The Ladder

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.

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.

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?

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.

Originally posted by PBE6
I got 125.

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?

Uh, what do you mean by "which is flush against the wall and the ground"?

Originally posted by twilight2007
Uh, what do you mean by "which is flush against the wall and the ground"?

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! 😕

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.

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.

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
🙂