The ladder

**TriKri** · December 8th 2006, 06:20 AM

A 10 meters long ladder is leaned against a wall and a 1 meter times 1 meter big box, as the picture illustrates. The width of the ladder is negligible. How high up on the wall does the ladder reach?

**ThePerfectHacker** · December 8th 2006, 08:43 AM

Look at the similar triangles.
(1 Blue dot and 2 red dots) and the triangle (1 Blue dot and 2 black dots).

Thus,
$\frac{h-1}{1}=\frac{h}{w}$

And, by Pythagorean theorem,
$w^2+h^2=100$

Now you can solve

**TriKri** · December 8th 2006, 08:57 AM

This gives me a fourth order equation. I don't know how to solve them.

**CaptainBlack** · December 8th 2006, 01:27 PM

Originally Posted by TriKri

This gives me a fourth order equation. I don't know how to solve them.

Take hackers equations and turn them into an interative scheme:

$ w_n=\frac{h_n-1}{h_n} $
$ h_{n+1}=\sqrt{100-w_n^2} $

Now put $h_0=9.9$ , $h_1=9.93794$ , $h_2=9.93799$ , $h_3=9.93799$ .

So $h \approx 9.93799$

(we choose an initial guess of $h=9.9$ by doing a rough drawing of the
arrangement and measuring the approximate height, but a rough guess of 9 will have
added just one more iteration to get this level of accuracy)

The scheme also produces an approimate value for $w\approx 1.11188$

RonL

**topsquark** · December 8th 2006, 01:41 PM

I would like to note that there is technically another solution: h = 1.11188 m, which the diagram (sort of) rules out. In this case the ladder is nearly horizontal rather than vertical. (I rarely take diagrams as being "to scale" or in the correct orientation unless I have to.)

-Dan

**TriKri** · December 8th 2006, 01:47 PM

Originally Posted by CaptainBlack

Take hackers equations and turn them into an interative scheme:

$ w_n=\frac{h_n}{h_n-1} $
$ h_{n+1}=\sqrt{100-w_n^2} $

Now put $h_0=9.9$ , $h_1=9.959509$ , $h_2=9.959454$ , $h_3=9.959454$ .

So $h \approx 9.959454$

(we choose an initial guess of $h=9.9$ by doing a rough drawing of the
arrangement and measuring the approximate height, but a rough guess of 9 will have
added just one more iteration to get this level of accuracy)

RonL

Nice! But that's cheating!

**CaptainBlack** · December 8th 2006, 02:14 PM

Originally Posted by TriKri

Nice! But that's cheating!

In what way, I could have guessed the solution anyway I wanted.

By the way for some reason the iterative solution appears to be wrong,
that I will have to look into

Fixed now. Note the value of w produced is topsquarks other solution.

RonL

**malaygoel** · December 27th 2006, 08:22 PM

I solved it without using numerical methods.
It is a bit tricky.

The ladder is 10m long and let at the point of contact with the box, it is divided into x(the upper part) and y(the lower part).{x+y=10}
You see that
$\frac{1}{x^2}+\frac{1}{y^2}=1$
$x^2 +y^2=x^2y^2$
$(x+y)^2=(xy)^2+2xy$
Let xy=b
$100=b^2+2b$
$b^2+2b-100=0$
since b is positive
$b(=xy)=\sqrt{101}-1$
$x(10-x)=\sqrt{101}-1$
solving for x, we get
$x=5\pm \sqrt{24-\sqrt{101}}$
Since we are looking for maximum height,
we have
$x=5+ \sqrt{24-\sqrt{101}}$
$h=\sqrt{x^2-1}+1$

Keep Smiling
Malay

**TriKri** · December 29th 2006, 01:42 PM

Wow, impressive! How did you know you should solve it like this? Is it some kind of standard equation in some branch of mathematics? I'm just wondering because I haven't read enough with math to be able to solve it myself... ^_^

**malaygoel** · December 30th 2006, 10:16 PM

Originally Posted by TriKri

Wow, impressive! How did you know you should solve it like this? Is it some kind of standard equation in some branch of mathematics? I'm just wondering because I haven't read enough with math to be able to solve it myself... ^_^

It just striked me.
I haven't studied fourth order equations, was trying to use the cube(its all sides are equal)

Keep Smiling
Malay

**benzi455** · December 31st 2006, 06:03 PM

That's some ladder problem. I didn't know it could get this hard :-P Man, my math class is wimpy XD

**TriKri** · January 1st 2007, 04:28 AM

Originally Posted by benzi455

That's some ladder problem. I didn't know it could get this hard :-P Man, my math class is wimpy XD

Have you showed them malaygoel's solution?

**TriKri** · January 6th 2007, 03:29 PM

Originally Posted by malaygoel

It just striked me.
I haven't studied fourth order equations, was trying to use the cube(its all sides are equal)

Keep Smiling
Malay

The cube? I haven't heard of that before, but not strange, regardning what I said before...

I was trying to make the equation more general by puting the sides of the bow to a and b, but I couldn't solve it you'r way since I couldn't get only xy left in the equation and no x or y. But I'm still impressed. Is this the way to solve third order equations?

**ThePerfectHacker** · January 6th 2007, 03:34 PM

Originally Posted by TriKri

Is this the way to solve third order equations?

Yes!

It was first made by Italian mathematicians: Tartalia, Cardano. I think they kept it secret.

I have memorized the original method for solving 3rd order equations

. And can up with my own which is simpler but not always works, 2 nights ago when I was falling asleep I was thinking about a techinique that will make my solution always workable did not try it yet.

Maybe, I will post it for thee.

**malaygoel** · January 18th 2007, 09:04 PM

Originally Posted by ThePerfectHacker

Yes!

It was first made by Italian mathematicians: Tartalia, Cardano. I think they kept it secret.

I have memorized the original method for solving 3rd order equations

. And can up with my own which is simpler but not always works, 2 nights ago when I was falling asleep I was thinking about a techinique that will make my solution always workable did not try it yet.

Maybe, I will post it for thee.

Trikri!!
i mistyped it.
I want to say I was trying use the square.
i don't know how to solve cubic equations.

Keep Smiling
Malay

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