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shubhamsrg
Group Title
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.
 one year ago
 one year ago
shubhamsrg Group Title
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.
 one year ago
 one year ago

This Question is Closed

mukushla Group TitleBest ResponseYou've already chosen the best response.2
dw:1345656175855:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
yes sir..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
r=1 here..from cosine law ..
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
thats right...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
but didn't u mention AB as sqrt( 2 + sqrt(2) )
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
ohh yes,,its sqrt( 2 + sqrt(2) ) and not 2 + sqrt(2) @mukushla
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
sorry...yes
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
dw:1345656382986:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
ohh wait,,i reach some where..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
thats right...
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
aha..so how do we continue ? AM,GM ? i dont think will help much ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
i think there are better methods...this one seems a little bit hard to get
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
@eliassaab ?
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
gradient ?
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Gradient
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
so gradient approximately means slope ?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i mean in layman language ?
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.1
It replaces derivative in functions of two variables,
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
yep,,that only,, so how do you make that conclusion from derivatives ?
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.1
Have you had calculus III?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
nops..am just high school pass sir,, :/
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.1
Ok. May be there is a more elementary method. Let us hope someone else will find it.
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
is AB parallel to CD?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
is this correct?\[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}\]
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
@panlac01 nops.. @mukushla i dont know,,how did you pull that off ?
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
I was just looking for possible entries. I didn't want to rely on the drawing above.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
to my understanding u can use Jensen's inequality here...
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
see if this might help http://www.artofproblemsolving.com/Wiki/index.php/Cyclic_quadrilateral
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
http://www.onlinemathlearning.com/quadrilateralcircle.html
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
\(f(x)=\sin x\) concave down in \((0,\pi)\) so according to the Jensen's inequality\[\frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \le f(\frac{x_1+x_2+...+x_n}{n})\] \[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}=\sin 75\]
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
muku, I like how you manipulated this one.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
altough i didnt understand most part of jenson;s inequality (tried wikipedia) , but still seems to be really useful,, thanks alot!
 one year ago

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
I think the key is in cyclic quadrilateral
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
well..given AB and angle AOB,,infinite cyclic quads ABCD can be made .. jenson;s inequality maybe is the easiest of the lot..
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
@panlac01 ... thank u man
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i got the same result using lagrange multipliers .. got a better understanding of the concept now.. thanks all ! :)
 one year ago
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