Cyclic quadrilateral

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Thanic Nur Samin
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Cyclic quadrilateral

Unread post by Thanic Nur Samin » Fri Jan 31, 2014 8:35 pm

Prove that, among the quadrilaterals inscribed in a circle, a square's area is the maximum.

Hint: Write the area formula in sines and notice for what $sin\theta$ is maximum.
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Shahriar Shakib
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Re: Cyclic quadrilateral

Unread post by Shahriar Shakib » Wed Mar 26, 2014 8:07 pm

Formula of the area of square is $ab\sin ϴ$(where $a$ and $b$ are the sides of a quadrilateral and $ϴ$ is the in between angle of those sides). Now,($\sin 90^\circ= 1$),and ($\sin 89^\circ = 0.999847695$) and ($\sin 91^\circ$ is also $0.999847695$).So $\sin 90^\circ$ is maximum and angle of a square is $90$ degree.So it is proved that a square's area is maximum in some quadrilaterals which are inscribed in a circle.

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Phlembac Adib Hasan
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Re: Cyclic quadrilateral

Unread post by Phlembac Adib Hasan » Thu Mar 27, 2014 2:29 pm

Let $ABCD$ be any cyclic quad inscribed in a circle with radius $R$. Suppose $\theta$ is the angle between the diagonals. We know $AC,BD\leq 2R$ and $\sin \theta \leq 1$. Therefore,
\[(ABCD)=\frac 1 2 AC\cdot BD \sin \theta \leq \frac 1 2 2R\cdot 2R\cdot 1=2R^2\]
With equality iff $AC=BD=2R$ and $\sin \theta =1$ (i.e. $\theta = 90^\circ$), in other words, when $ABCD$ is a square.
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*Mahi*
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Re: Cyclic quadrilateral

Unread post by *Mahi* » Sat Mar 29, 2014 1:09 pm

Or, the area is $\frac 1 2 r^2(\sin AOB+ \sin BOC + \sin COD + \sin DOA) \leq \frac 1 2 r^2 (4) = 2r^2$, with $O$ the center of circle $ABCD$.
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Thanic Nur Samin
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Re: Cyclic quadrilateral

Unread post by Thanic Nur Samin » Thu Apr 03, 2014 9:05 am

The red quadrilateral is a square.
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