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.
Cyclic quadrilateral
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Re: Cyclic quadrilateral
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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Re: Cyclic quadrilateral
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.
\[(ABCD)=\frac 1 2 AC\cdot BD \sin \theta \leq \frac 1 2 2R\cdot 2R\cdot 1=2R^2\]
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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Re: Cyclic quadrilateral
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
The red quadrilateral is a square.
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Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.