Another great mathematical problem: Quadrisection of a disc

[Anixx]

Along with the problem of squaring a circle and trisection of an angle, there is one more great problem: quarisection of a disc.

You have a disk and have to dissect it into four parts of equal area with three chords coming from the same point on the disc's boundary (one of these chords is a diameter).

What makes this problem impossible to solve with straightedge and compass?

 

[DaveC426913]

"quadrisection", for those of us who have to Google the problem.

 

[mathman]

Try Gptchat.

 

[fresh_42]

Along with the problem of squaring a circle and trisection of an angle, there is one more great problem: quarisection of a disc.

You have a disk and have to dissect it into four parts of equal area with three chords coming from the same point on the disc's boundary (one of these chords is a diameter).

What makes this problem impossible to solve with straightedge and compass?

In order to do this, you have to solve

$$
A= \dfrac{ \pi r^2}{4}= \dfrac{r^2}{2}(\alpha -\sin(\alpha)) \Longleftrightarrow \dfrac{\pi}{2}=\alpha -\sin(\alpha)
$$
which is something like $\alpha \approx 2.31$ and this number is nowhere even near a Galois extension of degree $2^n.$

 

[Anixx]

In order to do this, you have to solve
View attachment 329405
$$
A= \dfrac{ \pi r^2}{4}= \dfrac{r^2}{2}(\alpha -\sin(\alpha)) \Longleftrightarrow \dfrac{\pi}{2}=\alpha -\sin(\alpha)
$$
which is something like $\alpha \approx 2.31$ and this number is nowhere even near a Galois extension of degree $2^n.$

If we have a straightangle, compass and an angle of Dottie number available, can we divide a disk into arbitrary number of parts of equal area with chords?

What if we have only interval of Dottie number and no angle?

 

[fresh_42]

Things become completely different if additional tools can be used. IIRC then trisection becomes solvable with the help of an Archimedean spiral.

I don't know anything about the problem here with any auxiliary weapons. However, solving the equation for $\alpha$ looks rather difficult, even with additional tools. $\alpha - \sin(\alpha)$ is very inconvenient.