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shubhamsrg Group Title

what is the maximum length of rod that can pass through this L-shaped corridor :

  • one year ago
  • one year ago

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  1. shubhamsrg Group Title
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    |dw:1347549016758:dw|

    • one year ago
  2. shubhamsrg Group Title
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    its the same question which @siddhantsharan asked here : http://openstudy.com/study#/updates/504e2cd7e4b064acfea5f1ae there arent much people there,,and i liked this ques,,so i bring this here..

    • one year ago
  3. shubhamsrg Group Title
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    @mukushla @eliassaab

    • one year ago
  4. Mikael Group Title
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    |dw:1347549611645:dw| By SIMILARITY of Triangles [ three similar , right-angle triangle] we obtain (X+a)/(y+b) = x/b = a/y So ab = xy and bx + ba = yx + bx

    • one year ago
  5. Mikael Group Title
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    y = ab/ x Substitrute int Length^2 = (y+b)^2 + (x+a)^2 AND Make derivation , then equate to 0 and solve for x - here is your SOLUTION !!

    • one year ago
  6. Mikael Group Title
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    (ab/x + b)^2 + (x+a)^2 = a^2b^2/x^2 + b^2 + 2ab^2/x + x^2 + 2xa + a^2

    • one year ago
  7. shubhamsrg Group Title
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    well in the position you mentioned,,the rod cant move,so wont exit out.... |dw:1347549761719:dw|

    • one year ago
  8. Mikael Group Title
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    (a^b^2/x^3) * (-2) - 2ab^2/x^2 + 2x +2a = Derivative

    • one year ago
  9. Mikael Group Title
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    Now Deivative = o means the following should be solved

    • one year ago
  10. shubhamsrg Group Title
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    you're concept it wrong buddy,,see my last comment again,,and correct me if am wrong..

    • one year ago
  11. Mikael Group Title
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    |dw:1347549859821:dw| Well - sorry to be so direct. But you ARE wrong. Rotation is NO PROBLEM

    • one year ago
  12. Mikael Group Title
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    To be precise Rotatio + Simultaneous sliding of the ends

    • one year ago
  13. Mikael Group Title
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    You see - the interval (lets call it "stick") that is the solution is THE SHORTEST POSSIBLE OF ALL ANGLES. Thus, a little motion provides it with MORE SPACE , NOT LESS !

    • one year ago
  14. Mikael Group Title
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    Imagine a stick that fits into this space at "the most restrictive" angle. Well, it will obviously fit into positions at LES restrictive angles, wouldn't it ?

    • one year ago
  15. shubhamsrg Group Title
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    |dw:1347550109862:dw| for up ward movement,,downward movement should also be there,,which is not possible..

    • one year ago
  16. Mikael Group Title
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    |dw:1347550236581:dw|

    • one year ago
  17. shubhamsrg Group Title
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    sorry didnt get this figure..

    • one year ago
  18. Mikael Group Title
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    Firstly the LEFT END RISES, and the right end SLIDES into positions 3 and 3

    • one year ago
  19. Mikael Group Title
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    |dw:1347550435290:dw|

    • one year ago
  20. Mikael Group Title
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    |dw:1347550518272:dw|

    • one year ago
  21. shubhamsrg Group Title
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    it cant slide like that.. |dw:1347550476551:dw|

    • one year ago
  22. Mikael Group Title
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    Look forget - drawing for a second. If Some position is the SHORTEST interval - all its small motions will create SPACE AT BOTH ENDS or AT ONE END at least !

    • one year ago
  23. Mikael Group Title
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    |dw:1347550645409:dw|

    • one year ago
  24. Mikael Group Title
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    |dw:1347550693270:dw|

    • one year ago
  25. Mikael Group Title
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    Come on - after rotation the OLD SHORTEST STICK IS NOT FILLING THE NEW - LONGER SPACE

    • one year ago
  26. Mikael Group Title
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    |dw:1347550790801:dw|

    • one year ago
  27. Mikael Group Title
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    More SPACE - NO PROBLEM FITTING SHORTEST STICK !

    • one year ago
  28. Mikael Group Title
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    You need a stick that will move inside the SHORTEST place. Sooo just take the SHORTEST LENGTH - some infintisimal "epsilon"

    • one year ago
  29. shubhamsrg Group Title
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    you mean.. |dw:1347551005335:dw| ?

    • one year ago