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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.
 2 years ago
 2 years 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.
 2 years ago
 2 years ago

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

panlac01 Group TitleBest ResponseYou've already chosen the best response.0
http://www.onlinemathlearning.com/quadrilateralcircle.html
 2 years 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\[\frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \le f(\frac{x_1+x_2+...+x_n}{n})\] \[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}=\sin 75\]
 2 years ago

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
@panlac01 ... thank u man
 2 years 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 ! :)
 2 years ago
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