The picture probably explains my inquiry best.I require to discover a method to division a circle right into 3 parts of equal area with just 2 currently that intersect each other on the rundown of the circle.Also I must check, if whatever diameter is in between those lines, likewise splits circles with a different diameter into equal parts.And lastly, and also probably the most an overwhelming question: exactly how do I need to calculate the angle in between x lines the all crossing in one point, so that the circle is split into x+1 components with area = 1/(x+1) of the circle?I tried my best, yet couldn"t even discover a solitary answer or the appropriate strategy to tackle the question...

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You are watching: Divide a circle into 3 equal parts

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edited Mar 18 in ~ 20:50
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Andrei
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asked Mar 18 in ~ 20:40
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JonasHausJonasHaus
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$egingroup$
Given the edge $ heta$, split by the diameter comprise $B$, take into consideration the complying with diagram:

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$overlineBO$ is the line through the center and $overlineBA$ is the chord cutting off the lune who area us wish to compute.

The area of the one wedge subtended by $angle BOA$ is$$fracpi- heta2r^2 ag1$$The area of $ riangle BOA$ is$$frac12cdotoverbracersinleft(frac heta2 ight)^ extaltitudecdotoverbrace2rcosleft(frac heta2 ight)^ extbase=fracsin( heta)2r^2 ag2$$Therefore, the area the the lune is $(1)$ minus $(2)$:$$fracpi- heta-sin( heta)2r^2 ag3$$To obtain the area separated into thirds, us want$$fracpi- heta-sin( heta)2r^2=fracpi3r^2 ag4$$which method we desire to solve$$ heta+sin( heta)=fracpi3 ag5$$whose solution deserve to be achieved numerically (e.g. Use $M=fracpi3$ and also $varepsilon=-1$ in this answer)$$ heta=0.5362669789888906 ag6$$Giving us

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