Brain Twister: The Distance Between Two Walls

infinite number of solutions.

Or... It could be earlier than my normal brain operation time.

Nope. Only one.

is there something left out? Can the sticks cross at any point as long as they cross @ 12' perpendicular to the floor?

Looks like I'll need to get a cup of Java before i 'll loosen by brain Rust.

oops.. i see what i did wrong... I failed to recieve all the information... told you its too early for me today... somehow i missed: leans on the wall right/left.

--------------------> off to get coffee

+- 340' ...I think.

The answer must be less than 30' so that the smaller stick can lean on the other wall. I'll say 18'.

23.217'

Excel Solver???

Can you put in the Math

Nah, no math. I'm an engineer - I used CAD. Took about 5 min.

Are the sticks perpendicular to the walls (as viewed from above)? Are the walls parallel to each other?

"The walls and sticks are in the same plane" doesn't seem to make sense.

can you make a drawing?

refer to the drawing of Mike or John above and below your question

... or even this?

http://cr4.globalspec.com/thread/7412

Mike, well spotted - GA!

Doorman and Guest (#7), you may (I don't know) have the correct answer to the question Dedaelus intended to pose, but (as I think RVZ717 suggested in #1) the question actually posed has no solution.

Hi JohnDG,

the question Dedaelus intended to pose, but (as I think RVZ717 suggested in #1) the question actually posed has no solution.

I think we're talking that Dadaelus meant to/should have said "the two ladders are in the same plane which is perpendicular the the common plane of the walls". That, or some reasonable variation, is how I would have described it.

Thanks for the GA (I just remembered beating by brains out trying to work the algebra in the thread by JohnJohn you just linked)!

Mike

That is right and your diagram is also correct. I do not know how to get a sketch in here. Creates a problem to explain w/o a sketch

You can call me Ray. Easier.

The answer to http://cr4.globalspec.com/thread/7412 is very easy.

I do not know how to get a Diagram in here but there you go:

Your problem is:
A 32' rod from left corner leans on right wall
A 27' rod from right corner leans on left wall
Distance between two walls 24'
What is the perpendicular ht. from intersection to floor

let it be X

let the dist. from where 32 rod leans on right wall to the base be B

let the dist. from where 27 rod leans on left wall to the base be A
B=(32^2-24^2)^0.5 = 21.166
similarly
A=(27^2-24^2)^0.5 = 12.3693

let the horiz dist from left corner to the line coming down from the intersection on the base be: Y
so the horiz dist from right corner to the line coming down from the intersection on the base will be: (24-Y)
from similar triangles
Y/24 = X/B = X/21.1660 and
(24-Y)/24 =X/A = X/12.3693
solving for X and Y

Y = 8.8525 and
X = 7.8070

your answer for dist from intersection to base is X = 7.8070

If you want you can draw it and and check it out.

In my problem the dist from intersection to base is X is known.

you have to find the dist between the walls

No wonder your prof smiled at you

Yes. That is the diagram

For a general solution one can use the ladder lengths as A, and B and the Perpendicular distance from intersection to base as C. These are known.

Solve for length between walls X.

Please note that if you know X, A, and B; C can very easily be solved so the answer to this problem is unique

A similar problem is:

In an isoscles triangle prove that the equal angle bisectors are equal. Easy to prove.

but if put the other way:

If the angle bisectors are equal in a triangle, prove that the triangle is isoscles can only be done by reductio ad absurdum method.

Had lots of fun in school foxing the geometry teacher.

Aha, we had that one on CR4 some time ago too: after a lot of thought I came up with this "contradiction" proof:-

http://cr4.globalspec.com/comment/20093/Re-Isosceles-Triangle

I didn't get any feed back. Do you know a simpler method?

I think GUEST #7 is pretty close... I have 23'-3", but am still working the math.

Its 23'-2.601"

There is no ceiling - just two walls and a floor. Both sticks lean ON their respective walls not against them. The the overall length of the sticks shouldn't matter. They intersect at 12ft. The distance between the two walls at the base should therefore be 24ft

i think i got dumber reading that...

if your walls are 12 feet apart, your intersection will be alot higher then as if your walls were 24 feet across

put 2 straws together and find out.

ha ha ha...yO! ssinnett, it's OK - you now can get reading classes for free in most places. If your over 24!! This is an English class trick question - has little to do with math. More like geometry. ON vs AGAINST - no ceiling ?

just stop, you've made yourself sound even dumber

READ #19. The distance between the walls is 24 ft like I said. The perpendicular from base to intersect is 12 ft. how mush more do you need? Sorry, didn't mean to aggravate you. But I will stop anyway just to be courteous to you.

number 19 isnt the same as this one. lol

number 19 is talking about the link he posted, and the height for that one is 7 feet, not 12.

your still off.

I agree with 23.217'

I kinda cheated and used the microsoft excel solver to iterate the answer. I guessed a height up the wall by dividing 12 by some fraction. That fraction representing where the intersection occurs between the walls. (eg. 1 being the intersection at the wall and .5 being halfway between the two) For the other stick, I used 1 minus this fraction.

Using pythagorean theorem: distance between walls = sqrt(stick length^2 - calculated height^2)

I used the solver to minimize the differences between the calculated differences changing the intersection location.

I know there has to be a way to do it with trig. I tried a few times but gave up. Could anyone enlighten me with the real way to solve it?

Dan

23'

23.217 is the right answer.

The problem can be solved by either trignometry or similar triangles. But the final equation ends up in a complex form with underroot.

So iteration becomes the easy way to solve.

Assume angle A and B.

X be the distance between wall.

Cos A = x/40 Cos b= x/30

tanA= sqrt( 1-cos²A)/Cos A= Sqrt( 1-(x/40)²)/x/40

x= 12/Tan a +12/tanb

So you end up with a complex formula which can be solved by iteration.

Alternate method is use similar trianles.

Height at which stair touches the wall H1, H2

H1= sqrt( 40²-x²) h2=sqrt(30²-x²)

l1,l2 be the intersetion length

l1+l2= x

l1/x=12/h1 l2=12/h2

So you get an equation 1= 12/h1+12/h2
Iteration will give the solution.