A Touchy Situation

A ladder that is 4 ft long is leaning against a wall. Under the ladder ( up against the wall, on the floor) sits a cubical box with side length 1 ft. The ladder is simultaneously touching the edge of the box, the wall, and the floor.

What is the vertical distance from the top of the box to the top of the ladder?

This is the best I could do for a diagram. I think its sufficient to get the idea, if not let me know.

....../|
..../..|
../■|

@joe-shmo
Reveal Hidden Content

@Ponderable

You're awfully close with that ( as in decimal approximation range ), but in terms of the exact expression its substantially different.

My algebra is really rusty. 😛

@bigdoggproblem said
My algebra is really rusty. 😛

@joe-shmo said
This is a little tricky algebra... certainly its more involved than one might initially suspect.

A solution is still up for grabs!

Managed to form a fairly horrendous quadratic, which gives 2.761 or 0.362. If they are correct, I guess you can have a steep or shallow ladder?

@blood-on-the-tracks said
Managed to form a fairly horrendous quadratic, which gives 3.761 or 1.362. If they are correct, I guess you can have a steep or shallow ladder?

@blood-on-the-tracks said
Managed to form a fairly horrendous quadratic, which gives 2.761 or 0.362. If they are correct, I guess you can have a steep or shallow ladder?

@joe-shmo

Yes, my original solution was for how far up the wall the ladder reached. I edited to subtract the 1.

Should have read the question!

My method involved the gradient of the ladder and would take an age to type out!

@blood-on-the-tracks said
@joe-shmo

Yes, my original solution was for how far up the wall the ladder reached. I edited to subtract the 1.

Should have read the question!

My method involved the gradient of the ladder and would take an age to type out!

Let the vertical distance from the top of the box to the point were the top of the box touches the wall be "y"

Let the horizontal distance from the face of the box to the point were the ladder touches the floor be "x"

The length of the ladder L = 4 ft

The side length of the cubical box is 1 ft

From Pythagoras we have that:

( y + 1 )² + ( x + 1 )² = L² ...........Eq1

using similar triangles we also have that:

y = 1/x .................Eq2

Expand Eq1:

y² + 2y + 1 + x² + 2x + 1 = L²

( y² + x² ) + 2( x+y ) + 2 = L²

Notice the following:

( x + y )² = x² + y² + 2xy

x² + y² = ( x + y )² - 2xy ..................Eq3

Substitute Eq3 into Eq1

( x + y )² - 2xy + 2( x+y ) + 2 = L²

Notice from Eq2 that

2xy = 2x(1/x) = 2

Combine to create Eq4

( x + y )² + 2(x+y) - L² = 0

This is just a quadratic in ( x+ y )

Thus ( removing the negative solution as x+y > 0 ) ;

x+y = [ -2 + ( 4 - 4*1 * L² )^( ½ ) ] / 2 = 17^( ½ ) - 1 .................Eq5

From here we just re-substitute Eq2 into Eq5

1/y + y = 17^( ½ ) - 1

This is just another quadratic:

y² - ( 17^( ½ ) - 1 )*y + 1 = 0

y = [ ( 17^( ½ ) - 1 ) + ( ( 17^( ½ ) - 1 )² - 4 )^( ½ ) ]/2 ≈ 2.761 ft

and

y = [ ( 17^( ½ ) - 1 ) - ( ( 17^( ½ ) - 1 )² - 4 )^( ½ ) ]/2 ≈ 0.362 ft

And we are done! Please let me know if you you wish me to clarify anything in this ( its a bit challenging to read these equations in these forums - I've asked Russ for LaTeX capabilities...fingers crossed! )

@venda said
I was never really interested in Trigonometry,but out of curiosity ,I looked on the net to see if it was possible to find angles in right angled triangles when given just one side measurement and it talked about something called Pythagorean triples with equations (m² -n²😉 2mn and (m² +n²😉
I suspect the answer is in there somewhere?
Are you impressed with me using the ² sign you told me about ?
I don't know how the silly faces got on there .they were supposed to be )