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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

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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?

civonamzuk  Jun 11, 2015

#2
+19207
+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
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#1
+85958
+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
+19207
+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

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