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A square $DEFG$ varies inside equilateral triangle $ABC,$ so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$ and $G$ always lies on side $\overline{AC}.$ The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$ The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

Show that as square DEFG varies, the height of point D above line BC remains constant.

 Feb 7, 2019

Best Answer 

 #2
avatar+21848 
+8

A square $DEFG$ varies inside equilateral triangle $ABC,$
so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$
and $G$ always lies on side $\overline{AC}.$
The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$
The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

 

Show that as square DEFG varies, the height of point D above line BC remains constant.

 

\(\text{Let $\overline{DH}=h $ the height of point $D$ above line $\overline{BC}$ } \\ \text{Let ${\color{red}s} $ the side of the square $DEFG$ } \\ \text{Let $\overline{DF}=\sqrt{2}\color{red}s$ } \\ \text{Let $\angle EFD = 45^{\circ}$ } \\ \text{Let $\angle BFE = \alpha$ } \\ \text{Let $\angle CFG = 90^{\circ}-\alpha$ } \\ \text{The equilateral triangle $ABC$ has the angles $60^{\circ}$, $60^{\circ}$, $60^{\circ}$ and the sides $a$, $a$, $a$. }\)

 

In the rectangular triangle DHF the following applies:

\(\begin{array}{|rcll|} \hline \sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\sqrt{2}\color{red}s} \\\\ \sqrt{2}\sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\color{red}s} \quad | \quad \sqrt{2}\sin(\alpha + 45^{\circ}) = \sin(\alpha) + \cos(\alpha) \\\\ \sin(\alpha) + \cos(\alpha) & = & \dfrac{h}{\color{red}s} \\\\ \mathbf{h}& \mathbf{=} & \mathbf{{\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ \hline \end{array}\)

 

\(\text{Let $\overline{BC}=a $ is the side of the equilateral triangle $ABC$. }\)

\(\begin{array}{|rcll|} \hline a &=& \overline{BM} + \overline{MF} +\overline{FN} + \overline{NC} \\ && \overline{BM} = \dfrac{s\sin(\alpha)}{\tan(60^{\circ})} \\ && \overline{MF} = s\cos(\alpha) \\ && \overline{FN} = s\sin(\alpha) \\ && \overline{NC} = \dfrac{s\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& \dfrac{ {\color{red}s} \sin(\alpha)}{\tan(60^{\circ})} + {\color{red}s}\cos(\alpha) +{\color{red}s}\sin(\alpha) + \dfrac{{\color{red}s}\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& {\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha) + \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \dfrac{1}{ \tan(60^{\circ}) } \\ a &=& \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \mathbf{h={\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ a &=& h \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \tan(60^{\circ}) = \sqrt{3} \\ a &=& h \left( 1+\dfrac{1}{ \sqrt{3} } \right) \\ a &=& h \left( \dfrac{ \sqrt{3}+1}{ \sqrt{3} } \right) \\ \mathbf{h}& \mathbf{=} & \mathbf{a \left( \dfrac{ \sqrt{3} }{ 1+\sqrt{3}} \right)} \\ \hline \end{array}\)

 

So the height of point D above line BC remains constant.

 

laugh

 Feb 8, 2019
 #1
avatar
+1

Pick out the answer you like from these 3 links:

 

https://web2.0calc.com/questions/a-square-defg-varies-inside-equilateral-triangle
https://web2.0calc.com/questions/help_92263
https://web2.0calc.com/questions/help-needed-please-geometry

 Feb 7, 2019
 #2
avatar+21848 
+8
Best Answer

A square $DEFG$ varies inside equilateral triangle $ABC,$
so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$
and $G$ always lies on side $\overline{AC}.$
The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$
The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

 

Show that as square DEFG varies, the height of point D above line BC remains constant.

 

\(\text{Let $\overline{DH}=h $ the height of point $D$ above line $\overline{BC}$ } \\ \text{Let ${\color{red}s} $ the side of the square $DEFG$ } \\ \text{Let $\overline{DF}=\sqrt{2}\color{red}s$ } \\ \text{Let $\angle EFD = 45^{\circ}$ } \\ \text{Let $\angle BFE = \alpha$ } \\ \text{Let $\angle CFG = 90^{\circ}-\alpha$ } \\ \text{The equilateral triangle $ABC$ has the angles $60^{\circ}$, $60^{\circ}$, $60^{\circ}$ and the sides $a$, $a$, $a$. }\)

 

In the rectangular triangle DHF the following applies:

\(\begin{array}{|rcll|} \hline \sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\sqrt{2}\color{red}s} \\\\ \sqrt{2}\sin(\alpha + 45^{\circ}) &=& \dfrac{h}{\color{red}s} \quad | \quad \sqrt{2}\sin(\alpha + 45^{\circ}) = \sin(\alpha) + \cos(\alpha) \\\\ \sin(\alpha) + \cos(\alpha) & = & \dfrac{h}{\color{red}s} \\\\ \mathbf{h}& \mathbf{=} & \mathbf{{\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ \hline \end{array}\)

 

\(\text{Let $\overline{BC}=a $ is the side of the equilateral triangle $ABC$. }\)

\(\begin{array}{|rcll|} \hline a &=& \overline{BM} + \overline{MF} +\overline{FN} + \overline{NC} \\ && \overline{BM} = \dfrac{s\sin(\alpha)}{\tan(60^{\circ})} \\ && \overline{MF} = s\cos(\alpha) \\ && \overline{FN} = s\sin(\alpha) \\ && \overline{NC} = \dfrac{s\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& \dfrac{ {\color{red}s} \sin(\alpha)}{\tan(60^{\circ})} + {\color{red}s}\cos(\alpha) +{\color{red}s}\sin(\alpha) + \dfrac{{\color{red}s}\cos(\alpha)}{\tan(60^{\circ})} \\ a &=& {\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha) + \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \dfrac{1}{ \tan(60^{\circ}) } \\ a &=& \Big({\color{red}s}\sin(\alpha) +{\color{red}s}\cos(\alpha)\Big) \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \mathbf{h={\color{red}s}\sin(\alpha) + {\color{red}s}\cos(\alpha)} \\ a &=& h \left( 1+\dfrac{1}{ \tan(60^{\circ}) } \right) \quad | \quad \tan(60^{\circ}) = \sqrt{3} \\ a &=& h \left( 1+\dfrac{1}{ \sqrt{3} } \right) \\ a &=& h \left( \dfrac{ \sqrt{3}+1}{ \sqrt{3} } \right) \\ \mathbf{h}& \mathbf{=} & \mathbf{a \left( \dfrac{ \sqrt{3} }{ 1+\sqrt{3}} \right)} \\ \hline \end{array}\)

 

So the height of point D above line BC remains constant.

 

laugh

heureka Feb 8, 2019

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