Can such problem be solved?

Area A =area of the green segment.

x=-cos/(1-2cos)

y=sin/(1-2cos)

sin/(1-2cos)=(p-6a)/2

On 3/5/2025 at 8:32 AM, DimaMazin said:

Can such problem be solved?

Area A =area of the green segment.

x=-cos/(1-2cos)

y=sin/(1-2cos)

sin/(1-2cos)=(p-6a)/2

I don't think your equation is correct.

11 hours ago, Genady said:

I don't think your equation is correct.

Maybe you think sin and cos are sin(a) and cos(a)? No. sin is sin(p/2 +a) and cos is cos(p/2 +a)  as it showed on diagram.

y = cos(a)/(1+2sin(a))

cos(a)/(1+2sin(a))=(p-6a)/2

Edited March 7, 20251 yr by DimaMazin

3 hours ago, DimaMazin said:

y = cos(a)/(1+2sin(a))

cos(a)/(1+2sin(a))=(p-6a)/2

My starting equation is

\(\frac b 2 - \frac {\sin b \sin c} {2 \sin d} = \frac e 2 - \frac {\sin e} 2\)

where

\(b=2a\)
\(2c=\pi - (\frac {\pi} 2 + a)\) 
\(d = \pi - b - c\)
\(e = \frac {\pi} 2 - a\)

I don't know if it's the same as yours.

Seems we can exactly solve such problem:

a - sin(a) = (Pi - 2)/2 😛

On 12/10/2024 at 5:09 PM, Genady said:

a=1.8954942670...

a=2sin(a)

I have no higher education and I am janitor therefore I don't understand why we can not to exactly solve it relative to Pi. Can next method be correct?

I mistaken again.

How to solve next problem?

It appears that you have three unknowns: [math]x, y, \alpha[/math], and you have one equation [math]yx=\alpha[/math]. You need two more equations to solve the problem.

If I understand your sketch correctly, one equation could be that two red areas are equal, and another that the brown area equals sum of the red and the green areas.

If so, the rest is some trigonometry and some algebra.