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.

- shubhamsrg

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- anonymous

*

- anonymous

|dw:1345656175855:dw|

- shubhamsrg

yes sir..

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## More answers

- shubhamsrg

r=1 here..from cosine law ..

- anonymous

thats right...

- hartnn

but didn't u mention AB as sqrt( 2 + sqrt(2) )

- shubhamsrg

ohh yes,,its sqrt( 2 + sqrt(2) ) and not 2 + sqrt(2) @mukushla

- anonymous

sorry...yes

- shubhamsrg

|dw:1345656382986:dw|

- shubhamsrg

ohh wait,,i reach some where..

- shubhamsrg

|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 ?

- anonymous

thats right...

- shubhamsrg

aha..so how do we continue ? AM,GM ? i dont think will help much ?

- anonymous

i think there are better methods...this one seems a little bit hard to get

- shubhamsrg

hmm..

- shubhamsrg

@eliassaab ?

- anonymous

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.

- shubhamsrg

gradient ?

- anonymous

http://en.wikipedia.org/wiki/Gradient

- shubhamsrg

so gradient approximately means slope ?

- shubhamsrg

i mean in layman language ?

- anonymous

It replaces derivative in functions of two variables,

- shubhamsrg

yep,,that only,,
so how do you make that conclusion from derivatives ?

- anonymous

Have you had calculus III?

- shubhamsrg

nops..am just high school pass sir,, :/

- shubhamsrg

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 :)

- anonymous

Ok. May be there is a more elementary method. Let us hope someone else will find it.

- shubhamsrg

hmm..

- anonymous

is AB parallel to CD?

- anonymous

is this correct?\[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}\]

- shubhamsrg

@panlac01 nops..
@mukushla i dont know,,how did you pull that off ?

- anonymous

I was just looking for possible entries. I didn't want to rely on the drawing above.

- anonymous

to my understanding u can use Jensen's inequality here...

- anonymous

see if this might help
http://www.artofproblemsolving.com/Wiki/index.php/Cyclic_quadrilateral

- anonymous

http://www.onlinemathlearning.com/quadrilateral-circle.html

- anonymous

\(f(x)=\sin x\) concave down in \((0,\pi)\) so according to the Jensen's inequality\[\frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \le f(\frac{x_1+x_2+...+x_n}{n})\]
\[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}=\sin 75\]

- anonymous

muku, I like how you manipulated this one.

- shubhamsrg

altough i didnt understand most part of jenson;s inequality (tried wikipedia) , but still seems to be really useful,, thanks alot!

- anonymous

I think the key is in cyclic quadrilateral

- shubhamsrg

well..given AB and angle AOB,,infinite cyclic quads ABCD can be made .. jenson;s inequality maybe is the easiest of the lot..

- anonymous

@panlac01 ... thank u man

- shubhamsrg

i got the same result using lagrange multipliers .. got a better understanding of the concept now..
thanks all ! :)

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