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February 19th, 2014, 10:13 AM   #1
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Quadrilateral inscribed in a circle

Hello!
ABCD quadrilateral inscribed in a circle of radius R,prove that:
ABBCCDDA<=4R^4
Thank you!
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February 20th, 2014, 05:02 PM   #2
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Re: Quadrilateral inscribed in a circle

From the law of cosines,

a = 2r - 2rcos(w)
b = 2r - 2rcos(x)
c = 2r - 2rcos(y)
d = 2r - 2rcos(z)

with w + x + y + z = 360 degrees.

abcd = 16r^8(1 - cos(w))(1 - cos(x))(1 - cos(y))(1 - cos(z))

0 < (1 - cos(w))(1 - cos(x))(1 - cos(y))(1 - cos(z)) ? 1, but that seems difficult to prove.
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February 21st, 2014, 02:08 AM   #3
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Re: Quadrilateral inscribed in a circle

Thank you!
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February 21st, 2014, 06:49 AM   #4
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If a cyclic quadrilateral has sides of length a, b, c and d, and diagonals of length p and q,
abcdpq ? area(ABCD)(8R^4). You can Google this! It's in a book by Titu Andreescu and Dorin Andrica.

Now use the formula area(ABCD) = (pq/2)|sin(?)|, where ? is the angle between the diagonals of ABCD, and the fact that |sin(?)| ? 1.
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April 13th, 2014, 03:05 PM   #5
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Lemma: for $\displaystyle \alpha+\beta+\gamma+\delta= \pi $

$\displaystyle \sin \alpha \sin \beta \sin \gamma \sin \delta \leq \frac{1}{4}$

Proof: By AG inequality

$\displaystyle \sin \alpha \sin \beta \sin \gamma \sin \delta \leq (\frac{\sin \alpha+\sin \beta+\sin \gamma+\sin \delta}{4})^4 $

And by Jensen inequailty
$\displaystyle \sin \alpha+\sin \beta+\sin \gamma+\sin \delta}{4} \leq \sin \frac{\pi}{4}=\frac{\sqrt2}{2} $

so
$\displaystyle \sin \alpha \sin \beta \sin \gamma \sin \delta \leq (\frac{\sqrt2}{2})^4=\frac{1}{4}$


If we divide Quadrilateral ABCD in triangles ABC and ACD and we use sine law:
$\displaystyle AB=2R \sin \angle ACB$
$\displaystyle BC=2R \sin \angle BAC$
$\displaystyle CD=2R \sin \angle DAC$
$\displaystyle AD=2R \sin \angle ACD $

multiplying and knowing that $\displaystyle \angle ACB+\angle BAC+\angle DAC+\angle ACD=\pi $ we get the wanted inequality.

Skipjack's solution is obviously much nicer and elegant but this was the first that came to my mind.
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