## shubhamsrg 4 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.

1. anonymous

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2. anonymous

|dw:1345656175855:dw|

3. shubhamsrg

yes sir..

4. shubhamsrg

r=1 here..from cosine law ..

5. anonymous

thats right...

6. hartnn

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

7. shubhamsrg

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

8. anonymous

sorry...yes

9. shubhamsrg

|dw:1345656382986:dw|

10. shubhamsrg

ohh wait,,i reach some where..

11. 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 ?

12. anonymous

thats right...

13. shubhamsrg

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

14. anonymous

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

15. shubhamsrg

hmm..

16. shubhamsrg

@eliassaab ?

17. 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.

18. shubhamsrg

19. anonymous
20. shubhamsrg

so gradient approximately means slope ?

21. shubhamsrg

i mean in layman language ?

22. anonymous

It replaces derivative in functions of two variables,

23. shubhamsrg

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

24. anonymous

25. shubhamsrg

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

26. 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 :)

27. anonymous

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

28. shubhamsrg

hmm..

29. anonymous

is AB parallel to CD?

30. anonymous

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

31. shubhamsrg

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

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous
36. anonymous

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

37. anonymous

muku, I like how you manipulated this one.

38. shubhamsrg

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

39. anonymous

I think the key is in cyclic quadrilateral

40. shubhamsrg

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

41. anonymous

@panlac01 ... thank u man

42. shubhamsrg

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