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# Help!

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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 $\overline{BC}$ remains constant.

Please don't use sine law or cosine law

Nov 6, 2018

#1
+988
+3

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.

Nov 7, 2018

#1
+988
+3

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 Nov 7, 2018
#2
+102773
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Thanks GYangg

It is excellent to see this done without using trigonometry.

Now we have your solutions and Tiggsy's trig solutions which is great :)

Tiggsy's solution: