Offtopic: Math Puzzle
 

Hi,

I would like to challenge you all to a little puzzle that was given to me a couple of years ago. The picture below will explain everything, I think. Once you remember your geometry and algebra, it will be very easy to solve this with a computer. So here is the challenge: Dont use anything but a piece of paper and a pen. Try to come up with a solution that is exact, ie no 1.1123123...

Have fun!

P.S.: You should use x+1(1 being the size of the box) and y+1 for the variables, that will make it much easier.

[img]http://fb.design.fh-dortmund.de/matthias/bs/ladderproblem.gif[/img]

If you draw a circle centered at (-1, -1) with a radius of 5, where will it touches the axies at the positive sides?

Proposed Solution.
 

I may be wrong but can't this be solved with a little bit of manipulation to Pythagoras?

Hyp^2 = Adj^2 + Opp^2

Adj = (Hyp^2 - Opp^2)^.5

Adj = (5^2 - 1^2)^.5

Adj = 24^.5

Height of Ladder Against Wall = 1 + 24^.5

:exclaim:

Re: Proposed Solution.
 

[quote="S3SJK"]I may be wrong but can't this be solved with a little bit of manipulation to Pythagoras?[/quote]

Yes it can, but the fundamental thing to be answered here is does the ladder meet the wall at 3m above the floor or 4m? (Wall/Ladder/Floor is a 3,4,5 right triangle)

Re: Proposed Solution.
 

[quote="S3SJK"]
Adj = (Hyp^2 - Opp^2)^.5

Adj = (5^2 - 1^2)^.5

Adj = 24^.5

Height of Ladder Against Wall = 1 + 24^.5

:exclaim:[/quote]

Sorry. You are setting Opp == 1. This corresponds to the ladder sitting inside the box.

Re: Proposed Solution.
 

[quote="sdbardwick"]Yes it can, but the fundamental thing to be answered here is does the ladder meet the wall at 3m above the floor or 4m? (Wall/Ladder/Floor is a 3,4,5 right triangle)[/quote]

Again, a wrong answer. A 3/4/5 right triangle does not pass through the point (1,1). Thus in that position, the ladder will not touch the box.

But you have pointed out an interesting aspect. The problem has no unique solution. For every solution, there is a reflection that reverses the wall and floor. Only if the ladder were exactly 2^(3/2), would it lie at a 45 degree angle and have a unique solution.

Here's what I have so far.......

We make the ladder have the equation y=mx+b. It is bounded between the points (0,b) and (-b/m,0). We know the distance between these two points is 5. So sqrt( (0- -b/m)^2 + (b-0)^2) =5 . Simplify this, and we get (b/m)^2+b^2=25 . Now we apply the boundary condition that it passes through the point (1,1) . So 1= m+b , or m=1-b. You can substitute and get a quartic polynomial which you should be able to solve for b (the y-intecerpt, or height where the ladder hits). I've been unsuccessful in factoring it, but you should get 2 positive roots (because of the reflection mentioned above), and two more that are either negative solutions or imaginary. My quadratic came out to be b^4-2b^3-23b^2+50b-25=0 . Maybe someone who hasn't done roughly 10 hours of math the past 2 days can finish it off for me :? .

Re: Proposed Solution.
 

[quote="Wackerbarth"]
Again, a wrong answer. A 3/4/5 right triangle does not pass through the point (1,1). Thus in that position, the ladder will not touch the box.[/quote]

D'OH!!! must stop attempts to use brain when running on 2 hours sleep! :surprised:ops: Causes interesting hallucinations (like imagining 3 or 4 measurements)

Re: Topic...
 

How can anything in the Lounge be considered off-topic? :shock: :(

[quote="ebx"]If you draw a circle centered at (-1, -1) with a radius of 5, where will it touches the axies at the positive sides?[/quote]

Gee, nobody takes this as THE answer?

Connecting the meet points at X and Y to the tip of the box at (1, 1) gives you both solutions. When the ledder length is 2*sqrt(2), the two solutions reduce to 1. Shorter than 2*sqrt(2), not real sollutions.

Solution
 

we take x+1 and y+1 as the sides of the big triangle.
Then we compare the two smaller triangles(besause they are similar):
x/1=1/y
gives x*y=1

phythagoras:
(x+1)^2+(y+1)^2=5^2
x^2+2x+1+y^2+2y+1=25

then we substitute: y:=1/x
we get
x^2+2x+2/x+1/x^2=23

we now substitute:
x+1/x:=t (*)
t^2=x^2+2+1/x^2
x^2+1/x^2=t^2-2

t^2-2+2*t=23
gives the two solution
t1=-1+sqrt(26)
t2=-1-sqrt(26)

put t1 in equation (*) gives
x+1/x=-1+sqrt(26)
x^2-(-1+sqrt(26))*x+1=0
gives the two solutions
x1=(-1+sqrt(26))/2+sqrt((23-2*sqrt(26))/4)
x2=(-1+sqrt(26))/2-sqrt((23-2*sqrt(26))/4)

for t2 in (*) we get
x3=-(-1-sqrt(26))/2+sqrt((23+2*sqrt(26))/4)
x4=-(-1-sqrt(26))/2-sqrt((23+2*sqrt(26))/4)

x4=negative
x3=negative
x2=positive
x1=positive

we want to know 1+y=1+1/x=
x1=(1+sqrt(26))/2+sqrt((23-2*sqrt(26))/4)
x2=(1+sqrt(26))/2-sqrt((23-2*sqrt(26))/4)

so we get 2 solution x1 and x2 which are the possible heights of this triangle
q.e.d.