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# help

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Which is larger, the blue area or the orange area? Nov 14, 2019

#1
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Let the radius  of the orange circle   = r

So......its area  =  pi * r^2

We can find the side length,s, of the  equilateral triangle thusly

tan (30°)  =  r / [(1/2)s]

1/ √3 = r / [ (1/2)s ]

(1/ √3) ( 1/2)s  = r

s =  (2√3)  r   = [ √12  ] r

And the area of the equilateral  triangle  is   (  √3/ 4 ) ([√12]r)^2  = 3√3 r^2

So....the blue area inside the equilateral triangle  =

[area of equailateral triangle - area of small circle  ]  /  3

[√3 r^2  - (1/3) pi r^2 ]  =   r^2 [ √3 - pi/3]      (1)

The radius  of the larger circle  can be found as

√[ [√3r ]^2  + r^2 ]   =  √ [ 3r^2 + r^2 ]   =  2r

So....the area  of the  larger circle =  pi (2r)^2  =  4pi r^2

So   the area between the side of the equilateral triangle and the larger circle   is

[Area of larger circle  - area of equilateral triangle  ]  / 3  =

[ 4pi r^2  - 3√3r^2 ] / 3  =  r^2 [ (4/3)pi   - √3]    (2)

So the sum of  (1) and (2)   is  the sum of the blue areas

r^2  [ √3 - pi/3  + (4/3) pi - √3 ]   = pi r^2

So.....the orange and blue areas are equal   !!!!   Nov 14, 2019
#2
+2 Here is a different approach.The arc AC has measure $$\frac{1}{3}\cdot2\pi$$ and the angle AOB has measure half of the arc or $$\frac {\pi}{3}$$. If the radius of the larger circle is $$R$$, then the measure of OB, the radius of the smaller circle, is $$Rcos(\frac{\pi}{3})$$and so the orange region has area $$\pi(Rcos(\frac{\pi}{3}))^2=\frac{1}{4}\pi R^2$$ . The blue region, however, has area equal to $$\frac{1}{3}$$of the difference between area of the larger circle and the area of the smaller circle, i.e. $$\frac{1}{3}(\pi R^2-\frac{1}{4}\pi R^2) =\frac{1}{4}\pi R^2$$. So the two regions have the same area, each being $$\frac{1}{4}$$ of the area of the larger circle.

Nov 14, 2019
#3
+1

Thanks, Gadfly....I like that approach  !!!   Nov 14, 2019
#4
+2

Which is larger, the blue area or the orange area? $$\begin{array}{|rcll|} \hline {\color{orange}\text{orange}} +3\times {\color{blue}\text{blue}} &=& \pi r_{\text{circumcircle}}^2 \quad | \quad : {\color{orange}\text{orange}} \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { {\color{orange}\text{orange}} } \quad | \quad {\color{orange}\text{orange}} = \pi r_{\text{incircle}}^2 \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { \pi r_{\text{incircle}}^2 } \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 \\ 3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \right) \\ \\ && \boxed{\text{here, if triangle is equilateral: }\\ \mathbf{2\times r_{\text{incircle}} = r_{\text{circumcircle}}!!!} } \\\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ 2\times r_{\text{incircle}} } { r_{\text{incircle}} } \right)^2 -1 \right)\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(2^2 -1) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(3) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& 1 \\ \mathbf{ {\color{blue}\text{blue}} } &=& \mathbf{ {\color{orange}\text{orange}} } \\ \hline \end{array}$$ Nov 15, 2019
edited by heureka  Nov 15, 2019