Solving a Complex Math Problem: Real and Imaginary Solutions

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The math problem involves determining how high a 10-meter ladder reaches against a wall while resting on a cubic packing case. The key equations to solve the problem include the Pythagorean theorem, x² + y² = 10², and a condition derived from the ladder's interaction with the box, (1/x) + (1/y) = 1. The discussion highlights the complexity of the problem, noting that there are both real and imaginary solutions. Participants are working through the equations and relationships to find the correct values for x and y, emphasizing the need for a clear understanding of the geometric relationships involved. The conversation focuses on clarifying the steps needed to derive the necessary conditions for solving the problem.
Noir
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Homework Statement


Apparantley the following problem can't be solved with a graphics calculator and we have to state why this is, and give all the possible solutions.

A ladder AB leans against the side of a building. Where the wall meets the ground is a cubic packing case 1m x 1m x 1m. The ladder is adjusted so that it rests on the ground, touches the packing case and rests against the wall. If the ladder is 10 metres long, how far up the wall will the ladder reach.

That was the question and a diagram was drawn, except I can't scan it in at the moment. Thinking on the problem there is more than one answer to the problem. Apparantley there are real and imaginary solutions.


Homework Equations


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The Attempt at a Solution


Using similar triange theroy, I get the following; x.y = 1.
Using pascals formula; 1^2 = (x + 1)^2 + (y + 1)6=^2
Solving the two equations gives an x value of -1.88 and a y value of -0.53.
Except I know I've made an mistake above and don't know how to continue. Can anyone help me please?
 
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Hi Noir,

The length of the ladder is 10 m, so you should have x² + y² = 10². Use the fact that the ladder must touch the box to get another condition, and solve for x and y.

Hint: The second condition is (1/x) + (1/y) = 1.
 
Thanks for the reply dx, sorry for my late response. Thanks for helping me but I have a quick question, how do you get the 1/x + 1/y = 1 relationship - It sounds stupid, and I know it works , but I seem to be missing a step in my head.
 
The distance from the tip of the ladder on the x-axis to the edge of the box touching the x-axis is (x-1). The slope of the ladder (towards the y-axis) is y/x. Therefore, the height of the ladder at x = 1 is h = (x-1)(y/x). What must this height be if the ladder touches the box at this point?
 

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