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A pair of the identical circles are overlapping each other. Their diameter (D) is 20cm, and the distance between the centers is 10cm. In the grey area, it is possible to inscribe only two  golden rectangles - different in size.

Can you find the area of a larger one?

 

Image result for rectangle inscribed in a overlap of 2 circles

civonamzuk  Jun 11, 2015

Best Answer 

 #2
avatar+18715 
+15

A pair of the identical circles are overlapping each other. Their diameter (D) is 20cm, and the distance between the centers is 10cm. In the grey area, it is possible to inscribe only two  golden rectangles - different in size.

Can you find the area of a larger one?

 

$$\small{\text{$\boxed{\varphi=\dfrac{1+\sqrt{5}}{2}} \qquad \Delta_c=10~cm\qquad D = 20~cm$}}\\\\
\mathrm{circle:~~}
(\vec{x})^2= (\frac{D}{2})^2\\\\
\mathrm{line:~~}
\binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}=\vec{x}\\\\
\mathrm{golden~ rectangle:~~}
\small{\text{$
\binom{ \frac{width}{2} }{ \frac{height}{2} } = \lambda\binom{1}{\varphi} ~~ \mathrm{~or~}~~\mathrm{~width~} = 2\lambda ~~ \mathrm{~and~} ~~ \mathrm{~height~} = 2\varphi\lambda
$}}\\
\mathrm{ratio~}=\frac{height}{width}=\frac{2\varphi\lambda
}{2\lambda } = \varphi\\\\
\boxed{\mathrm{~area~}~~=
\mathrm{~width~} \cdot \mathrm{~height~} = 4 \varphi\lambda^2}$$

 

$$\\\mathrm{solve~~} \lambda :\\\\
\small{\text{$
\begin{array}{rcl}
\left[
\binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}
\right]^2
=(\vec{x})^2 &=& (\frac{D}{2})^2\\\\
\left[
\binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}
\right]^2
&=& (\frac{D}{2})^2\\\\
\binom{\frac{\Delta_c}{2}}{0}\cdot \binom{\frac{\Delta_c}{2}}{0}
+
2\lambda\binom{\frac{\Delta_c}{2}}{0}\cdot \binom{1}{\varphi}
+ \lambda^2 \binom{1}{\varphi}\cdot \binom{1}{\varphi}
&=& (\frac{D}{2})^2\\\\
\frac{\Delta_c^2}{4}+\lambda\Delta_c+\lambda^2(1+\varphi^2)
&=&\frac{D^2}{4}\\\\
(1+\varphi^2)\lambda^2 + \Delta_c\lambda - \frac{D^2-\Delta_c^2}{4}&=& 0\\\\
\mathbf{\lambda }&\mathbf{=}&
\mathbf{
\dfrac{
\sqrt{\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c } { 2(1+\varphi^2) }
}
\end{array}
$}}$$
 

 

$$\small{\text{$
\begin{array}{rcl}
\mathrm{~area}
= 4 \varphi\lambda^2
&=& 4 \varphi\
\left[
\dfrac{
\sqrt{\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c } { 2(1+\varphi^2) }
\right]^2\\\\
\mathrm{~area}
&=& 4 \varphi\
\dfrac{
\left[ \sqrt{
\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2} { 4(1+\varphi^2)^2 }\\\\
\mathrm{~area}
&=& \varphi\
\dfrac{
\left[ \sqrt{
\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2} { (1+\varphi^2)^2 }\\\\
\mathrm{~area}
&=& \dfrac{ \varphi } { (1+\varphi^2)^2 }
\left[ \sqrt{
\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad 1+\varphi^2 = 2+\varphi = \sqrt{5}\varphi\\\\
\mathrm{~area}
&=& \dfrac{ \varphi } { 5\varphi^2 }
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \\\\
\mathrm{~area}
&=& \dfrac{ 1} { 5\varphi }
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad \dfrac{1}{\varphi} = \varphi -1 \\\\
\mathrm{~area}
&=& \dfrac{ \varphi -1 } { 5 }
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2
\end{array}
$}}$$

 

$$\boxed{
\mathrm{~area}
= \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2
}$$

 

$$\small{\text{$
\begin{array}{rcl}
\mathrm{~area}
&=& \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad \Delta_c = 10 \qquad D = 20 \\\\
\mathrm{~area}
&=& \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
100+ (2+\varphi )(400 - 100) } -10 \right]^2 \\\\
\mathrm{~area}
&=& \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
100+ (2+\varphi )300 } -10 \right]^2 \\\\
\mathrm{~area}
&=& 0.12360679775\cdot
\left[ 34.4297864737 -10 \right]^2\\\\
\mathrm{~area}
&=& 0.12360679775\cdot 596.814467151\\\\
\mathrm{~area}
&=& 73.7703251354 \mathrm{~cm^2}
\end{array}
$}}$$

 

heureka  Jun 12, 2015
Sort: 

2+0 Answers

 #1
avatar+78754 
+15

Here's a possible orientation......https://www.desmos.com/calculator/rjfefnmdaw

 

The largest golden rectangle will have its vertces at about (±3.37611,  ±5.46267)    ....and its area will be.....

 

2(3.37611) * 2(5.46267)  ≈ 73.77 cm^2

 

The other golden rectangle will have its vertices  at about (±4.58932, ±2.83636)  ....and it's area will be...

 

2(4.58932)*2(2.83636)  ≈ 52.068 cm^2

 

 

CPhill  Jun 11, 2015
 #2
avatar+18715 
+15
Best Answer

A pair of the identical circles are overlapping each other. Their diameter (D) is 20cm, and the distance between the centers is 10cm. In the grey area, it is possible to inscribe only two  golden rectangles - different in size.

Can you find the area of a larger one?

 

$$\small{\text{$\boxed{\varphi=\dfrac{1+\sqrt{5}}{2}} \qquad \Delta_c=10~cm\qquad D = 20~cm$}}\\\\
\mathrm{circle:~~}
(\vec{x})^2= (\frac{D}{2})^2\\\\
\mathrm{line:~~}
\binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}=\vec{x}\\\\
\mathrm{golden~ rectangle:~~}
\small{\text{$
\binom{ \frac{width}{2} }{ \frac{height}{2} } = \lambda\binom{1}{\varphi} ~~ \mathrm{~or~}~~\mathrm{~width~} = 2\lambda ~~ \mathrm{~and~} ~~ \mathrm{~height~} = 2\varphi\lambda
$}}\\
\mathrm{ratio~}=\frac{height}{width}=\frac{2\varphi\lambda
}{2\lambda } = \varphi\\\\
\boxed{\mathrm{~area~}~~=
\mathrm{~width~} \cdot \mathrm{~height~} = 4 \varphi\lambda^2}$$

 

$$\\\mathrm{solve~~} \lambda :\\\\
\small{\text{$
\begin{array}{rcl}
\left[
\binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}
\right]^2
=(\vec{x})^2 &=& (\frac{D}{2})^2\\\\
\left[
\binom{\frac{\Delta_c}{2}}{0}+\lambda\binom{1}{\varphi}
\right]^2
&=& (\frac{D}{2})^2\\\\
\binom{\frac{\Delta_c}{2}}{0}\cdot \binom{\frac{\Delta_c}{2}}{0}
+
2\lambda\binom{\frac{\Delta_c}{2}}{0}\cdot \binom{1}{\varphi}
+ \lambda^2 \binom{1}{\varphi}\cdot \binom{1}{\varphi}
&=& (\frac{D}{2})^2\\\\
\frac{\Delta_c^2}{4}+\lambda\Delta_c+\lambda^2(1+\varphi^2)
&=&\frac{D^2}{4}\\\\
(1+\varphi^2)\lambda^2 + \Delta_c\lambda - \frac{D^2-\Delta_c^2}{4}&=& 0\\\\
\mathbf{\lambda }&\mathbf{=}&
\mathbf{
\dfrac{
\sqrt{\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c } { 2(1+\varphi^2) }
}
\end{array}
$}}$$
 

 

$$\small{\text{$
\begin{array}{rcl}
\mathrm{~area}
= 4 \varphi\lambda^2
&=& 4 \varphi\
\left[
\dfrac{
\sqrt{\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c } { 2(1+\varphi^2) }
\right]^2\\\\
\mathrm{~area}
&=& 4 \varphi\
\dfrac{
\left[ \sqrt{
\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2} { 4(1+\varphi^2)^2 }\\\\
\mathrm{~area}
&=& \varphi\
\dfrac{
\left[ \sqrt{
\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2} { (1+\varphi^2)^2 }\\\\
\mathrm{~area}
&=& \dfrac{ \varphi } { (1+\varphi^2)^2 }
\left[ \sqrt{
\Delta_c^2+ (1+\varphi^2)(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad 1+\varphi^2 = 2+\varphi = \sqrt{5}\varphi\\\\
\mathrm{~area}
&=& \dfrac{ \varphi } { 5\varphi^2 }
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \\\\
\mathrm{~area}
&=& \dfrac{ 1} { 5\varphi }
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad \dfrac{1}{\varphi} = \varphi -1 \\\\
\mathrm{~area}
&=& \dfrac{ \varphi -1 } { 5 }
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2
\end{array}
$}}$$

 

$$\boxed{
\mathrm{~area}
= \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2
}$$

 

$$\small{\text{$
\begin{array}{rcl}
\mathrm{~area}
&=& \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
\Delta_c^2+ (2+\varphi )(D^2 - \Delta_c^2) } -\Delta_c \right]^2 \qquad |\qquad \Delta_c = 10 \qquad D = 20 \\\\
\mathrm{~area}
&=& \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
100+ (2+\varphi )(400 - 100) } -10 \right]^2 \\\\
\mathrm{~area}
&=& \left( \dfrac{ \varphi -1 } { 5 } \right)
\left[ \sqrt{
100+ (2+\varphi )300 } -10 \right]^2 \\\\
\mathrm{~area}
&=& 0.12360679775\cdot
\left[ 34.4297864737 -10 \right]^2\\\\
\mathrm{~area}
&=& 0.12360679775\cdot 596.814467151\\\\
\mathrm{~area}
&=& 73.7703251354 \mathrm{~cm^2}
\end{array}
$}}$$

 

heureka  Jun 12, 2015

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