Ladder and box (and wall)

• November 18th 2006, 05:37 PM
TriKri
Ladder and box (and wall)
A ladder with the length 10 m is leaned against a box (all sides has the length1 m) placed against a wall. But the ground is so slippery that the ladder falls ower the bow and leans against the wall as well. How high up on the wall does the ladder reach?

I had tried to solve this but only get a fourth order equation, and I don't know how to solve any equation of order higher than two. It seems easy but it turns out it isn't. Is the equation solvable?

By the way, my previous thread about modular arithmetic and equation systems wasn't finished, though it was marked as if it was, I happen to click on the "thank you" button by mistake. I would appreciate if someone took a look at it again, or maybe unmarked it as a finished thread. Please? :)
• July 31st 2008, 06:03 PM
Serena's Girl
Reviving an old problem
This is how I visualize the situation:

http://farm4.static.flickr.com/3271/...a5454f.jpg?v=0

The value of (x + 1) indicates how high up the wall the ladder reaches, while the value of (y + 1) indicates the distance of the end of the ladder from the wall.

So set up an equation based on the similarity of the two gray triangles:

$
x = \frac {1} {y}
$

And then set up an equation using the Pythagorean theorem:

$
(x + 1)^2 + (1 + y)^2 = 100
$

Combine the two equations, and you end up with:

$
x^4 + 2x^3 - 98x^2 + 2x + 1 = 0
$

It is a fourth-order equation, and I do not know how to solve this algebraically. However, in this situation, I prefer using numerical methods. Specifically, for this problem, I choose to use Newton-Raphson method together with MS Excel (Wiki article here: Newton's method - Wikipedia, the free encyclopedia).

Newton-Raphson method is an iterative method, where:

$
x_{n+1} = x_n - \frac {f(x_n)} {f'(x_n)}
$

where

$
f(x) = x^4 + 2x^3 - 98x^2 + 2x + 1
$

$
f'(x) = 4x^3 + 6x^2 - 196x + 2
$

The goal is to make f(x) = 0. Iterating with an initial value of x = 1, I obtained x = 0.111881932. Iterating with an initial value of x = 10, I obtained x = 8.937993689.

So before sliding down, the ladder reaches about 9.94 m up the wall. After sliding down, it reaches up only 1.11 m up the wall.
• July 31st 2008, 07:04 PM
Soroban
Hello, TriKri!

I too got a fourth-degree equation, too.
But there is a way around it . . .

[quote]A 10-m ladder is leaned against a box (1 x 1 x 1 m) placed against a wall.
But the ground is so slippery that the ladder falls over the box and leans against the wall.
How high up on the wall does the ladder reach?
Code:

                              |                               *                           *  |y               10    * - - - *                   *  |  1  |               *      1|      |1           *          |      |       * - - - - - - - * - - - *               x          1

From the large right triangle: . $(y+1)^2 + (1+x)^2 \:=\:10^2$ .[1]

The two smaller right triangles are similar.
. . So we have: . $\frac{1}{x} \:=\:\frac{y}{1} \quad\Rightarrow\quad x \:=\:\frac{1}{y}$ .[2]

Substitute [2] into [1]: . $(y+1)^2 + \left(1 + \frac{1}{y}\right)^2 \:=\:100$

and we have: . $y^2 + 2y + 1 + 1 + \frac{2}{y} + \frac{1}{y^2} \;=\;100$

. . $\left(y^2 + 2 + \frac{1}{y^2}\right) + \left(2y + \frac{2}{y}\right) - 100 \:=\:0$

. . $\left(y + \frac{1}{y}\right)^2 + 2\left(y + \frac{1}{y}\right) - 100 \:=\:0$

Let $u \:=\:y+\frac{1}{y}$
. . and we have the quadratic: . $u^2 + 2u - 100 \:=\:100$

Use the Quadratic Formula to solve for $u$,

. . then back-substitute and solve for $y.$

• July 31st 2008, 08:00 PM
Serena's Girl
Wow, that is so cool! *bows down to Soroban*(Bow)
• August 1st 2008, 11:43 AM
TriKri
Congratulations to both of you! You showed two different ways to solve the problem in and both where rigth. Am I supposed to deliver some virtual medal to you now?