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

1. mukushla Group Title

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2. mukushla Group Title

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3. shubhamsrg Group Title

yes sir..

4. shubhamsrg Group Title

r=1 here..from cosine law ..

5. mukushla Group Title

thats right...

6. hartnn Group Title

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

7. shubhamsrg Group Title

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

8. mukushla Group Title

sorry...yes

9. shubhamsrg Group Title

|dw:1345656382986:dw|

10. shubhamsrg Group Title

ohh wait,,i reach some where..

11. shubhamsrg Group Title

|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 Group Title

thats right...

13. shubhamsrg Group Title

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

14. mukushla Group Title

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

15. shubhamsrg Group Title

hmm..

16. shubhamsrg Group Title

@eliassaab ?

17. eliassaab Group Title

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 Group Title

19. eliassaab Group Title
20. shubhamsrg Group Title

so gradient approximately means slope ?

21. shubhamsrg Group Title

i mean in layman language ?

22. eliassaab Group Title

It replaces derivative in functions of two variables,

23. shubhamsrg Group Title

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

24. eliassaab Group Title

25. shubhamsrg Group Title

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

26. shubhamsrg Group Title

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 Group Title

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

28. shubhamsrg Group Title

hmm..

29. panlac01 Group Title

is AB parallel to CD?

30. mukushla Group Title

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

31. shubhamsrg Group Title

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

32. panlac01 Group Title

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

33. mukushla Group Title

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

34. panlac01 Group Title

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

35. panlac01 Group Title
36. mukushla Group Title

$$f(x)=\sin x$$ 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 Group Title

muku, I like how you manipulated this one.

38. shubhamsrg Group Title

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

39. panlac01 Group Title

I think the key is in cyclic quadrilateral

40. shubhamsrg Group Title

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 Group Title

@panlac01 ... thank u man

42. shubhamsrg Group Title

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