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ABCD is a quadrilateral inscribed in a circle. given that AB = sqrt( 2 + sqrt(2) ) and AB subtends an angle of 135degrees at the center O of the circle. Find maximum possible area of ABCD.

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yes sir..

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Other answers:

r=1 here..from cosine law ..
thats right...
but didn't u mention AB as sqrt( 2 + sqrt(2) )
ohh yes,,its sqrt( 2 + sqrt(2) ) and not 2 + sqrt(2) @mukushla
ohh wait,,i reach some where..
|dw:1345656422768:dw| area of each triangle = sin(angle) r^2 /2 so area will be maximum when sina + sin b +sin c is maximum given that a+b+c = 225 !! hmm..was that right ?
thats right... how do we continue ? AM,GM ? i dont think will help much ?
i think there are better methods...this one seems a little bit hard to get
Let O be the center of the circle. Let r be its radius. Let s=angle AOD , t= angle COD, and \( \frac {5 \pi}4 - s -t=\)angle COB. We need to maximize the sum of the areas of the triangles AOD, COD and COB. This sum is \[ \frac 1 2 r^2 \sin(s) +\frac 1 2 r^2 \sin(t) +\frac 1 2 r^2 \sin\left( \frac {5 \pi}4 - s -t\right) =\\ \frac 1 2 r^2\left( \sin(s) + \sin(t) + \sin\left( \frac {5 \pi}4 - s -t\right)\right) \] One needs to maximize \[ f(s,t)= \sin(s) +\sin(t) + \sin\left( \frac {5 \pi}4 - s -t\right) \] Using gradient we find that f(s,t) has a maximum when \[ s=t= \frac {5 \pi}{12} \] Which means when the three triangles are equal.
gradient ?
so gradient approximately means slope ?
i mean in layman language ?
It replaces derivative in functions of two variables,
yep,,that only,, so how do you make that conclusion from derivatives ?
Have you had calculus III? just high school pass sir,, :/
you method might be legitimate but there ought to be a high school level solution..its that standard ques only..hmm really sorry,,though i appreciate your efforts :)
Ok. May be there is a more elementary method. Let us hope someone else will find it.
is AB parallel to CD?
is this correct?\[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}\]
@panlac01 nops.. @mukushla i dont know,,how did you pull that off ?
I was just looking for possible entries. I didn't want to rely on the drawing above.
to my understanding u can use Jensen's inequality here...
see if this might help
\(f(x)=\sin x\) concave down in \((0,\pi)\) so according to the Jensen's inequality\[\frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \le f(\frac{x_1+x_2+...+x_n}{n})\] \[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}=\sin 75\]
muku, I like how you manipulated this one.
altough i didnt understand most part of jenson;s inequality (tried wikipedia) , but still seems to be really useful,, thanks alot!
I think the key is in cyclic quadrilateral
well..given AB and angle AOB,,infinite cyclic quads ABCD can be made .. jenson;s inequality maybe is the easiest of the lot..
@panlac01 ... thank u man
i got the same result using lagrange multipliers .. got a better understanding of the concept now.. thanks all ! :)

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