# Two points on circle resulting in 5 equal regions

What values of $Z_1$ and $Z_2$ make the five regions of the unit circle, shown below, equal in area? $\overline{Z_1}$ and $\overline{Z_2}$ are conjugates of $Z_1$ and $Z_2$; in other words they lie directly across the real axis from their counterparts.

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Is it possible to get the source? – Aryabhata Dec 22 '10 at 3:45
@Moron, I don't know what you mean by the source. – I. J. Kennedy Dec 22 '10 at 5:17
By source I mean, did you get it from somewhere? If so, where? Or did you make it up yourself? If so, what motivated it? – Aryabhata Dec 22 '10 at 5:18
It's my own problem. I felt sure there would be a solution but apparently not! – I. J. Kennedy Dec 22 '10 at 15:27

The problem is overdetermined/impossible. The values of $Z_1$ and $Z_2$ are forced, as in Ross's answer, but the segments with area $\frac{2\pi}{5}$ are not divided into regions of equal area.

Let $0<\alpha<\pi$ be the solution to $\frac{\alpha-\sin\alpha}{2}=\frac{\pi}{5}$ (so $\alpha\approx 2.11314$). $\arg(Z_2)=\pi-\frac{\alpha}{2}$.

Let $0<\beta<\pi$ be the solution to $\frac{\beta-\sin\beta}{2}=\frac{2\pi}{5}$ (so $\beta\approx 2.8248$). $\arg(\overline{Z_1})=\arg(Z_2)-\beta=\pi-\frac{\alpha}{2}-\beta$.

Any point on the line segment joining $Z_2$ and $\overline{Z_1}$ is a linear combination of these two points, $t\cdot Z_2+(1-t)\overline{Z_1}$ with $0<t<1$. The intersection of that line segment and the line segment joining $\overline{Z_2}$ and $Z_1$ has imaginary part $0$. Solving for $t$ in terms of $\alpha$ and $\beta$ yields $t=\frac{1}{2}\csc\frac{\beta}{2}\sec\frac{\alpha+\beta}{2}\sin(\frac{\alpha}{2}+\beta)$ ($t\approx 0.436382$). The location of the point of intersection of the two lines is at $-\cos\frac{\beta}{2}\sec\frac{\alpha+\beta}{2}+0i$ ($\approx 0.20166$).

Now, the area of the region determined by this point of intersection, $Z_1$, $\overline{Z_1}$, and the circle is the area of the segment of the circle plus the area of the triangle determined by those three points. The area of the segment of the circle is $\frac{2\arg(Z_1)-\sin(2\arg(Z_1))}{2}=\frac{1}{2}(\alpha+2\beta-2\pi-\sin(\alpha+2\beta))$ $\approx 0.241854$; the area of the triangle is $|\sin^2(\frac{\alpha}{2}+\beta)\tan\frac{\alpha+\beta}{2}|\approx 0.361977$ (found using shoelace method with the three vertices). The total area of the region is thus $\approx 0.603831<0.628319\approx\frac{\pi}{5}$.

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We have segment $Z_2\overline{Z_2}$ is $\frac15$ of the area. The area of a segment is $\frac{\theta-\sin(\theta)}{2}$. Setting this to $\frac{\pi}{5}$ gives the angle at the center to be $\theta$=2.113139. Segment $Z_1\overline{Z_1}\overline{Z_2}$ then has area $\frac{2\pi}{5}$, so the angle is 2.824797. So $Z_2=\exp{(\pi-1.056569)i}$ and $Z_1=\exp{(.74890i)}$

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Does this actually give five equal areas? I think the problem is overconstrained. – Peter Shor Dec 22 '10 at 4:56
I think it works. Clearly you can make $Z_2\overline{Z_2}$ be 1/5 of the circle and centering it on $\pi$ makes them conjugate. Then just swing $Z_1$ around to make the two segments (which are symmetric) be 2/5 of the circle. – Ross Millikan Dec 22 '10 at 5:16
@Peter Shor: Isaac proves you are right. – Ross Millikan Dec 23 '10 at 20:32