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shubhamsrg

  • 3 years ago

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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  1. mukushla
    • 3 years ago
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    *

  2. mukushla
    • 3 years ago
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    |dw:1345656175855:dw|

  3. shubhamsrg
    • 3 years ago
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    yes sir..

  4. shubhamsrg
    • 3 years ago
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    r=1 here..from cosine law ..

  5. mukushla
    • 3 years ago
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    thats right...

  6. hartnn
    • 3 years ago
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    but didn't u mention AB as sqrt( 2 + sqrt(2) )

  7. shubhamsrg
    • 3 years ago
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    ohh yes,,its sqrt( 2 + sqrt(2) ) and not 2 + sqrt(2) @mukushla

  8. mukushla
    • 3 years ago
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    sorry...yes

  9. shubhamsrg
    • 3 years ago
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    |dw:1345656382986:dw|

  10. shubhamsrg
    • 3 years ago
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    ohh wait,,i reach some where..

  11. shubhamsrg
    • 3 years ago
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    |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 ?

  12. mukushla
    • 3 years ago
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    thats right...

  13. shubhamsrg
    • 3 years ago
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    aha..so how do we continue ? AM,GM ? i dont think will help much ?

  14. mukushla
    • 3 years ago
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    i think there are better methods...this one seems a little bit hard to get

  15. shubhamsrg
    • 3 years ago
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    hmm..

  16. shubhamsrg
    • 3 years ago
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    @eliassaab ?

  17. eliassaab
    • 3 years ago
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    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.

  18. shubhamsrg
    • 3 years ago
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    gradient ?

  19. eliassaab
    • 3 years ago
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    http://en.wikipedia.org/wiki/Gradient

  20. shubhamsrg
    • 3 years ago
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    so gradient approximately means slope ?

  21. shubhamsrg
    • 3 years ago
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    i mean in layman language ?

  22. eliassaab
    • 3 years ago
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    It replaces derivative in functions of two variables,

  23. shubhamsrg
    • 3 years ago
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    yep,,that only,, so how do you make that conclusion from derivatives ?

  24. eliassaab
    • 3 years ago
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    Have you had calculus III?

  25. shubhamsrg
    • 3 years ago
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    nops..am just high school pass sir,, :/

  26. shubhamsrg
    • 3 years ago
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    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 :)

  27. eliassaab
    • 3 years ago
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    Ok. May be there is a more elementary method. Let us hope someone else will find it.

  28. shubhamsrg
    • 3 years ago
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    hmm..

  29. panlac01
    • 3 years ago
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    is AB parallel to CD?

  30. mukushla
    • 3 years ago
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    is this correct?\[\frac{\sin a+\sin b+\sin c}{3} \le \sin \frac{a+b+c}{3}\]

  31. shubhamsrg
    • 3 years ago
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    @panlac01 nops.. @mukushla i dont know,,how did you pull that off ?

  32. panlac01
    • 3 years ago
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    I was just looking for possible entries. I didn't want to rely on the drawing above.

  33. mukushla
    • 3 years ago
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    to my understanding u can use Jensen's inequality here...

  34. panlac01
    • 3 years ago
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    see if this might help http://www.artofproblemsolving.com/Wiki/index.php/Cyclic_quadrilateral

  35. panlac01
    • 3 years ago
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    http://www.onlinemathlearning.com/quadrilateral-circle.html

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

  37. panlac01
    • 3 years ago
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    muku, I like how you manipulated this one.

  38. shubhamsrg
    • 3 years ago
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    altough i didnt understand most part of jenson;s inequality (tried wikipedia) , but still seems to be really useful,, thanks alot!

  39. panlac01
    • 3 years ago
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    I think the key is in cyclic quadrilateral

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

  41. mukushla
    • 3 years ago
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    @panlac01 ... thank u man

  42. shubhamsrg
    • 3 years ago
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    i got the same result using lagrange multipliers .. got a better understanding of the concept now.. thanks all ! :)

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