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

 

Can you do this without trig, my school hasn't taught it yet. 

 Aug 7, 2018

Best Answer 

 #1
avatar+972 
+2

The first step in order to tackle this problem is to draw another square enclosing square DEFG.

 

First, we prove that \(WXYZ\) is actually a square: 

\( \because\overline{ED}=\overline{DG}=\overline{GF}=\overline{FE}\because\angle{Z}=\angle{Y}=\angle{YXW}=\angle{ZWX}\because\angle{DEZ}+\angle{WEF}=90º,\angle{DEZ}+\angle{ZDE}=90º\Rightarrow\angle{ZDE}=\angle{WEF}.\)

\(\text{Using the same reasoning, we get:} \angle{ZDE}=\angle{WEF}=\angle{DGY}=\angle{GFX}.\)
\(\therefore\text{By AAS congruency:} \triangle{ZDE}\cong\triangle{YGD}\cong\triangle{XFG}\cong\triangle{WEF}.\)

 

From this, we get

\(\overline{ZE}+\overline{EW}=\overline{ZD}+\overline{DY}=\overline{YG}+\overline{GX}=\overline{FX}+\overline{FW}, \)which simplifies to \(\overline{ZW}=\overline{ZY}=\overline{YX}=\overline{XW}.\)

Therefore \(WXYZ \) is a square.

Since \(\triangle ABC\) is equilateral,\( \angle B=60º. \because \triangle BEW\) is a 30-60-90 triangle, \(\frac{\overline{EW}}{\overline{BW}}=\sqrt3.  \text{Same goes with } \triangle GXC, \frac{\overline{GX}}{\overline{XC}}=\sqrt3.\)
\(\text{If }\overline{EW}=x \text{ and } \overline{GX}=y,\text{ we get }\overline{BW}=\frac{x}{\sqrt3} \text{ and } \overline{XC}=\frac{y}{\sqrt3}.\)

If the equilateral triangle's side length is \(a, a=\overline{BW}+\overline{WF}+\overline{FX}+\overline{XC}=\frac{x}{\sqrt3}+y+x+\frac{y}{\sqrt3}.\)

After simplifying, we get \(x+y=\frac{3-\sqrt3}{2}a.\)

\(\because \overline{WX}=x+y \therefore x+y=\overline{DH}.\)

Since in any case, \(\overline{DH}=\frac{3-\sqrt3}{2}a, \) the length remains consistent.

 Aug 7, 2018
 #1
avatar+972 
+2
Best Answer

The first step in order to tackle this problem is to draw another square enclosing square DEFG.

 

First, we prove that \(WXYZ\) is actually a square: 

\( \because\overline{ED}=\overline{DG}=\overline{GF}=\overline{FE}\because\angle{Z}=\angle{Y}=\angle{YXW}=\angle{ZWX}\because\angle{DEZ}+\angle{WEF}=90º,\angle{DEZ}+\angle{ZDE}=90º\Rightarrow\angle{ZDE}=\angle{WEF}.\)

\(\text{Using the same reasoning, we get:} \angle{ZDE}=\angle{WEF}=\angle{DGY}=\angle{GFX}.\)
\(\therefore\text{By AAS congruency:} \triangle{ZDE}\cong\triangle{YGD}\cong\triangle{XFG}\cong\triangle{WEF}.\)

 

From this, we get

\(\overline{ZE}+\overline{EW}=\overline{ZD}+\overline{DY}=\overline{YG}+\overline{GX}=\overline{FX}+\overline{FW}, \)which simplifies to \(\overline{ZW}=\overline{ZY}=\overline{YX}=\overline{XW}.\)

Therefore \(WXYZ \) is a square.

Since \(\triangle ABC\) is equilateral,\( \angle B=60º. \because \triangle BEW\) is a 30-60-90 triangle, \(\frac{\overline{EW}}{\overline{BW}}=\sqrt3.  \text{Same goes with } \triangle GXC, \frac{\overline{GX}}{\overline{XC}}=\sqrt3.\)
\(\text{If }\overline{EW}=x \text{ and } \overline{GX}=y,\text{ we get }\overline{BW}=\frac{x}{\sqrt3} \text{ and } \overline{XC}=\frac{y}{\sqrt3}.\)

If the equilateral triangle's side length is \(a, a=\overline{BW}+\overline{WF}+\overline{FX}+\overline{XC}=\frac{x}{\sqrt3}+y+x+\frac{y}{\sqrt3}.\)

After simplifying, we get \(x+y=\frac{3-\sqrt3}{2}a.\)

\(\because \overline{WX}=x+y \therefore x+y=\overline{DH}.\)

Since in any case, \(\overline{DH}=\frac{3-\sqrt3}{2}a, \) the length remains consistent.

GYanggg Aug 7, 2018
 #2
avatar+98042 
+1

Very nice....!!!!

 

 

cool cool cool

CPhill  Aug 7, 2018
 #3
avatar+972 
+1

Thank you! smiley

GYanggg  Aug 7, 2018
edited by GYanggg  Aug 7, 2018

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