On Area percentage -1

In the Discussion I will generalise a case one of which is in this question .

In gmJOGC\parallel_{gm} JOGC assuming JC=CG\overline{JC}=\overline{CG} the red\red{red} area is 100π(COGsin2(COG)sin(2COG))%\frac{100}{\pi}(\angle COG - \frac{\sin^2(\angle COG)}{\sin(2\angle COG)})\% of the whole circle(all the angles are measured in radians only)

Proof:

Let us denote COG=α\angle COG=\alpha overall the proof (to make things easier to write for me)

And let JC=CG=a;OC=r\overline{JC}=\overline{CG}=a;\overline{OC}=r

Now as JOCGJOCG is a parallelogram \therefore JOCGJOG=CGD\overline{JO} || \overline{CG} \Rightarrow \angle JOG=\angle CGD

Also as the diagonal of a parallelogram bisects its vertex angles 2α=CGD{2\alpha=\angle CGD}

In CGD\triangle CGD : sin(CGD)=sin(2α)=CDa\sin(\angle CGD)=\sin(2\alpha)=\frac{\overline{CD}}{a} CD=asin(2α)\Rightarrow \overline{CD}=a\sin(2\alpha) In COD\triangle COD : sin(α)=CDr\sin(\alpha)=\frac{\overline{CD}}{r} CD=rsin(α)\overline{CD}=r\sin(\alpha) rsin(α)=asin(2α)\Rightarrow r\sin(\alpha)=a\sin(2\alpha) r=asin(2α)sin(α)\Rightarrow r=a\frac{\sin(2\alpha)}{\sin(\alpha)} Area of gmJOGC=A1=OG×CD=a×asin(2α)=a2sin(2α)||_{gm}JOGC=A_1=\overline{OG}\times\overline{CD}=a\times a\sin(2\alpha)=a^2\sin(2\alpha)

Area of circle =A2=πr2=πa2sin2(2α)sin2(α)=A_2=\pi r^2=\pi a^2\frac{\sin^2(2\alpha)}{\sin^2(\alpha)}

Area of the sector JOG2πA=2α2πA=απA=απ(πa2sin2(2α)sin2(α))=a2αsin2(2α)sin2(α)\frac{\angle JOG}{2\pi}A=\frac{\cancel{2}\alpha}{\cancel{2}\pi}A=\frac{\alpha}{\pi}A=\frac{\alpha}{\cancel{\pi}}(\cancel{\pi} a^2\frac{\sin^2(2\alpha)}{\sin^2(\alpha)})=a^2\alpha \frac{\sin^2(2\alpha)}{\sin^2(\alpha)}

Area of the red\red{red} region =AA1=A-A_1

Percentage of the area to the circles area=AA1A2×100=(a2αsin2(2α)sin2(α)a2sin(2α))×100sin2(α)πa2sin2(2α)=100π(αsin2(α)sin(2α))=\frac{A-A_1}{A_2}\times 100=(a^2\alpha \frac{\sin^2(2\alpha)}{\sin^2(\alpha)}-a^2\sin(2\alpha))\times \frac{100\sin^2(\alpha)}{\pi a^2\sin^2(2\alpha)}=\boxed{\frac{100}{\pi}({\alpha}-\frac{\sin^2(\alpha)}{\sin(2\alpha)})}

Bonus:I have proved it for the case when all sides of the parallelogram are equal, you can find a formula for general case quadrilateral

Note by Zakir Husain
3 months, 1 week ago

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1 vote

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@Zakir Husain, sin(α)=CDrrCD\sin{(\alpha)} = \dfrac{\overline{CD}}{r} \ne \red{\dfrac{r}{\overline{CD}}}

Mahdi Raza - 3 months, 1 week ago

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Improved that typing mistake!

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain: "I have proved it for the case when all sides of the parallelogram are equal, you can find a formula for general case", I don't see which line have you assumed that a=asin(2α)a = a \sin{(2 \alpha)}

The proof is very elaborate and on a general case only.

Mahdi Raza - 3 months, 1 week ago

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In the second line assumed that JC=CG=a\overline{JC}=\overline{CG}=a

Zakir Husain - 3 months, 1 week ago

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Ohh, didn't catch that. I was just reading and cross-checking the proof

Mahdi Raza - 3 months, 1 week ago

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Great!

A Former Brilliant Member - 3 months, 1 week ago

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What the hell? You linked this page???

Páll Márton (no activity) - 2 months, 1 week ago

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Oh! a big mistake. But now I have edited it!

Zakir Husain - 2 months, 1 week ago

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