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 #2
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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?

 

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

 

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

 

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

 

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

 

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

 

12.06.2015