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Both circles have a radius r=1. Can you make these 3 regions, A B and C, to occupy equal areas?

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I'm sorry, but this is the wrong question. The 'right'() question is here:>

civonamzuk  May 21, 2015

#1
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Both circles have a radius  r=1.  Can you make these 3 regions,  A  B  and  C, to occupy the equal areas ?

$$\small{\text{ \rm{central~angle}= \varphi_\rm{rad} }}$$

$$\small{\text{ \begin{array}{rcl} \rm{area}_B &=& 2\cdot\left[ \pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\ \rm{area}_A &=& \rm{area}_B=\rm{area}_C\\\\ \rm{area}_{circle_1} = \rm{area}_{circle_2} &=& \pi r^2\\ \rm{area}_{circle_1} - \rm{area}_B &=& \rm{area}_B \\ \rm{area}_{circle_1} &=& 2\cdot \rm{area}_B \\ \pi r^2 &=& 2\cdot \rm{area}_B \\ \pi r^2 &=& 2\cdot 2\cdot\left[ \pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\ \pi r^2 &=& 2\cdot r^2 \cdot \varphi_{\rm{rad}} - 2 \cdot r^2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : r^2\\ \pi &=& 2 \cdot \varphi_{\rm{rad}} - 2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : 2\\ \frac{\pi}{2} &=& \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } \\\\ \end{array} }}$$

$$\boxed{ \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } = \frac{\pi}{2} }$$

The Newton-Raphson method in one variable is implemented as follows:

Given a function ƒ defined over the reals x, and its derivative ƒ',

we begin with a first guess x0 for a root of the function f.

Provided the function satisfies all the assumptions made in the derivation of the formula,

a better approximation x1 is

$$\small{\text{  x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)} \,. }}$$

$$\varphi_{\rm{rad}_1} = \varphi_{\rm{rad}_0} - \dfrac{ \varphi_{\rm{rad}_0} -\sin{ ( \varphi_{\rm{rad}_0} ) } - \frac{\pi}{2} } {1-\cos{ (\varphi_{\rm{rad}_0}) } }$$

Iteration:

$$\small{\text{ \begin{array}{rcl} \varphi_{\rm{rad}_0} &=& 2\\ \varphi_{\rm{rad}_1} &=& 2.33901410590 \\ \varphi_{\rm{rad}_2} &=& 2.31006319657 \\ \varphi_{\rm{rad}_3} &=& 2.30988146730 \\ \varphi_{\rm{rad}_4} &=& 2.30988146001 \\ \cdots \\ \varphi_{\rm{rad}} = 2.30988146001 \\ \end{array} }}$$

$$\small{\text{ \begin{array}{rcl} \rm{area}_B &=& 2\cdot\left[ \pi r^2 \cdot \frac{\varphi_\rm{rad} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_\rm{rad} )} \right] \qquad | \qquad r=1 \\ \rm{area}_B &=& 2\cdot\left[ \pi \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot \sin{(\varphi_{\rm{rad}} )} \right] \\ \rm{area}_B &=& \varphi_{\rm{rad}} - \sin{(\varphi_{\rm{rad}} )} \\\\ \rm{area}_B = 2.30988146001 -\sin{ ( 2.30988146001 )} &=& \frac{\pi}{2}\\ \rm{area}_A = \rm{area}_C &=& \pi r^2 - \rm{area}_B \qquad | \qquad r=1 \\ \rm{area}_A = \rm{area}_C &=& \pi - \frac{\pi}{2} = \frac{\pi}{2} \end{array} }}$$

heureka  May 21, 2015
#1
+19653
+13

Both circles have a radius  r=1.  Can you make these 3 regions,  A  B  and  C, to occupy the equal areas ?

$$\small{\text{ \rm{central~angle}= \varphi_\rm{rad} }}$$

$$\small{\text{ \begin{array}{rcl} \rm{area}_B &=& 2\cdot\left[ \pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\ \rm{area}_A &=& \rm{area}_B=\rm{area}_C\\\\ \rm{area}_{circle_1} = \rm{area}_{circle_2} &=& \pi r^2\\ \rm{area}_{circle_1} - \rm{area}_B &=& \rm{area}_B \\ \rm{area}_{circle_1} &=& 2\cdot \rm{area}_B \\ \pi r^2 &=& 2\cdot \rm{area}_B \\ \pi r^2 &=& 2\cdot 2\cdot\left[ \pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\ \pi r^2 &=& 2\cdot r^2 \cdot \varphi_{\rm{rad}} - 2 \cdot r^2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : r^2\\ \pi &=& 2 \cdot \varphi_{\rm{rad}} - 2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : 2\\ \frac{\pi}{2} &=& \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } \\\\ \end{array} }}$$

$$\boxed{ \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } = \frac{\pi}{2} }$$

The Newton-Raphson method in one variable is implemented as follows:

Given a function ƒ defined over the reals x, and its derivative ƒ',

we begin with a first guess x0 for a root of the function f.

Provided the function satisfies all the assumptions made in the derivation of the formula,

a better approximation x1 is

$$\small{\text{  x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)} \,. }}$$

$$\varphi_{\rm{rad}_1} = \varphi_{\rm{rad}_0} - \dfrac{ \varphi_{\rm{rad}_0} -\sin{ ( \varphi_{\rm{rad}_0} ) } - \frac{\pi}{2} } {1-\cos{ (\varphi_{\rm{rad}_0}) } }$$

Iteration:

$$\small{\text{ \begin{array}{rcl} \varphi_{\rm{rad}_0} &=& 2\\ \varphi_{\rm{rad}_1} &=& 2.33901410590 \\ \varphi_{\rm{rad}_2} &=& 2.31006319657 \\ \varphi_{\rm{rad}_3} &=& 2.30988146730 \\ \varphi_{\rm{rad}_4} &=& 2.30988146001 \\ \cdots \\ \varphi_{\rm{rad}} = 2.30988146001 \\ \end{array} }}$$

$$\small{\text{ \begin{array}{rcl} \rm{area}_B &=& 2\cdot\left[ \pi r^2 \cdot \frac{\varphi_\rm{rad} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_\rm{rad} )} \right] \qquad | \qquad r=1 \\ \rm{area}_B &=& 2\cdot\left[ \pi \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot \sin{(\varphi_{\rm{rad}} )} \right] \\ \rm{area}_B &=& \varphi_{\rm{rad}} - \sin{(\varphi_{\rm{rad}} )} \\\\ \rm{area}_B = 2.30988146001 -\sin{ ( 2.30988146001 )} &=& \frac{\pi}{2}\\ \rm{area}_A = \rm{area}_C &=& \pi r^2 - \rm{area}_B \qquad | \qquad r=1 \\ \rm{area}_A = \rm{area}_C &=& \pi - \frac{\pi}{2} = \frac{\pi}{2} \end{array} }}$$

heureka  May 21, 2015