010, 09:14 PM
simplependulum
A 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 PM
Sudharaka
Quote:

Originally Posted by simplependulum
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,

$\displaystyle If~B\hat{D}F=C\hat{D}E=\alpha~and~BF=x,~CE=y$

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 AM
undefined
Quote:

Originally Posted by Sudharaka
Dear simplependulum,

$\displaystyle AB=AC~;~since~ABC~is~eqilateral$

$\displaystyle A\hat{B}C=A\hat{C}B=60^{o}~;~since~ABC~is~eqilater al$

$\displaystyle BD=CD~;~since~BFD~and~CED~traingles~are~congruent$

Hence the traingles ABD and ADC are congruent. Therefore, $\displaystyle A\hat{D}B=A\hat{D}C=90^{o}$

Hence all the sides of the traingle ABC is equal to x+y.

If I interpreted correctly, the side length is x+y if and only if x = y, and the triangles you said are congruent are only similar in general. I have not solved this problem.
• Jun 20th 2010, 07:06 AM
Sudharaka
Quote:

Originally Posted by undefined
If I interpreted correctly, the side length is x+y if and only if x = y, and the triangles you said are congruent are only similar in general. I have not solved this problem.

Dear undefined,

Thanks. That is a mistake I have'nt seen. I cannot take BD=CD since it is only valid if x=y. Thanks again for pointing that out. I have edited my previous post with a new approach.
• Jun 20th 2010, 07:19 PM
simplependulum
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 .