Curse

Sjeng · Mar 25, 2009, 03:29 PM // 15:29

Here's a little brain teaser for those math wizzkids out there. I've been trying to get it, but after 1 hour of writing formulas and not really getting anywhere, I thought I'd post it here. It's probably really easy once you know what to do, but for the life of me, i'm stumped...

There's TWO ladders in an alley. One is 5 meters long, the other is 4 meters long. They're standing opposite eachother against each wall. They cross each other at a point that's 1.2 meters above the ground. The bases of the ladders are in the corners where the walls meet the floor.
Now the question is: how wide is the alley?

Good luck!

Painbringer · Mar 25, 2009, 03:42 PM // 15:42

3.2 meters

it just a guess

Sjeng · Mar 25, 2009, 03:53 PM // 15:53

No, I need a solution. How would you calculate it. formulas! Not just an answer. It's driving me crazy xD

I used this:

If the 5m ladder starts in the left corner, and the 4m ladder in the right corner, then I named the distance between the walls "X".
X1 is the distance from the left corner to the point on the floor where they cross.
X2 is the distance from the point on the floor where they cross to the right corner.

X=X1+X2

Then I named the height of ladder 4m against the left wall "Y1" and the height of ladder 5m against the right wall "Y2"

And as we know: in a rightsided triangle, a²+b²=c².
That means X²+Y2²=5²=25 and X²+Y1²=4²=16
That means X²=16-Y1²=25-Y2²
and we know that X²=(X1+X2)²=X1²+2(X1)(X2)+X2²

I did some more reasoning with Z1 being the length of the ladder 4m from base to crossing point, and Z2 being the length of the ladder 5m from base to crossing point.
(the rest of the legths are then 4-Z1 and 5-Z2)

Then:
X1²= (5-Z2)²-(Y2-1.2)² and X1²=Z1²-1.2²
X2²= (4-Z1)²-(Y1-1.2)² and X2²=Z2²-1.2²

Then my paper get extremely cluttered, *sigh*. This is killing me lol.

I pwnd U · Mar 25, 2009, 04:48 PM // 16:48

I assume that you will probably need to know the angles that they have with the pavement. That way you could use the Pythagorean Theorem and sin/cos/tan to find out the other sides to the triangle, since each ladder would form a triangle with the all and the pavement. Also you know one angle will be 90 degrees (between the wall and pavement). Figure out one other angle and you can figure out the answer.

(Also would be helpful to know what level of school this problem is for)

Arduin · Mar 25, 2009, 05:06 PM // 17:06

Quote:

Originally Posted by I pwnd U

Also would be helpful to know what level of school this problem is for

Probably one of his personal Demons. I'll see what I can do

Hmm.. my math skills are kinda rusty...

Winterclaw · Mar 25, 2009, 06:26 PM // 18:26

I've forgotten all of geometry so I can't help you much.

Quote:

They're standing opposite eachother against each wall.

Does that mean they are standing directly across from each other such that they'd hit each other when you lowered them. If that's the case the height of the triangle is 1.2 and the hypotenuse of them are 5 and 4.

5 = (1.2^2 + x^2) ^ 1/2
4 = (1.2^2 + y^2) ^ 1/2

Halmyr · Mar 25, 2009, 07:30 PM // 19:30

I wanna say 3 m

so For the 5 m ladder anyways.
a^2 + b^2 = c^2
a^2 + b^2 = 25
9 + 16 = 25
3^2 + 4^2 = 5^2

This is the simplest answer I could find....and more out of a bit of logic then anything else....see if that gives you any result....but I'l take no offence if you ignore this completly too

zelgadissan · Mar 25, 2009, 07:48 PM // 19:48

^^ I don't see any way you can say b=4 on the 5m hypotenuse triangle.

Anyway, it's been a while since I mathed, but I can't find a way to solve this. To solve a triangle you have to have at least three of the six pieces of information (three sides, three angles). I'm not seeing any way to get more than two for any triangle in the drawing.

Somebody PM Rahja

Zak Mcracken · Mar 25, 2009, 08:28 PM // 20:28

I got ~3.414m as the width of the alley, which is the solution to:
(((16/(x^2))-1)^(1/2))*((25-x^2)^(1/2))/((((25/(x^2))-1)^(1/2))-(((16/(x^2))-1)^(1/2)))-((25-x^2)^(1/2))-1.2=0

I didn't really want to solve this manually so I just graphed it.
As for getting to that point, I set up two lines:
f(x)=m_1x
g(x)=m_2x+b

f is the 1st line, which passes through the origin, where the y-axis can be seen as the left wall of the alley, this line I decided to be the ladder of length 4.
Then g is line of the other ladder, starting at b height and intercepting the x-axis at -b/m_2, which is thereby the width of the alley.
I also declared x_0 to be the point where: f(x_0)=g(x_0)=1.2

From there it's mostly finding a solution for each variable in terms of x_n, which I thought to be the width of the alley. What I wrote down was pretty messy and such, but I'll list of most of the equations I came up with, skipping steps because It's ugly to write math stuff.

x_n=-b/m_2
x_0(m_2-m_1)+b=0 or x_0=-b/(m_2-m_1)
b=(25-(x_n)^2)^(1/2)
m_1=(16/(x_n)^2-1)^(1/2)
m_2=(16/(x_n)^2-1)^(1/2)

Then after alot of equation synthesis and such I came to that large thing at the top. Which I rechecked ish, and the result seems/feels correct, but I would be interested in the real/exact answer if it's available to you.

In addition, I tried to do this with geometry/triangles and stuff at first, but I ran into too many variables.

Well, hope most of this seems coherent.

Savio · Mar 26, 2009, 12:37 AM // 00:37

The answer is 3.56, not sure what the above poster is doing. Not sure what I did either but the answer seems to hold up. I may write up something later.

Zak Mcracken · Mar 26, 2009, 01:27 AM // 01:27

heh, I suppose I don't know what I'm doing either, but I got a value that sounded right, so I just stuck with that. It's pretty likely that I subbed incorrectly somewhere as I was using a large system.

Would like to see how you did it your way, I couldn't think of a good way to do it purely geometrically.

Savio · Mar 26, 2009, 02:11 AM // 02:11

http://i44.tinypic.com/14ucnlv.png

1. Pythagorean theorem with both larger triangles.
2. Ratios with similar triangles.
3. Pythagorean theorem with both smaller triangles.
4. Put them all into one big equation and use a calculator/program to solve for x.

The only difficult part would be solving that last equation, to set it up you only need to know the Pythagorean theorem and similar triangles.

AidinSwiftarrow · Mar 26, 2009, 03:38 AM // 03:38

Wow. My brain hurts.

Zak Mcracken · Mar 26, 2009, 03:43 AM // 03:43

Yea, the similar triangle part seems to be what I was missing. I went back and changed and subbed and what not and got the same formula for x. My way is certainly less clear though, suppose I just like the Cartesian plane a good bit.

Well, in any case it was solved, so yay.

pumpkin pie · Mar 26, 2009, 04:05 AM // 04:05

open illustrator.

draw 2 lines, 1x 5cm 1x 4cm
drag a guide to be the base
drag another line to be at 1.2cm height from the base guide
rotate 5cm line at base 45 degree. (rough guess)
rotate 4cm line at base -45 degree.
align both 4cm and 5 cm line on the right with the alignment tool
draw a guide to the left side at end of 5cm line
drag 4cm line on the left side to alight with the left guild line.
now rotate 4cm line until the right point of the line align with the 5cm right side and see if they cross at the 1.2cm point lol

measure the distance of the wide of the space created which is 35.4cm

convert that to meter you get 3.54 lol voila.

Sjeng · Mar 26, 2009, 10:46 AM // 10:46

Well people, Savio is correct. Pumpkin kinda cheated, and didn't get the correct answer either. Thanks for that image Savio! That really helps!

Savio wins thread!

Painbringer · Mar 26, 2009, 01:31 PM // 13:31

I forgot to mention my guess was based off my vast knowledge of Kaineng City Engineering

............... Please somebody give those people a straight ruler