1. Standard memberuzless
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    10 Jun '08 19:231 edit
    A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.


    Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.


    EDIT: I believe this question is looking for the shortest ladder length possible.
  2. Standard memberforkedknight
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    10 Jun '08 19:27
    Originally posted by uzless
    A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.


    Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.
    I think I know what you are trying to say, but what exactly do you mean by "just touches a box".

    Does the box need to be underneath the ladder?
  3. Standard memberuzless
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    10 Jun '08 19:291 edit
    Originally posted by forkedknight
    I think I know what you are trying to say, but what exactly do you mean by "just touches a box".

    Does the box need to be underneath the ladder?
    Yes. Sorry, can't draw pictures in this thing.
  4. Standard memberPBE6
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    10 Jun '08 19:51
    Originally posted by uzless
    A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.


    Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.


    EDIT: I believe this question is looking for the shortest ladder length possible.
    I got 125.
  5. Standard memberwolfgang59
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    10 Jun '08 20:402 edits
    Originally posted by PBE6
    I got 125.
    Intuitively the ladder should be at an angle of 45 degrees. (yes/no?)

    That would mean it would be a distance of 91 units from wall (27+64)

    Pythagoras gives us 128.7 for the ladder length.

    Presumably wrong! 😕
  6. SubscriberAThousandYoung
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    11 Jun '08 00:49
    Originally posted by forkedknight
    I think I know what you are trying to say, but what exactly do you mean by "just touches a box".

    Does the box need to be underneath the ladder?
    The ladder touches the corner that is not in contact with the wall or the ground.
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    11 Jun '08 04:30
    Uh, what do you mean by "which is flush against the wall and the ground"?
  8. Standard memberuzless
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    11 Jun '08 04:55
    Originally posted by twilight2007
    Uh, what do you mean by "which is flush against the wall and the ground"?
    the angle where the wall meets the ground is 90 degrees and the box is touching the ground and the wall at the same time.
  9. Standard memberuzless
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    11 Jun '08 04:55
    Originally posted by wolfgang59
    Intuitively the ladder should be at an angle of 45 degrees. (yes/no?)

    That would mean it would be a distance of 91 units from wall (27+64)

    Pythagoras gives us 128.7 for the ladder length.

    Presumably wrong! 😕
    you can't make the two opposite corners 45 degrees if the box is taller than it is wide.

    You're on the right track though.
  10. Joined
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    11 Jun '08 05:061 edit
    I have a quartic to solve to find tan theta, I'm getting there...
    Right all i had to do was differentiate it. I get that the gradient of the ladder is 4/3. So therefore the length is 125.
  11. Standard memberwolfgang59
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    11 Jun '08 15:43
    Originally posted by uzless
    you can't make the two opposite corners 45 degrees if the box is taller than it is wide.

    You're on the right track though.
    I can put a box of ANY dimensions benetah a ladder at an angle of 45degrees.
  12. Backside of desert
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    12 Jun '08 02:20
    Q = length of ladderr
    u = hypotonuse of lower triangle
    v = Hypotonuse of upper triangle
    x = angle between ladder and wall

    u = 64 sec x
    v = 27 csc x
    Q = 64 sec x + 27 csc x
    dQ = 64 sec x tan x - 27 csc x cot x
    0 = 64 sec x tan x - 27 csc x cot x
    64 sec x tan x = 27 csc x cot x
    64 (tan x)^2 = 27 cot x
    64 (tan x)^3 = 27
    (tan x)^3 = 27 / 64
    tan x = 3 / 4
    x ~ 26.86 degrees
    sec x = 10 / 8
    csc x = 10 / 6
    Q = 64 sec x + 27 csc x
    Q = 64 * 10 / 8 + 27 * 10 / 6
    Q = 640 / 8 + 270 / 6
    Q = 80 + 45
    Q = 125
    🙂
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