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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

#2
+25480
+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.

Feb 8, 2019

#1
+1

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

Feb 7, 2019
#2
+25480
+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.

heureka Feb 8, 2019