/sci/ - Science & Math

[math]\sqrt{\frac{\pi-1}{1-\pi}}[/math]

>>9406822
[math]\sqrt{\frac{\pi-2}{4-\pi}}[/math]

>>9406826
It's actually sqrt{(pi-2)/(4-pi)}. Notice how (pi-1)/(1-pi)=-1 so the square root would be i.

>>9406822
red = r1^2 * (4 - Pi)

blue = r2^2 * Pi - r2^2 * 2

(
(2*r2)^2 = s^2 + s^2
4 * r2^2 = 2 * s^2
2 * r^2 = s^2

AsmallSquare = s^2
)

red = r1^2 * (4 - Pi)
blue = r2^2 * Pi - r2^2 * 2

Just for fun:
r1^2 * 4 + r2^2 * 2 = r2^2 * Pi + r1^2 * Pi
(r1^2 * 2 + r2^2) * 2 = (r2^2 + r1^2) * Pi

r1 = sqrt(RB / (4-Pi))
r2 = sqrt(RB / (Pi - 2))

r1/r2 = sqrt( (RB / (4-Pi)) * ((Pi - 2) / RB) )
r1/r2 = sqrt((Pi - 2) / (4-Pi))

>>9406847
Set r1 = 50 cm

RB = 50^2 * (4 - Pi) = ~2146 cm2
r2 = sqrt(50^2 * (4 - Pi) / (Pi - 2)) = ~43.36cm
d2 = 2*r2 = ~86.71 cm

r1-r2 = ~6.64 cm
A = r1^2 = 10,000 cm2 = 1 m2
As = s^2 = ~3760 cm2
A1 = r1^2 * Pi = ~7854 cm2
A2 = r2^2 * Pi = ~5905 cm2

>>9407052
engineer detected.

>>9406822
[eqn]\sqrt{\frac{\pi - 2}{4-\pi}}[/eqn]

>>9407285
[eqn]\sqrt{\frac{\pi - 2}{4-\pi}}\approx 1.1532112482813995893145190280706[/eqn]

>>9406822
>>9406822
Let [math]A[/math] equal the area of the red and blue space, which are given as equal to each other. Let [math]l[/math] equal the large circle's radius and let [math]s[/math] equal the small circle's area.
Then [math]A = 2\dot l^2 - \pi\dot l^2[/math] Factoring out [math]l^2[/math] from the right side, we have [math]A = l^2(4-\pi)[/math]. Divide both sides by [math](4-\pi)[/math] and take the square root to solve for [math]l = \sqrt{\frac{A}{(4-\pi}}[/math].
Repeating the process for s, we have [math]A=\pi\dot s^2 - ((2\dot\sqrt{2})\dot s)^2[/math] which, when we factor out [math]s^2[/math] gives us [math]A=(s^2)\dot(\pi - 2)[/math]. Divide both sides by [math]\pi-2[/math] and take the square root and we have [math]s=\sqrt\frac{A}{\pi-2}}[/math]. Plug in these values for the original ratio in question and simplify and you have [math]\sqrt{\frac{\pi-2}{4-\pi}}[/math]

>>9406822
>>9406822
Let [math]A[/math] equal the area of the red and blue space, which are given as equal to each other. Let [math]l[/math] equal the large circle's radius and let [math]s[/math] equal the small circle's area.
Then [math]A = 2\dot l^2 - \pi\dot l^2[/math] Factoring out [math]l^2[/math] from the right side, we have [math]A = l^2(4-\pi)[/math]. Divide both sides by [math](4-\pi)[/math] and take the square root to solve for [math]l = \sqrt{\frac{A}{(4-\pi}}[/math].
Repeating the process for s, we have [math]A=\pi\dot s^2 - ((2\dot\sqrt{2})\dot s)^2[/math] which, when we factor out [math]s^2[/math] gives us [math]A=(s^2)\dot(\pi - 2)[/math]. Divide both sides by [math]\pi-2[/math] and take the square root and we have [math]s=\sqrt\frac{A}{\pi-2}}[/math]. Plug in these values for the original ratio in question and simplify and you have [math]\sqrt{\frac{\pi-2}{4-\pi}}[/math]

>>9406822
Let [math]A[/math] equal the area of the red and blue space, which are given as equal to each other. Let [math]l[/math] equal the large circle's radius and let [math]s[/math] equal the small circle's area.
Then [math]A = 2\dot l^2 - \pi\dot l^2[/math] Factoring out [math]l^2[/math] from the right side, we have [math]A = l^2(4-\pi)[/math]. Divide both sides by [math](4-\pi)[/math] and take the square root to solve for [math]l = \sqrt{\frac{A}{(4-\pi}}[/math].
Repeating the process for s, we have [math] A=\pi \dot s^2 - ( ( 2\dot\ sqrt{2} )\ dot s)^2 [/math] which, when we factor out [math]s^2[/math] gives us [math]A=(s^2)\dot(\pi - 2)[/math]. Divide both sides by [math]\pi-2[/math] and take the square root and we have [math]s=\sqrt\frac{A}{\pi-2}}[/math]. Plug in these values for the original ratio in question and simplify and you have [math]\sqrt{\frac{\pi-2}{4-\pi}}[/math]

why can't i understand any of this

>>9407294
[eqn] \sqrt{ \frac{ \pi - 2}{4- \pi}} \approx 1.1532112482813995893145190280706[/eqn]

>>9406822
Let [math]A[/math] equal the area of the red and blue space, which are given as equal to each other. Let [math]l[/math] equal the large circle's radius and let [math]s[/math] equal the small circle's area.
Then [math]A = 2\cdot l^2 - \pi\cdot l^2[/math] Factoring out [math]l^2[/math] from the right side, we have [math]A = l^2(4-\pi)[/math]. Divide both sides by [math](4-\pi)[/math] and take the square root to solve for [math]l = \sqrt{\frac{A}{(4-\pi)}}[/math].
Repeating the process for s, we have [math]A=\pi\cdot s^2 - ( (2\cdot \sqrt{2} ) \cdot s)^2[/math] which, when we factor out [math]s^2[/math] gives us [math]A=(s^2)\cdot(\pi - 2)[/math]. Divide both sides by [math]\pi-2[/math] and take the square root and we have [math]s = \sqrt{\frac{A}{\pi-2}}[/math]. Plug in these values for the original ratio in question and simplify and you have [math]\sqrt{\frac{\pi-2}{4-\pi}}[/math]

[eqn] \sqrt{ \frac{ \pi - 2}{4- \pi}} \approx 1.1532112482813995893145190280706 [/eqn]

>>9406822
>red area = blue area
Is that even possible?

>>9406822
[eqn]\sqrt{ \frac{ \pi - 2}{4- \pi}} \approx 1.1532112482813995893145190280706[/eqn]

>>9406822
Let [math]A[/math] equal the area of the red and blue space, which are given as equal to each other. Let [math]l[/math] equal the large circle's radius and let [math]s[/math] equal the small circle's area.
Then [math]A = 2\cdot l^2 - \pi\cdot l^2[/math] Factoring out [math]l^2[/math] from the right side, we have [math]A = l^2 \cdot (4- \pi) [/math]. Divide both sides by [math](4-\pi)[/math] and take the square root to solve for [math] l = \sqrt{\frac{A}{(4-\pi}}[/math].
Repeating the process for s, we have [math]A = \pi \cdot s^2 - ( ( 2 \cdot \sqrt{2} ) \dot s)^2[/math] which, when we factor out [math] s^2 [/math] gives us [math] A = (s^2) \cdot ( \pi - 2) [/math]. Divide both sides by [math]\pi-2[/math] and take the square root and we have [math] s = \sqrt{\frac{A}{\pi-2}}[/math]. Plug in these values for the original ratio in question and simplify and you have [math] \sqrt{\frac{\pi-2}{4-\pi}}[/math]

>>9406833
>>9406838
>>9407285
>>9407294
>>9407418
>>9407423
>>9407426
>using a computer to solve a problem this easy
>>9406847
>>9407052
>>9407415
>>9407425
plebs
>>9407427
/thread

>>9407283
I just wanted to get a feel for the squares (using the formulas I derived).

>>9407427
I see what you did there and I like it.
Elegant solution.

Approximately 1.1532 (4dp). I'm a physicist so I don't do exact form.

But here it is anyway: (((((2*Pi)-4)/(4-Pi))/2)^(1/2))=R/r

Presh Talwalkar here

>>9407432
EZ (quick) maths

Eazy