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?

2. 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

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

4. 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

5. 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

6. 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!

7. 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

8. 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

9. 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... ^_^

10. 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

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

12. 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?

13. 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?

14. 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.

15. 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