# Another great mathematical problem: Quadrisection of a disc

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• Anixx
In summary, the problem of quadrisection of a disc involves dissecting a disk into four equal parts with three chords coming from the same point on the disc's boundary, one of which is a diameter. This problem is impossible to solve using only a straightedge and compass due to the difficulty of solving the equation involved. Additional tools may make the problem solvable, but it remains a challenging problem.
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?

Last edited:

topsquark and weirdoguy
Try Gptchat.

DaveC426913
Anixx said:
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.##

dextercioby, Anixx and berkeman
fresh_42 said:
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?

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.

Last edited:
dextercioby

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