Another great mathematical problem: Quadrisection of a disc

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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.
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Anixx
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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?
 
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"quadrisection", for those of us who have to Google the problem. :wink:
 
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Try Gptchat.
 
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  • #4
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
1689717515785.png

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