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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: 7,968 Thanks: 1152 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: 21,036 Thanks: 2273 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. Thanks from crom
 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.

### dorin andrica

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