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February 19th, 2014, 10:13 AM  #1 
Newbie Joined: Feb 2014 Posts: 26 Thanks: 0  Quadrilateral inscribed in a circle
Hello! ABCD quadrilateral inscribed in a circle of radius R,prove that: AB•BC•CD•DA<=4R^4 Thank you! 
February 20th, 2014, 05:02 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,943 Thanks: 1132 Math Focus: Elementary mathematics and beyond  Re: Quadrilateral inscribed in a circle
From the law of cosines, a² = 2r²  2r²cos(w) b² = 2r²  2r²cos(x) c² = 2r²  2r²cos(y) d² = 2r²  2r²cos(z) with w + x + y + z = 360 degrees. a²b²c²d² = 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. 
February 21st, 2014, 02:08 AM  #3 
Newbie Joined: Feb 2014 Posts: 26 Thanks: 0  Re: Quadrilateral inscribed in a circle
Thank you!

February 21st, 2014, 06:49 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
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. 
April 13th, 2014, 03:05 PM  #5 
Senior Member Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 
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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