Solving a Complex Math Problem: Real and Imaginary Solutions

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Homework Help Overview

The problem involves a ladder leaning against a wall, with a cubic packing case at the base. The original poster states that the problem cannot be solved using a graphics calculator and mentions the existence of real and imaginary solutions. The ladder's length is given as 10 meters, and the task is to determine how high it reaches up the wall.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the ladder's length and its position relative to the packing case and wall. There are attempts to derive equations based on geometric relationships, including the use of similar triangles and conditions involving the ladder's slope.

Discussion Status

Some participants have provided hints and additional conditions to consider, such as the equation relating x and y. Questions about the derivation of certain relationships indicate ongoing exploration of the problem's setup and assumptions.

Contextual Notes

The original poster acknowledges a mistake in their calculations and expresses uncertainty about how to proceed. There is a mention of needing to clarify the relationship between the variables involved in 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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